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Thread: the best way to solve x+ bx = c

  1. #1 the best way to solve x+ bx = c 
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    Please consider the depressed cubic equation when b is positive i.e.:

    x + 3x = +/-149,036 . The equation has only one root [x = +/-53].
    Is there a method to find quickly this solution with only a few operations? The same applies when b is negative in the region where there is only one solution:
    x - 3x = +/- 148718
    what is , as today, the best general algorithm, method?
    Thanks.

    Btw, do you know that if c is an up-to-9 digit integer you can find the solution straight away, using only logics?


    Last edited by logic; April 14th, 2012 at 12:23 AM.
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    This equation has one real and two complex roots, generally speaking. It is actually possible to solve x^3+bx+c=0 exactly and analytically:

    x^3+bx=c - Wolfram|Alpha

    So the best algorithm would probably be to use these general solutions and simply insert the values for b and c. I would imagine that actually finding these analytical expressions via algebra is fairly time consuming ( I haven't tried ).


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    The OP asks for an algorithm, not for a print-out of an online calculator.There is a sort of algorithm, though I'm not sure I (or anyone els) would describe it as such.

    So suppose that is a polynomial over the field . Any such that is called a zero (or root) for .

    So that the element is a zero of iff . (The pipe, or vertical bar, means divides exactly). This is easy to see........

    If then there must exist some polynomial of lower degree than that of call it such that . Now .

    As a generalization, the zeros of any polynomial may be written as where has no zeros in and the are the multiplicities of the zeros

    I offer 2 simple examples (arguing backwards, as it were). Suppose , This doesn't factorize, yet we see by simple inspection that are zeros. So although the polynomial above doesn't factorize, I may still have that so that (edited by author)

    Or, suppose that . This does factorize, as , so that with multiplicity 1, and with multiplicity 2.

    Hmm....this came out a bit more technical than I had hoped. But, hey!! This is a mathematics forum, live with it
    Last edited by Guitarist; April 13th, 2012 at 07:01 AM.
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    Quote Originally Posted by Guitarist View Post
    (The pipe, or vertical bar, means divides exactly).
    Just as a curious reader I'm wondering what "divides exactly" exactly means? And how it is different from usual division?

    iff
    I sort of dislike that shortening as to some readers it just looks like a typo, when logically it has a very significant meaning. No worries though, just saying.
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  6. #5  
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    Pretty sure it means a divisor that leaves a zero remainder.
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  7. #6  
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    Quote Originally Posted by Markus Hanke View Post
    This equation has one real and two complex roots, generally speaking. It is actually possible to solve x^3+bx+c=0 exactly and analytically:...
    Could you please explain what you mean?
    If b is positive there is only one root (Khayym solution) and complex numbers are not necessary: the solution is always exact where x is an integer or has a finite number of decimals,
    Quote Originally Posted by Guitarist View Post
    The OP asks for an algorithm, not for a print-out of an online calculator.There is a sort of algorithm
    Hmm....this came out a bit more technical than I had hoped. But, hey!! This is a mathematics forum, live with it

    Thanks, Guitarist, but what about Cardano, isn't it an algorithm?
    Is it still the best method, so far, as it requires only 12 operations: x= n-m, where n = (c+sqrt(c+b/27))/2 and m = a-c ?

    (It is not bad , but , as I mentioned, you can find the exact solution with less operations, and in most cases, with no operation at all.)
    Last edited by logic; April 18th, 2012 at 01:48 AM.
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    Quote Originally Posted by logic View Post
    Could you please explain what you mean?
    What I meant was that in general this equation has three roots, but only one of them is real, whereas the other two roots are complex numbers. If you restrict the solutions to the set of real numbers only, then you are correct in saying that there is only one solution.

    If you follow the link provided you will see the three general solutions as expressions over the coefficients (b,c).

    Guitarist, you are of course right - what I gave wasn't an algorithm. What you presented in post 3 is the proper way to do this, however, it mightn't be the most practical. For example, implementing your algorithm in computer code would be most challenging...
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  9. #8 Iterate! 
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    We know that there are explicit analytical solutions, and a practical answer to the question - ask any Mathematical software, e.g. Mathematica, and get an answer in a fraction of second.

    However, I think author meant different task. First, we only want a case with unique real solution. That is why we want b>0, to be sure that is monotonously increasing.

    Let us take also c>0 (c<0 case is obtained by x -> -x ) and assume that we do not want use the explicit formula for solution.

    First, we have to estimate a magnitude of solution. Take , solution is smaller than this number. In fact, if , gives already a good approximation.

    Then, if solution is not necessarily integer and b satisfies condition above (in the example of the author this is so), we can iterate . We can try to invent a better decomposition for iterations, but this is a case by case study.

    If solution is expected to be integer, try step by step all integers smaller than , starting from the largest possible integer. For the provided example, already the first guess gives an answer!

    You can improve your technique if you first solve equation over some finite field. That is, you can check whether solution is even, is divisible by 3 etc.
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    Quote Originally Posted by Markus Hanke View Post
    If you restrict the solutions to the set of real numbers only, then you are correct in saying that there is only one solution. If you follow the link provided ...
    I am aware that that is the current opinion, Markus, but I refute it from the point of view of logics : if, historically, the "casus irriducibilis" had not occurred, we would not be here to discuss of imaginary solutions (and numbers). It seems obvious that a method that can solve polynomials without discriminants (positive or negative) does not need the complex numbers.
    If you look at the x+bx=c in any function explorer/ plotter you can see that there is no value of y (c) where there is more than one root (why real or imaginary?). The function is as monotonic as x=c, would you ever say that x has two imaginary solutions?
    Let's stick to the Law of parsimony.

    I do not know if you have examined the method I quoted, it gives the exact solution for every c when b>0 and one of the 2-3 solutions when b < 0.

    Moreover, in most cases, you need not waste time even with 12 operations: give me a number with 3 decimals with a few digit missing ( to make sure I am not cheating) and I'll give you the exact solution in no time, using only logics [knowing, of course the times table] :

    x = 84.6; x + 7x = 606087.936. If you give me , for example, 606,xxx.x36 , I can give you the exact solution using a simple algorithm.

    p.s. is there a term to indicate an integer or number with finite numbers of decimals?
    Last edited by logic; April 18th, 2012 at 01:53 AM.
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    Leaving aside the towering arrogance of this post, it is quite wrong. There is a fundamental theorem that states the for any polynomial of degree there are exactly zeros, if you include multiplicities.

    An algebraic proof is fiendish, but try this....

    If is true, and if we write this in the extremely ugly form and if we write the degree of a polynomial as , say, then must be equal to , otherwise the equality fails.

    Of course, it is allowed to fail, but then an alternative must be offered. That is why the algebraic proof is hard.

    Personally I dispute this can be done using logic's "logics"
    Last edited by Guitarist; April 13th, 2012 at 09:44 AM.
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  12. #11  
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    Quote Originally Posted by logic View Post
    ..in most cases, you need not waste time even with 12 operations: give me a number with 3 decimals with a few digit missing ( to make sure I am not cheating) and I'll give you the exact solution in no time, using only logics knowing, of course the times table :
    If you give me , for example, 606,xxx.x36 , I can give you the exact solution using a simple algorithm.
    Quote Originally Posted by Guitarist View Post
    Leaving aside the towering arrogance of this post, it is quite wrong.
    There is a fundamental theorem that states the for any polynomial of degree there are exactly zeros, if you include multiplicities.
    An algebraic proof is fiendish, but Of course, it is allowed to fail, but then an alternative must be offered. That is why the algebraic proof is hard.

    Personally I dispute this can be done using logic's "logics"
    I cannot see 'towering arrogance' on my side, Guitarist. The post is not arrogant. Stating the primacy of logics (not logic's logics) can, neither, be the arrogance of logics, as it is well known that it is the foundation of math and of the universe.

    I do not understand your attitude, either, but will not qualify it.
    I made this post to verify the state of the art, to see if Cardano's is still, after so many centuries, the best way to solve a depressed cubic : the title of my thread asks: "the best way to solve x+bx=c"?.
    You keep ignoring this and offer me fancy methods that you, yourself, say may fail.
    I was not obliged to offer you an alternative, as I am just a logician, but I did, an alternative that is not allowed to fail.
    You dispute that something can be done, you do not prove your proposition, you do not challenge mine.

    Where is your logics? why be rude?

    The fundamental theorem is only a general frame, you should read it properly" nth power/n zeros" (it's understood : at most): if all powers are odd and b >0, as Vel has already confirmed, there is only one solution.

    BTW, the wolfram site does show an algorithm.
    Last edited by KALSTER; April 19th, 2012 at 05:33 AM. Reason: fixed quote tags
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  13. #12  
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    Guitarist, you are giving general theory about polynomials over general fields, this is not an algorithm for solving a practical problem. Algorithm is something that you can teach computer to do.

    Logic, you did not precise what do you mean by "an operation". Is taking a square root counts as one operation? If yes, then your answer with 12 operations is probably the best way to solve the problem if b and c are really arbitrary, though I do not know a theorem.

    But let us modify your question. You have to write the program on the computer which solves your equation in the fastest way possible (and the winner takes 1 million dollars
    ). Then I am certain that Cardano formula is not the fastest way. How you are going to teach computer to take fractional powers? Certainly through Taylor expansion or similar, which is much more than one operation.

    If you think a little bit how you implement any algorithm in practice (including Cardano formula), it will be very similar to the iteration procedure I described.

    Your "pure logic" allowing to find solution actually also works similarly to iteration reasoning. It is based on two facts
    1) You know that solution is nice: rational and with not too large numerators and denominators. Btw, how do you know this?
    2) If relative error in the knowledge of c is , then relative error in the knowledge of
    is of order , if b is small enough. And then of course, if c is large and answer is nice you can the answer.

    If we have some polynomial equation and we know that solution is nice enough, there is a PSLQ algorithm to find it. A variant of PSLQ will win $$$.

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    Quote Originally Posted by Vel View Post
    Logic, you did not precise what do you mean by "an operation". Is taking a square root counts as one operation? If yes, then your answer with 12 operations is probably the best way to solve the problem if b and c are really arbitrary, though I do not know a theorem.

    Thanks for your brilliant post, Vel. That's sound reasoning. Let's start from the beginning:

    Question: The best way to solve x + 12x = 63 ?Answer: IMHO :
    k = [63 + sqrt ( 63
    + 4 * 12 : 27)] : 2 = 64 ; n = k - 63 = 1; sqrt 64 = 4; sqrt 1 = 1, x = 4 - 1 = 3. The (only) solution of the cubic is x = 3
    13 operations +,sqrt,x, +, x,
    x x , : , : , -, sqrt, sqrt, - , (counting ^3 two ops).
    Guitarist, would you agree that this is the correct way to answer a post?

    This is not 'my answer' anyway, Vel, it is called Cardano's method, and is centuries old; then you have Lagrange, trigonometric/hyberbolic, geometric, Khayym etc.; wolfram sometimes gives you a glimpse of the underlying algorithm , which starts with : cuberoot(2b:3)/(cuberoot(sqrt (3)*sqrt(4b+.....)
    I am not a specialist and would have taken me ages to evaluate all these procedures, so I asked for your advice.


    Your "pure logic" allowing to find solution actually also works similarly to iteration reasoning. It is based on two facts
    1) You know that solution is nice:.....
    2) If relative error in the knowledge of c is .....
    .


    Now, suppose you ask me to solve x + 7x = 606xxx,xx6
    My logic is not "iteration", I get the solution straight away, and it is not 'pure logic' knowing a few rules by heart, the times table, and performing simple basic operations , like 28-25, 12 * 4 etc... in my mind.
    Of course, if you give me a 30-digit number, or x has infinite decimals, I cannot get the solution directly, and must use a simple iterative algorithm and a pocket calculator to get the required precision in a second or two. But even with pencil and paper the method would require only a few operations to reach a 10-12-digit accuracy, probably less ops than Cardano's.
    It can be useful with quintic equations: x^5 + (2x^4) +3x+ x = 30617606.25024, (which I heard have no solution , do you know of a formula for solving that?), as yet.( you can get x = 31.4 immediately.)
    Last edited by logic; April 14th, 2012 at 04:15 AM.
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    Quote Originally Posted by logic View Post
    If you look at the x+bx=c in any function explorer/ plotter you can see that there is no value of y (c) where there is more than a solution, real or imaginary. The function is as monotonic as x=c, would you ever say that x has two imaginary solutions?
    Let's stick to the Law of parsimony.
    What exactly are you implying ? Are you saying to me that a cubic function of the form x^3+bx=c always has only one root ?
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  16. #15  
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    Ok, so for you taking square root is as elementary operation as + or *. This is practical for most people who lives today because what you will do, you will take calculator and press or , etc on it and get immediate answer for an operation.

    But suppose you live in the time of Newton, and you want to compute square root. Then you will immediately realize that it is a hard task, except if you have a table of answers (which is the same as having calculator). So what I want to say is that among the formulas that guarantee the exact answer, Cardano formula can indeed have the least amount of operations in the sense that you defined. But in practice, there are faster ways to get the answer.

    If you want to play a little bit, below I write a formula which is not exact, I can easily provide examples when it will fails. But it will work perfectly for all the examples you wrote, and it is much faster then Cardano:



    Before answering me, please take calculator and check!

    What I did, is just a Taylor epxansion. In Newton times this was the main method of solving such kind of equations.

    Once more, I don't pretend that this is exact, but it gives answers with sufficient enough precision for all your examples.


    For equations of degree 5 or higher there is no solution in radicals (1/n roots and +,-,*,/) for arbitrary coefficients in equation, it is a result of the Galois theory. But approximate formulas using Taylor expansion can be written. So, for equation , formula below gives excellent precision for your example (which contains a mistake btw):

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    Quote Originally Posted by Markus Hanke View Post
    What exactly are you implying ? Are you saying to me that a cubic function of the form x^3+bx=c always has only one root ?
    Yes , if b >0, it has the same shape of x, of x^5, x^5+x+x, etc.. isnt' it so?
    Could you please show me the other roots of x+ 12 x =63?
    Last edited by logic; April 14th, 2012 at 04:16 AM.
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    Quote Originally Posted by Vel View Post
    1) Ok, so for you taking square root is as elementary operation as + or *. ...But suppose you live in the time of Newton...,
    ............
    2)Once more, I don't pretend that this is exact, but it gives answers with sufficient enough precision for all your examples.
    For equations of degree 5 or higher there is no solution in radicals ....
    3).....(which contains a mistake btw):

    1)If you can take a cube root, a fortiori you can take a square root, I am talking, of course of x integer or with finite decimals.
    You can solve x= 277729 without a pencil, Vel, and I have made a fresh thread implying you can make a powerful algorithm using only one operation and reach, just with a pencil 10-20-decimal precision in a short time with only a handful of operations.
    Newton's algorithm is less powerful than Babylon or Bashkali, (you can adapt Babylon to the cube root: 5 ops vs. Newton 7), and Taylor is less powerful than Newton and possibly the worst.
    Anyway, I was looking for the best method, the one with fewer operations: less operations is good in any age, if you are using a calculator or a pencil, makes no difference. I cannot get your point.
    2) I am always talking of exact solutions, when x has finite n of digits.

    Now since you are so kind and competent , tell me please:
    a) what do you call a number like 38.5, 745.987654321 etc...
    can I identify such a number saying n= x or x?
    b) could you show me (practically, with an example, as I did) how you solve exactly: x-bx = c, when there are 3 roots?
    c)or exactly: x^5 + bx+ cx =d?
    3)) sorry, what mistake?
    Thanks a lot
    Last edited by logic; April 14th, 2012 at 03:35 AM.
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    Quote Originally Posted by logic View Post
    Yes , if b >0, it has the same shape of x, of x^5, x^5+x+x, etc.. isnt' it so?
    Could you please show me the other roots of x+ 12 x =63?
    This polynomial has three roots :





    It is actually easy to show that every cubic polynomial of the form x^3+bx=c always has three roots, one real one and two imaginary ones, so long as b > 0.
    You are right though in saying that only one of them is real.
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    Quote Originally Posted by Markus Hanke View Post
    You are right though in saying that only one of them is real.
    Yes, Markus, that is what I mean, on a plotter you see only one intersection you cannot show anything else, the rest is just (literally) imagination!

    There is only one root, there is only one value of x that corresponds to the value of y (= c)
    Last edited by logic; April 18th, 2012 at 02:16 AM.
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    Quote Originally Posted by logic View Post
    Quote Originally Posted by Markus Hanke View Post
    You are right though in saying that only one of them is real.
    Yes, Markus, that is what I mean, on a plotter you see only one intersection , the rest is just (literally) imagination!
    Agreed.
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    Quote Originally Posted by Markus Hanke View Post
    Agreed.
    Have you caught a glimpse of the formula at wolfram?
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    Quote Originally Posted by logic View Post
    Quote Originally Posted by Markus Hanke View Post
    Agreed.
    Have you caught a glimpse of the formula at wolfram?
    No, I used MAPLE to get the roots, but I did use WA to get the general expressions referenced in post 2.
    I am more of a physics guy than a maths guy, so for me the end result is more important than the algorithm used to arrive at it. Therefore I have no problem taking a shortcut via Wolfram or CAM software.
    Guitarist will of course disagree, he is the classic "maths guy", and he is very good at that.
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    Quote Originally Posted by Markus Hanke View Post
    I am more of a physics guy ...so ....
    I am glad to hear that, Markus, I was just looking for one . Could you , please, tell me what instances are there in physics of a cubic function x+x+x ?
    Can you find one where more than one root is possible?
    While we are at it, tell me, if you have time to spare, can you find imaginary roots in x = 343?
    Thanks.
    Last edited by logic; April 18th, 2012 at 02:18 AM.
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    Quote Originally Posted by logic View Post
    Could you , please, tell me what instances are there in physics of a cubic function x+x+x ?
    I cannot immediately think of any, which isn't to say they don't exist. Why are you asking this ? I fail to see the relevance to this thread.

    Can you find one where more than one root is possible?
    I'm not sure what you mean by this. x^3+x^2+x has again one real and two complex roots.

    While we are at it, tell me, if you have time to spare, can you find imaginary roots in x = 343?
    Of course - yet again, you have one real ( x=7 ) and two imaginary roots :

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    Quote Originally Posted by Markus Hanke View Post
    . Why are you asking this ?
    I do not know if you are willing to discuss theoretical issues, as you said you are more of a physics guy. If you are, I'll be delighted to explain.
    Thanks for your attention, Markus, much obliged.
    Last edited by KALSTER; April 19th, 2012 at 05:30 AM. Reason: fixed quote tags
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  27. #26  
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    Quote Originally Posted by logic View Post
    I am talking, of course of x integer or with finite decimals.
    As you see, it was not clear to nobody who replied to your post.

    Quote Originally Posted by logic View Post
    what do you call a number like 38.5, 745.987654321 etc...
    These are rational numbers but I would write them in the form and .
    If you write for a math guy 38.5, he would immediately think about real number known with 3-digit precision only (so this number can be 38.500000000 or 38.5000000120 or even 38.51).

    So it is better to use the word "rational number". Since in your case this rational number is of type integer/10 or integer/100 etc, let us redefine x=y/10 (or /100, or other number, depending how many decimals after point you have) and speak only about an equation on y which has the unique integer solution.

    If you have an algebraic equation and you know that it has an integer solution, then the most powerful method for today to find an answer is PSLQ algorithm (and its cousins). It is not an algorithm which gives you explicit formulas like Cardano. It is a quickly convergent (to an integer) numerical method which uses mathematically justified guesses. Many amazing relations in number theory were found in this way, and they were proved only later. This is what is called experimental mathematics.

    Quote Originally Posted by logic View Post
    less operations is good in any age, if you are using a calculator or a pencil, makes no difference. I cannot get your point.
    My point is that square root and cubic roots is a more time consuming operation than others, so we have to try to reduce number of times we use them. It is impossible to avoid cubic root. My proposal is a way to guess the exact answer by taking only one cubic root. Explicit example you proposed:
    . in this example. Therefore a good approximation (using calculator, but only once) . If I know that the answer is integer, then the first guess will be , then I check that this is correct. The check is simple. I don't need compute 53^3+3 53, but I make estimation of a mistake I could make. It is , which is much smaller than . So 53 is the only integer in the range predicted by estimated mistake. I can make another simple check to be confident: The last digit of 53 is 3, therefore the last digit of is 6, which is correct.

    Quote Originally Posted by logic View Post
    Now, suppose you ask me to solve x + 7x = 606xxx,xx6
    My logic is not "iteration", I get the solution straight away
    So, can you please explain your logic to get solution right away? I applied my method described above and I claim that there is no rational solution to this equation which can be expressed in a decimal representation with at most 4 integers after point.



    Quote Originally Posted by logic View Post
    b) could you show me (practically, with an example, as I did) how you solve exactly: x-bx = c, when there are 3 roots?
    I showed above. Sign of b is inessential for my method when c is large.

    In your practically explained example you used Cardano formulas, and you was lucky to be able take square roots explicitly. In general case you will face square roots which is more difficult to take. Try to apply Cardano formulas to:



    Actually a question we can try to answer is the following: Is there a way to write solution which contains only such roots that can be evaluated to rational numbers, if it is known that x is integer.

    Quote Originally Posted by logic View Post
    c)or exactly: x^5 + bx+ cx =d?
    Galois theorem states that this is impossible for arbitrary b,c,d (well, strictly speaking I know only the theorem when there is also term present. But it is known how to analyze any given polynomial equation and to conclude whether or not it can be solved in radicals.).

    Quote Originally Posted by logic View Post
    3))what mistake?
    x=31.4=314/10 is not a solution to the equation: x^5 + (2x^4) +3x+ x = 30617606.25024
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    Quote Originally Posted by logic View Post
    Quote Originally Posted by Markus Hanke View Post
    . Why are you asking this ?
    I do not know if you are willing to discuss theoretical issues, as you said you are more of a physics guy. If you are, I'll be delighted to explain.
    Thanks for your attention, Markus, much obliged.
    I am more of a physics guy, but still happy to discuss theoretical issues. It's just that I have clear limits when it comes to the intricate details of maths...anyway, if you have something you want to discuss, then bring it on
    Last edited by KALSTER; April 19th, 2012 at 05:30 AM. Reason: fixed quote tags
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    Quote Originally Posted by Markus Hanke View Post
    I have clear limits when it comes to the intricate details of maths..
    .anyway, if you have something you want to discuss, then bring it on
    All right, Markus, you surely have no limits, because I know 1 /10 of the math you know and can manage. It is not about the intricate details of math,
    but about the simple tenets of logics. I mentioned the basic Law of parsimony.
    Cubics with b,..>0 have one solution, cubics with b<0 or x (etc...even powers) have more than one solution in an infinitesimal region of the y-axis:
    let's take a plotter and peruse the curve of a cubic function: you have
    - 1 solution for every value of x or y, or
    - 1 root everywhere, 2 roots at the (what do you call them: flexes, extrema? please correct my terminology) and 3 between them: a microscopic region [ and only when (b/3)< (-c/2)]. Why warp reality, and state that every cubic has three roots? The roots are always real, there are no imaginary roots.
    I think, in principle, that it is extremely dangerous to call a root 'imaginary' only because you are not able to find the root of -1,-4 ...etc.

    The cubic that brought about the "casus irriducibilis" : x -15x = 4, is a splendid example, with elementary logic (x=16-1=15: 4=1*x ) you find the solution. x- 20x =25 (x=25 -5=20: 25=5*5 ) and so on
    There are many ways, method, algorithms to find all the roots of a cubic, you do not need complex, imaginary numbers , because there are no imaginary roots.
    Can you show me some of what you deem the most difficult cases of cubics?

    I'll pause to wait for your comments and check if you want to go on.

    P.S.could you , please, make a drawing of a cubic x+ bx + cx = d and its depressed equivalent plotted together?
    Last edited by logic; April 18th, 2012 at 02:25 AM.
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    Quote Originally Posted by Vel View Post

    1)As you see, it was not clear to nobody who replied to your post.
    2) So, can you please explain your logic to get solution right away?
    3) which is more difficult to take. Try to apply Cardano formulas to:

    4)x=31.4=314/10 is not a solution to the equation: x^5 + (2x^4) +3x+ x = 30617606.25024
    I answer quickly some points, as I have to go now, I'll gladly discuss in depth all questions at another time:
    1)I was referring to [9-digit] rationals only when I said you can get the root immediately I stated it clearly in OP.
    2) Let's check if it works : take any rational you like xx.x as the root of x5+ x+x and give me from the result only 4-5 digits out of the 13-15 : first 3-4 and last. No formula can get the solution, so you'll be sure logics (not my logic, please) is more powerful
    3) please note I am not a supporter of Cardano. The method is valid for that cubic too, I see no problem, probably tou cannot work it out because your calculator has only ten digits. But the case is so simple that you do not need Cardano to find 51, it is patent , too easy.
    4) sorry, a slip of the finger (I did not change glasses from the PC to the calculator) 30617386.45024. I hope I got it right this time. Now if you give me 3061xxxx.xxxx4, you'll get the solution

    If you are interested, Vel, please say your view in the discussion I'm having with Markus.
    Thank you for your attention
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    Quote Originally Posted by logic View Post

    2) Let's check if it works : take any rational you like xx.x as the root of x5+ x+x and give me from the result only 4-5 digits out of the 13-15 : first 3-4 and last. No formula can get the solution, so you'll be sure logics (not my logic, please) is more powerful
    ...
    4) sorry, a slip of the finger (I did not change glasses from the PC to the calculator) 30617386.45024. I hope I got it right this time. Now if you give me 3061xxxx.xxxx4, you'll get the solution

    If you are interested, Vel, please say your view in the discussion I'm having with Markus.
    Thank you for your attention
    2) You never told what you are doing explicitly. But whatever it is, it is based on the fact that changing say xx.3 to xx.4 drastically changes the magnitude of x^5, it affects numbers in first 3-4 digits of answer, which you know... Explanation for this: if you make a fair estimate about x, the mistake is very small, of order 1/x, it is smaller than 0.1, so it affects no digits in xx.x. Then you need to guess this last digit, if not yet done by first 3-4 digits of answer. That is last digit of answer which helps.

    4) Actually it was x^5+3x^3+x=306**.... which you was solving.

    About appearance of cubic (and any polynomial equations) in physics - there are a lot. Simplest nice example: Van-der-Waals equation of state is cubic in volume, and there is nice interpretation about possibility of gas/water phase transition. If polynomial in V is monotonously increasing (one real solution) then there is no phase transition, if there is exists a region with three real solutions - there is phase transition.

    Complex numbers certainly appear in physics. Already in description of electric circuits. And quantum mechanics is impossible without them. It is true that we cannot observe complex numbers, but it is wrong that we don't use them on intermediate states. Actually we observe only integer numbers. All the rest is invention for our convenience.
    Last edited by KALSTER; April 19th, 2012 at 05:29 AM. Reason: fixed quote tags
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    Vel: I admire your patience, but I give up on this guy, save for one thing.

    I tried to explain this before, to no avail. It is a fact known to every schoolboy (though he will almost certainly not be able to prove it), and here I will shout from sheer exasperation:

    ANY POLYNOMIAL OF DEGREE HAS AT LEAST AND AT MOST ZEROS (ROOTS), THOUGH THEY MAY NOT BE DISTINCT.

    \end shouting.

    Consider the polynomial equation given as an example: . First note that errors may very well occur (in our thinking) unless we realize this is in reality .

    There was an earlier claim that there was a requirement that here . This is false. All that is required is that i.e is rational (assuming that is rational - a reasonable but not terribly interesting assumption)

    Which leads us to a nice Definition: A number is called algebraic iff (poor brody!) it is a zero of a polynomial with rational coefficients, else it is called transcendental.

    Notice, as has been repeatedly pointed out, algebraic numbers are in general complex. The fact that some algebraic numbers are "apparently" real causes no difficulties, as we are always free to set via the inclusion map that sends whenever an element in the complex field is written as

    As to the obvious questions: what are the roots of a polynomial with irrational coefficients, or even do such beasts exist. And if they do, are their roots transcendental, or equivalently (or perhaps not), are the transcendentals the roots of any polynomial?

    To these questions, I have not an effing clue, and as I am busy drinking beer on this Sat night, have no intention of even thinking about them
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    Quote Originally Posted by logic View Post
    Why warp reality, and state that every cubic has three roots? The roots are always real, there are no imaginary roots.
    That is incorrect. It is mathematical fact that every polynomial of degree n has exactly n roots, though these n roots might not necessarily be all different, as pointed out already by Guitarist. This is called the "Fundamental Theorem of Algebra" :

    Fundamental theorem of algebra - Wikipedia, the free encyclopedia

    It is clear that some or all of these roots might be complex, which is fine, because the set of complex numbers is well defined in a mathematical sense. Of course depending on the problem the polynomial relates to, the complex roots may not be allowed, or may not be physically meaningful, but they are always well defined mathematically.

    I think, in principle, that it is extremely dangerous to call a root 'imaginary' only because you are not able to find the root of -1,-4 ...etc.
    A root x is called "imaginary" when it is an element of the complex numbers with Im(x) <> 0. As simple as that.

    P.S.could you , please, make a drawing of a cubic x+ bx + cx = d and its depressed equivalent plotted together?
    What are b,c,d ? Without values for these the function cannot be plotted. Also, what do you mean by "depressed equivalent" ?
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    Quote Originally Posted by Markus Hanke View Post
    1) That is incorrect. It is mathematical fact that every polynomial of degree n has exactly n roots,
    2) A root x is called "imaginary" when it is an element of the complex numbers with Im(x) <> 0. As simple as that.
    3)What are b,c,d ? Without values for these the function cannot be plotted. Also, what do you mean by "depressed equivalent" ?
    1), 2) Thanks, Markus for your help. You said you are willing to get entangled in theoretical, epistemological issues, that is exactly what I meant: what is a matematical fact, is it like the notorious Anselm's ontological proof?
    Math is just an arbitrary, symbolic, axiomatic system of equivalences and tautologies. You can prove whatever you want, if you choose the axioms. Axioms are not dogms, are questionable, and if they do not respect reality are useless or even risible.
    I know the things [you and Guitarist and] official math states, I am not that ignorant: the point is that I question them.

    There is something that I ignore , and that is why I asked you if in Nature you can find a phenomenon which has imaginary roots. I asked eminent Profs what are complex numbers, and they shrugged their shoulders: nobody really knows, but axcept them because they "work".

    Now I ask you two questions (a,b), Markus
    a) what is the use of imaginary numbers, apart from covering the incapacity of finding the sqrt od -1,-4... etc? I already asked once: if casus irriducibilis had not occurred, would we be here to talk about imaginary roots?
    Of course the fundamental theorem was made to measure, to justify the shortcomings of mathematicians. I hope someone will not take this as arrogance, it is only the truth.
    If the cubic had been properly solved, would we say today that x=343 has more than one root : 7?
    Is this incorrect?

    b) 2) What does, in reality, mean that 7/2(...) is a root that is a solution to x=343? What is the use of it, anyway, even if it is just a figment? Woudn't math be more rational without it?
    x-7x=-6 (casus irreducibilis)has three roots :-3,1,2 ,
    one math site (sosmath.com) says:"...this is a paradox: though the root is real, formula contains imaginary numbers!...there will never be an improvement of the formula which avoids complex numbers"...that is preposterous, they stopped searching!
    What's the truth? Is it a paradox or a distortion of reality?
    Why dont' they find a method that has no negative discriminants, or the solution to sqrt -1?
    Why don't they look for a method that doesn't require discriminants at all?

    3) x+3x+4x+2=15650 is a cubic, x+x= 15650 is its depressed equivalent? has the latter the same roots as the original x+ax...equation?
    . If you want to discuss it further, you may plot the equation x-7x = -6, ( a depressed cubic with an imaginary root) and the original non-depressed cubic function? x+ ax.....?
    Could you tell me what are the most difficult equations?
    Last edited by logic; April 18th, 2012 at 02:46 AM.
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    Quote Originally Posted by Vel View Post
    2) You never told what you are doing explicitly..
    I simply do not want to spoil your satisfaction, you nearly found it by yourself.
    It's so simple!
    Quote Originally Posted by Vel View Post
    1)These are rational numbers....
    (..2.My point is that square root and cubic roots is a more time consuming operation)
    I'll discuss one issue per post:

    1) The problem is that some rationals have finite (1/8) and some infinite decimals :1/7
    When you meet them in decimal form you cannot recognise them: 1/7 + 3*1/7 gives a result you find truncated in some form, and even if it were not you would not recognize it.
    How can you solve the equation?


    P.S.(2) I do not know if I got it right, but you want to say that a method that does not use sqrt is preferable anyway?
    Can you show me the operations you need with x+12x=63?
    Last edited by logic; April 15th, 2012 at 11:48 PM.
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    Oh dear, do get so very tired of saying this:

    logic I have read all your posts here, and remind you of the following:

    This is a mathematics subforum. You seem to be unaware of the following......

    1. Mathematics is NOT a natural science id est it is under no obligation whatever to describe the natural world

    2. To the extent that a theory of the natural world can be posed in mathematical terms, then if the mathematics is wrong the theory is wrong

    3. If the "natural world theory" can be proved experimentally, say, we had pretty fast find some mathematics to justify this. History shows this can always be done.

    Look, you claim you are not being arrogant. Well, if suggesting that 4 centuries of the finest mathematical minds are wrong and logic is right is not arrogance, then I obviously don't understand the meaning of the word.

    In other circumstances, I would close this thread except for the fact that the valued members Vel and Markus seem to be enjoying it

    Please take my comments to heart
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    Quote Originally Posted by logic View Post
    ...I hope someone will not take this as...
    Quote Originally Posted by Guitarist View Post
    Oh dear, do get so very tired of saying this:
    This is a mathematics subforum. You seem to be unaware of the following......
    Yees it is, and I posed a technical question, Guitarist, "how do you solve x+ 12x = 63 with less operations"?.

    You have started giving answers which do not answer the question, and now accuse me of rambling.
    If you are so clever, why don't you give me a strictly technical answer?
    Thank you!

    Please, say in which subforum we can discuss these issues

    P.S. (after 15 centuries the finest minds hadn't found the negative numbers. Now after 20 centuries the finest minds are still not able to find the sqrt of -4 and a satisfactory way of solving a cubic and a quintic.
    What is your problem? Is remarking this being arrogant?)
    Last edited by logic; April 16th, 2012 at 03:02 AM.
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    Quote Originally Posted by logic View Post
    You said you are willing to get entangled in theoretical, epistemological issues,
    Actually, I only said I was willing to get involved in a discussion about theoretical issues, which I understood to relate to mathematics, not epistemology.

    what is the use of imaginary numbers, apart from covering the incapacity of finding the sqrt od -1,-4... etc?
    They are a natural extension of the body of the real numbers, and greatly simplify the treatment of many problems. Simple example - expressing the trigonometric functions in terms of e. In physics, they play a vital role in quantum mechanics, and lead to real observational predictions, which are all experimentally verified. See Schroedinger Equation, as a simple example.

    Is this incorrect?
    Yes. You know perfectly well that sort(-1) is not solvable except in terms of the imaginary unit.

    What does, in reality, mean that 7/2(...) is a root that is a solution to x=343? What is the use of it, anyway, even if it is just a figment? Woudn't math be more rational without it?
    It only means that, when plotting the function in the complex plane, you get an additional two roots which do not lie on the real axis. Simple ?!

    Why dont' they find a method that has no negative discriminants,
    Because it doesn't exist. Maths are maths, you can't pick and choose what you like and what you dislike. If some of the roots are complex, then so what ? The complex numbers are well defined, and well understood.

    x+3x+4x+2=15650 is a cubic, x+x= 15650 is its depressed equivalent? has the latter the same roots as the original x+ax...equation?
    Mathematically these are two distinct polynomials, so in general they will not have the same roots, however, it is of course possible that some or all of the roots coincide if the coefficients are chosen accordingly.

    If you want to discuss it further, you may plot the equation x-7x = -6, ( a depressed cubic with an imaginary root)
    The roots of this are all real, there are no imaginary roots :

    x&#xfffd;-7x = -6 - Wolfram|Alpha

    Could you tell me what are the most difficult equations?
    "Difficult" is a subjective term.
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    Quote Originally Posted by logic View Post
    what is the use of imaginary numbers...
    You seem to be under the rather naive delusion that because real numbers are called "real" and complex numbers have an "imaginary" components that the former are somehow more, well, real. That is just nonsense.

    I can't answer for phsyics (not being a physicist) but in engineering, we regularly use high order polynomials and all the roots are complex. One component or the other may be 0, but that is rarely relevant. More significant is where the roots lie on the complex plane (e.g. their relationship to the unit circle). Now if you think that whether a piece of hardware works or not is "non physical" then I think you are really stretching things.

    what are complex numbers
    Well, it is obvious what they are. Do you mean what they "mean" or what do they represent? If so, then as with the natural numbers, the answer is equally obvious: it depends.

    The natural numbers don't "exist" as physical objects. We can use them to represent certain classes of things. Ditto the more general case of complex numbers.

    why call it imaginary
    An accident of history.
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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    Quote Originally Posted by Markus Hanke View Post
    They are a natural extension of the body of the real numbers, and greatly simplify the treatment of many problems...
    That's it, that is the point I can't get across:... problems they cannot solve in a better way, what if they could?

    It takes a long time to express my views, and prove that wrong, Markus, and moderator would have a chance to close this thread. It would be a pity, as I want to discuss the technical aspects of my thread.
    You see, to take square root Newton's [finest mind] produced an algorithm that requires many operations, but you can find an algorithm that requires only one.
    The same applies to any (math) problem.

    I'll answer briefly the other points, but I am prepared do discuss them at lenght in any other subforum you choose.

    You know perfectly well that sort(-1) is not solvable except in terms of the imaginary unit.
    1) No, I do not, I do not accept it . You cannot prove that. As I said this is a very dangerous non-scientific attitude, if you accept that attitude, you stop researching
    "What is the use of it, anyway"...?
    It only means that, when plotting the function in the complex plane, you get an additional two roots which do not lie on the real axis. Simple ?!
    2)That does not answer my question, why can't we say that x= n has only one solution?why should I plot it into the complex plane?What are the negative consequences if I refuse to?.
    You keep forgetting the I am questioning the complex plane, anyway
    Because it doesn't exist.
    Maths are maths, you can't pick and choose what you like and what you dislike. If some of the roots are complex, then...t ? The complex numbers are...,
    3)That is dogmatic, too. No one can say or prove that a priori.
    The bolded proposition is a simple fallacy "begging the question". They are is the issue being questioned
    Mathematically these are two distinct polynomials, so in general they will not have the same roots,....
    4) is this a reliable site ? :The Geometry of the Cubic Formula, and this :Cubic function - Wikipedia, the free encyclopedia
    If it is as you say, why reduce/depress (which is correct?) a cubic?

    The roots of this are all real, there are no imaginary roots :x&#xfffd;-7x = -6 - Wolfram|Alpha)
    5) see link quoted above
    "Difficult" is a subjective term.
    6)... one that can justify the existence of complex numbers and imaginary roots.

    P.S. Please answer the main question that can make me change my stance:
    7) if they could take the square root of -1, would you still need or use complex numbers?. If yes, please tell me for what?
    I cannot find it anywhere!
    Thanks you for your attention and patience, Markus! (I hope I did not sound arrogant. Was Kopernicus, Galilei ...arrogant?)
    Last edited by logic; April 16th, 2012 at 05:58 AM.
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    Quote Originally Posted by Strange View Post
    1)You seem to be under the rather naive delusion that because real numbers are called "real" and complex numbers have an "imaginary" components that the former are somehow more, well, real.
    2)in engineering, we regularly use high order polynomials and all the roots are complex.Well, it is obvious what they are. Do you mean what they "mean" or what do they represent? If so, then as with the natural numbers, the answer is equally obvious: it depends.
    3)The natural numbers don't "exist" as physical objects
    4)An accident of history.
    Thanks ,Strange, for your post
    1) I am not
    2) Can you please give me a case in which is not involved?
    3) Of course
    4) Pity
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    Quote Originally Posted by logic View Post
    2) Can you please give me a case in which is not involved?
    I don't understand the question.

    In the engineering fields I am familiar with (or, least, was several decades ago ) all the roots of a polynomial are significant and all are complex. Clearly, if it is a polynomial of 3rd order, there will be three roots. If the order is 100 then there will be 100 roots. All complex, all equally real, all equally important, and all equally physical.

    I'm not sure what your point is.
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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    Quote Originally Posted by Strange View Post
    I don't understand the question.
    ..., if it is a polynomial of 3rd order, there will be three roots.... If
    I'm not sure what your point is.
    My point is the same as in the other posts: if we find a simple way to solve those polynomials, do we need imaginary numbers?
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    Quote Originally Posted by logic View Post
    My point is the same as in the other posts: if we find a simple way to solve those polynomials, do we need imaginary numbers?
    Yes, because the roots are points on the complex plane. To put it another way, they each encode/convey two pieces of information. In signal processing this might be amplitude and phase, for example.
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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    Quote Originally Posted by logic View Post
    No, I do not, I do not accept it . You cannot prove that. As I said this is a very dangerous non-scientific attitude, if you accept that attitude, you stop researching
    Right then, since you do not wish to accept the existence of complex numbers, why don't you show me a real number the square of which is -1.
    I am looking forward to your soonest reply. And no philosophical excuses please, just a straight simple answer to a straight simple question.
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    Quote Originally Posted by logic View Post
    My point is the same as in the other posts: if we find a simple way to solve those polynomials, do we need imaginary numbers?
    Yes.
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    Quote Originally Posted by logic View Post
    2)That does not answer my question, why can't we say that x= n has only one solution?why should I plot it into the complex plane?What are the negative consequences if I refuse to?.
    If you are designing a system (e.g. an signal processor or, I believe, a control system although I am not an expert on control theory) and you ignore the solutions which do not lie on the real axis then your system will not work. Your filter will oscillate, your robot will rip itself apart, etc.
    As simple as that.
    brody likes this.
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    Quote Originally Posted by logic View Post
    You see, to take square root Newton's [finest mind] produced an algorithm that requires many operations, but you can find an algorithm that requires only one.
    And what is this algorithm?
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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    Quote Originally Posted by Markus Hanke View Post
    You know perfectly well that sort(-1) is not solvable except in terms of the imaginary unit.
    Quote Originally Posted by logic
    No, I do not, I do not accept it . You cannot prove that.
    I am reasonably sure he can, but just in case........

    Write for some REAL

    The usual "rules" for the multiplication of Real numbers dictate that

    1. is positive for any

    2. is positive for any real

    3. is negative for any real and any

    So .

    But the usual "rules" dictate that refer to a Real number multiplied BY ITSELF.

    So . But the only Real number for which is could even for second considered true is and if we can forget the sign and write whci by the cancellation law for the Real numbers becomes

    But for all Real which implies that which in turn implies that which is clearly absurd for Real which in turn implies that

    This, if you never encountered it before, is called "proof by absurdity"
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    Hi everybody,

    I have to support Logic in the point that his inquiry was about the best algorithm to find a solution to the polynomial equation, if it is known that the solution is integer (or very special rational number). So, discussion of complex numbers if off-topic, strictly speaking. Btw, Logic seems to have his own algorithm to quickly search for solutions, and he never disclosed it to challenge us. I am certain that it is all about estimating a magnitude of the answer, as I explained in detail in my posts.

    But I have to support everybody else that you, Logic, should admit importance of complex numbers (in more general context than your post). You had been given a plenty examples of the usefulness by each who posted here, and I can add one more which will be closer to you: you should know properties of complex numbers to prove Galois theorem (which is true also in the case when solution itself is real).

    And since we already have this discussion, let me add my 5 cents. The complete truth is that there are only two objects we really need: 0 - absence of something and 1 - presence of something. All the numbers can be derived step by step from these two. So, I can think about 1 as a dash on the paper. If I add one more dash, I will define this as 2 dashes, in this way I define 3,4,... integer number of dashes. When I play a little bit, I will understand usefulness of substraction operation, and earlier or later I will arrive to understanding that I cannot subtract say 2, from 1. Then, to resolve the problem I define -1=1-2 and I will find that it is a useful definition when I ought more money than I posses... To compute surface of something, it will be useful to introduce multiplication, and to divide apples between friends - division. Then I will soon realize that 1 cannot be divided by n and I will define a new number that I will call 1/n (piece of cake). Thus I will get what is called rational numbers. Then I will need to compare surfaces and invent the Pithagorian theorem, which a quadratic equation and I will get an idea of taking square roots. Then I will soon realize that is not a rational number, therefore I will define... well, you know (and you know that these numbers are useful). If I ask whether the ratio of length of a circle to its diameter can be expressed in terms what was defined until now (algebraic numbers), a theorem was proven that it is impossible. So let us define this ratio as a new number called . What I can do with all these numbers, I can compare the magnitude of them, so I can show that < . Then, I will introduce for my convenience notation which literally means that and , and precision can be increased by adding more decimals. Then, to be certain that I am ready to describe any new number like , I will just develop general decimal notation with infinitely many digits. This notation, which defines what is called Real numbers, includes everything I found until now, and even more.

    Can be I happy now? No, because as I could not evaluate in terms of rational numbers, I cannot evaluate in terms of Real numbers (because by definition, square root of x is a number that squares to x, and Guitarist explained exhaustively that there is no such real number when x=-1) . So, as you can guess now, I define a new object , which has a property . What is unusual about this object? It cannot be compared with real number. It is no sense to ask whether some real number or some real number, because existence of this relation would immediately imply that is a real number itself (from definition of real numbers hinted above). But let us ask that I can do with any other operation: add to it numbers, multiply by numbers, take powers... . This demand will lead to the definition of generic complex number , and to our utter satisfaction we do not need to invent anything more because field of complex numbers is algebraically closed. This means that all the operations you did with real numbers, including powers and square roots (and even more: logarithms, exponents, in fact taking any smooth function), you can do also with complex numbers, and to find the result you will never need to define new objects. Already this is a great reason to work with complex numbers, but besides that you find them extremely useful in mathematics, physics, engineering...

    So as we now so advanced in idea to define something new, let us ask what else useful (or useless ) can we define? Many things in fact. So, if we want to define general rotation of our 3-dimensional space, to produce a graphical card, we need to learn how to work with matrices which are just collection of numbers with some rules how to work with this collection. If we want to go fancy and exotic, this collection can be called quaternion ring. It was just one example, many objects can be constructed. But they are not called numbers usually because some operations with them are not the same as with numbers. For instance, multiplication of quaternions is defined, but for some elements (however, some people do say "quaternionic numbers", which I think is abuse).

    The final conclusion is that in Nature there is only 0 and 1. And you can describe everything, using them only! But certainly our life-time is limited, so probably we will not do that but go and define some collective objects (natural, rational, real, complex numbers) and will deal with them.
    Last edited by Vel; April 16th, 2012 at 12:40 PM.
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    This reminds me of people who argue that we still might find a repeating sequence in the decimal expansion of pi. Just arguing away without a clue.
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    Quote Originally Posted by logic View Post
    Please consider the depressed cubic equation when b is positive i.e.: x + 3x = +/-149,036 . The equation has only one root [x = +/-53]. Is there a method to find quickly this solution with only a few operations?
    Quote Originally Posted by Guitarist View Post
    , I would close this thread except for the fact that the valued members Vel and...
    Quote Originally Posted by Vel View Post
    ..1).his inquiry was about the best algorithm to find a solution to the polynomial equation
    2)you, Logic, should admit importance of complex numbers
    3)I cannot evaluate in terms of Real numbers .... So, as you can guess now, I define a new object], which has a property . ...
    I must congratulate you, Vel: you absolutely deserve to be considered a valued Member after a handful of posts!
    You are right, after 50 posts I haven't received the shred of an answer to my query!.
    I will *avoid further sterile discussion on imaginary numbers.

    I ask you all to present your *solutions to x+ 12x = 63. Any method is welcome, Taylor, Complex n, approximations ...anything that will yield x = 3 as its root. I hope Vel shows in this example the method (hinted in post #15) that is faster than Cardano.
    , Thanks Vel.
    I must stress on one point. The algorithm (a formula is an algorithm used only once, isn't it?) is valid for all numbers: if you start with a number (I'llcall it: n. ) with n decimals you'll find the solution: n. No approximation, you can always find the original number.
    * Edit: solution: any formula, method, algorithm that finds the exact root [x=3], showing the individual operations!



    *If someone is interested and moderator agrees, I am prepared to continue the discussion in separate posts or in a fresh thread. Thank you, all!
    P.S. I'll add a brief comment to your brilliant post:
    2)I have conceded usefulness, I refuted that imaginary numbers is the best (let alone the only) makeshift to your incapacity to solve
    . If the motorway from Bonn to Berlin is blocked, there are a billion alternative routes you can take to reach Berlin, but if you choose to reach your goal via the South Pole I do have a right to criticize you. Finest minds did not find a better route probably as they thought, like you, that there is no other way; this is my stance, probably I could not make myself understood.
    3) That's a brilliant exposition, Vel, but I know that: I have thoroughly studied the problem; moreover, from a technical point of view, complex numbers do not even completely solve the problem, because they shift the problem from -1 to

    If, as someone said, Complex numbers are useful apart from taking the root of negative numbers, I'll be glad to learn what that is, and tell you if you can find a better solution.
    I already asked what happens if a refuse to extend x to the complex field, but got no reply.
    Last edited by logic; April 17th, 2012 at 08:16 AM.
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    Quote Originally Posted by logic View Post
    Math is just an arbitrary, symbolic, axiomatic system of equivalences and tautologies. You can prove whatever you want, if you choose the axioms....
    Quote Originally Posted by Markus Hanke View Post
    theoretical issues, which I understood to relate to mathematics, not epistemology.
    Quote Originally Posted by Markus Hanke View Post
    why don't you show me areal number the square of which is -1.....
    If you are interested, start a separate thread and we can discuss all you want, Markus, but you ought to be prepared to discuss the logical foundations of your [arbitrary*] system. If it is flawed, you get yourself into a cul-de-sac.
    You should at least be prepared to question all your basic definitions of number, zero, rational, real etc. and abandon poetic licenses and figments (such as : 0 is a number,.. x^0 = 1...). If you start with false or flawed axioms you get in deep, irreversible trouble.
    At physicsforums they prove that 0.9999999...... = 1 , is that an idiocy of theirs or is it an acknolwedged conclusion (mathematical fact, as you put it)? if that is official math and you agree, I hope you realize that, by that, you are implying that 2=1=836= ... that is sheer chaos!
    I hope I made myself clear, this time.

    *(for example) you surely know that current set theory (the logical foundations of math) accepts the "axiom of choice" which produces the Banach-Tarsky paradox that proves that 2=1
    Last edited by logic; April 17th, 2012 at 04:50 AM.
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    Quote Originally Posted by logic View Post
    At physicsforums they prove that 0.9999999...... = 1 , is that an idiocy of theirs or is it an acknolwedged conclusion (mathematical fact, as you put it)? if that is official math and you agree, I hope you realize that, by that, you are implying that 2=1=836= ... that is sheer chaos!
    I hope I made myself clear, this time.
    You have made your ignorance of mathematics quite clear, thanks.
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    Quote Originally Posted by logic View Post

    2)I have conceded usefulness, I refuted that imaginary numbers is the best (let alone the only) makeshift to your incapacity to solve
    . If the motorway from Bonn to Berlin is blocked, there are a billion alternative routes you can take to reach Berlin, but if you choose to reach your goal via the South Pole I do have a right to criticize you. Finest minds did not find a better route probably as they thought, like you, that there is no other way; this is my stance, probably I could not make myself understood.
    You have failed to answer my question in post 44. Not using complex numbers, what is the square root of -1, or any negative number ?

    If, as someone said, Complex numbers are useful apart from taking the root of negative numbers, I'll be glad to learn what that is, and tell you if you can find a better solution.
    Quantum mechanics. Many of its elemental laws in the non-relativistic limit are described via the Schroedinger Equation, which is a complex-valued partial differential wave equation. It's solutions allow us to derive probability densities for observables. The results are in perfect agreement with observational evidence.
    If you can suggest a better wave equation for quantum mechanics that does not need complex numbers of any kind, then please show us. I would be interested to see it.
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    Quote Originally Posted by logic View Post
    Quote Originally Posted by logic View Post
    Math is just an arbitrary, symbolic, axiomatic system of equivalences and tautologies. You can prove whatever you want, if you choose the axioms....
    Quote Originally Posted by Markus Hanke View Post
    theoretical issues, which I understood to relate to mathematics, not epistemology.
    Quote Originally Posted by Markus Hanke View Post
    why don't you show me areal number the square of which is -1.....
    If you are interested, start a separate thread and we can discuss all you want, Markus, but you ought to be prepared to discuss the logical foundations of your [arbitrary*] system. If it is flawed, you get yourself into a cul-de-sac.
    You should at least be prepared to question all your basic definitions of number, zero, rational, real etc. and abandon poetic licenses and figments (such as : 0 is a number,.. x^0 = 1...). If you start with false or flawed axioms you get in deep, irreversible trouble.
    At physicsforums they prove that 0.9999999...... = 1 , is that an idiocy of theirs or is it an acknolwedged conclusion (mathematical fact, as you put it)? if that is official math and you agree, I hope you realize that, by that, you are implying that 2=1=836= ... that is sheer chaos!
    I hope I made myself clear, this time.

    *(for example) you surely know that current set theory (the logical foundations of math) accepts the "axiom of choice" which produces the Banach-Tarsky paradox that proves that 2=1
    It is mathematical fact that 0.9999... = 1.
    Do you want me to proof it to you ? It is very elementary, via the convergence theorem of geometric series. Leonard Euler has done this back in 1770.
    Any elementary maths textbook will give the proof for this, but I can spell it out here if you need it.
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    Quote Originally Posted by logic View Post
    I refuted that imaginary numbers is the best (let alone the only) makeshift to your incapacity to solve [/I].[I]
    You have asserted that but provided no alternative nor evidence for an alternative. That is not a refutation, just a hollow claim.
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    Hey, what about the higher order sets of numbers, like hypercomplex numbers, quaternions etc etc ? Any thoughts ?
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    And the transfinite cardinals ...
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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    logic, your basic claim here is that the Fundamental Theorem of Algebra ( any polynomial of degree n has exactly n roots, which are not necessarily all different ) is incorrect.
    Can you present formal, rigorous mathematical proof of your claim.
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    I don't see the problem here. The complex numbers are just R2, equipped with (a,b) + (c,d) = (a+c,b+d) and (a,b)*(c,d) = (ac-bd,ad-bc). If you have a better solution to solving roots of negative number then chances are it can be shown to be equivalent, and if not then why don't you present it.
    Also, a square root of i is (1 + i)/(sqrt(2)), still a complex number. There is no shift of the problem, its actually completely solved. The square root of a complex number is always another complex number.
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    Quote Originally Posted by Markus Hanke View Post
    If, as someone said, Complex numbers are useful apart from taking the root of negative numbers, I'll be glad to learn what that is, and tell you if you can find a better solution.
    Quantum mechanics. Many of its elemental laws in the non-relativistic limit are described via the Schroedinger Equation, which is a complex-valued partial differential wave equation. It's solutions allow us to derive probability densities for observables. The results are in perfect agreement with observational evidence.
    If you can suggest a better wave equation for quantum mechanics that does not need complex numbers of any kind, then please show us. I would be interested to see it.
    To further Markus' point i believe Maxwell formulated his equations of electromagnetism in terms of Quaternions before Heaviside revised them in terms of vector notation, they are also used in the study of Dynamic Systems to determine the behaviour of such systems near equilibrium, problems involving capacitors, resistors and inductors are simplified by the use of functions that require complex numbers. In mathematics valuable tools such as Fourier Transforms, which require complex numbers, are useful in solving many partial differential equations, often just as easily as other methods.

    Undoubtedly Maxwell's equations are easier to manage using vectors and vector calculus, but the same results might be obtainable if one is careful enough with their use of Quaternions. (So there's one example where they weren't the easiest way to solve problems) But when determining the types of equilibrium points of linear dynamic systems (or linearised non-linear dynamic systems) we can determine the Eigenvalues of the operators that define the system, if said Eigenvalues are:
    • Real, then the equilibrium is a node and the trajectories tend directly towards or away from equilibrium.
    • Complex, then the trajectories of the system will spiral towards or away from the equilibrium point.
    Finding this information can be useful when the equations of the system do not have an analytic solution, since it's not that difficult to determine equilibrium behaviour, and these behaviours can change when we change the parameters of the problem. (A bifurcation may be present)

    Since eigenvalues are the roots of specific polynomial functions this serves as an example that ignoring complex solutions to polynomial equations can lead to entirely different predictions of real world phenomenon. (something i think strange mentioned earlier)
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    Quote Originally Posted by logic View Post
    At physicsforums they prove that 0.9999999...... = 1 , is that an idiocy of theirs or is it an acknolwedged conclusion (mathematical fact, as you put it)? if that is official math and you agree, I hope you realize that, by that, you are implying that 2=1=836= ... that is sheer chaos!
    I would like to see you show that 0.99999...=1 implies that 2=1. As far as I know you need to assume a contradiction to end up with that result.
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    Quote Originally Posted by logic View Post
    I ask you all to present your solutions to x+ 12x = 63.
    Already been fully answered in post 18. Why do we need to repeat ourselves ?
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    Quote Originally Posted by logic View Post
    *(for example) you surely know that current set theory (the logical foundations of math) accepts the "axiom of choice" which produces the Banach-Tarsky paradox that proves that 2=1
    The Banach-Tarsky "paradox" is something from STG ( set-theoretic geometry ), and has nothing whatsoever to do with real or complex numbers. In particular, it does not prove that 2=1 - it is really quite embarrassing that you are even suggesting such a thing. BTP applies only to certain geometric objects, not to numbers.
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    Mod says Sorry, but I am losing patience

    I have 3 choices - leave it be, shut it down, or move it to New Ideas.... As I am your servant not your master, you (collectively) decide.

    Before you do let me make a coupla observations:

    There seems little appetite in this thread for formal mathematics of the sort I have contributed here. Weird for a Math forum

    Straying into Physics territory, as some have done, is not what we do here, at least not primarily

    logic's insistence that the Reals are "real" and the complex numbers are a mathematical "fiction" is inconsistent with his later assertion that , since the Reals are constructed by Cauchy as the limit of the sequence formed from partial sums of rationals (or by Dedekind cuts - the result is the same )
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    I am answering here to Logic's original post about integer/rational solutions, and I am not intended to discuss any other types of solutions, this can be done in other threads (or by reading textbooks).

    Answer below is a practical application of theoretical thoughts explained in my previous posts. And to be sportive, I will use only my head and paper. My solution is not unique operation, but number of simple operations. I would count not number of operations, but the total time needed to perform them. Awaiting for Logic's unique operation solving equation.

    First simple example is solution to x + 12x = 63. Important: somebody secretly told me that a solution we are looking for is integer!

    1) Estimation of the answer: It is clear that and (I cannot avoid taking cubic roots, but I restrict to some simple knowledge like what is n^3 for n=1,2,3....,10). So x can be 3,2,1.
    2) Already from the previous step one can guess . But let us be systematic and solve cubic equations in the field of integers mod 3 (i.e. I don't distinguish numbers n and m, if n-m is divisible by 3). Then , , , so equation is reduced to , so it is clear that solution is



    But this was too trivial and I will consider a more complicated example



    The number is exact (after its last digit, 4, all decimals are 0). Important: somebody secretly told me that is integer!

    1) Estimation of the answer: . Magnitude of coefficients in front of and show that this is very-very good estimation. I do not know table of 5th powers of integers, but common sense tells that is in between 1 and 5. So let us make few tries: , (everybody knows because this is number of bytes in 1KB). Conclusion: our answer is larger than 30 and is smaller than 40, and closer to 30.

    2) Solving equation in the field of integers mod 2,3 and 10.

    First, introduce integer number , and write down equation for :



    Solving this equation mod 2 relies on the fact mod 2, by using it we quickly reduce equation to the statement that is even.
    Solving this equation mod 3 relies on the fact that mod 3, then 300 =0 mod 3, 10000=1 mod 3. To find mod 3 we will use the fact that when dividing by 3, residue of number is equal to the residue of sum of it coefficients: so mod 3 we have , so our equation becomes



    which tell us that , and since is even it should be

    Finally we solve the equation mod 10 (find last digit). Since 300 =0 and 10000=0, we need to compute only , but , which is related to the fact that . The conclusion is that the last digit of is the same as of , i.e. 4.

    3) Final refining.

    How many numbers between 300 and 400, which of the form , and with the last digit ? The first such number is , and then we have to add to find all others, which are and . My initial numerical estimation is in favor for , and we will confirm it by making a better estimation about . Let us see whether it is larger or smaller then 33. . Because , but , we conclude that , so it should be





    Remark: I was lazy to make a more precise evaluation of , but of course it is possible and you can find , which already gives you an answer! To make life more interesting, I choose to not increase a precision in computation, but to find what is a residue of , when dividing by 2,3, 10. Though I made "more" operations, it might be a faster way, at least sometimes.

    Probably I am done with this thread, I would still like to hear from Logic his logic.
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    Quote Originally Posted by Guitarist View Post
    There seems little appetite in this thread for formal mathematics of the sort I have contributed here. Weird for a Math forum
    True, but how do you convince someone that the complex numbers are no less real than the real numbers, if he doesn't even accept the Fundamental Algebraic Theorem as valid ?
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    I'll wait for others answers before replying to Vel.
    In the meanwhile I'd like to put the record straight because my words have been twisted : I never said complex/imaginary numbers (CN) are figments or that reals are real. I said that the roots you cannot show on a curve, ona plotter, are figments of your imaginations. And nobody, has yet, has been able to tell me what is the use of the two imaginary roots of x=n..
    ..that would cancel all my objections, I am only too eager to learn.
    If nobody can do that, try to find another root to x=8 the same way as I do:
    11111111 =8 11 11| | 11 11 =8 2+2| |2+2=8 2x2| |2x2 =82x(2x2)=8 2x2=8 2 = 8

    My original position was to refute the necessity of CN in this thread, and that has been acknowledged.
    Next I stated, and confirm, that if we find a good method for any polynomial, we do not need at all CN, and that is undeniable.

    I had to wait 65 posts to get a sensible response to my question, is that my fault?
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    Quote Originally Posted by logic View Post
    I said that the roots you cannot show on a curve, ona plotter, are figments of your imaginations.
    They are normally plotted on the complex plane. There is a limited amount of information that can be represented on a 2D graph so different representations are used for different purposes. Change the axes of your graph/plot and you will only be able to show the imaginary solutions and not the real ones.

    And nobody, has yet, has been able to tell me what is the use of the two imaginary roots of x=n..
    Well, you have been given several examples of the use of other polynomials. Are you interested in the use of that one specifically?

    I had to wait 65 posts to get a sensible response to my question, is that my fault?
    Of course. You asked a rather poorly worded question and then threw in all sorts of false and irrelevant information. It was quite hard to see what your central point was. And, of course, the factual inaccuracies in some of your ramblings had to be corrected.
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    Quote Originally Posted by Strange View Post
    They {the roots of an arbitrary polynomial} are normally plotted on the complex plane.
    I am totally confused! What are you "plotting" against what?

    As far as I am aware, only functions admit of "plots" of the sort you seem to be referring to. As far as I can see, there can be no graph on any imaginable plane of a polynomial against its roots.

    Are you suggesting there is a function that sends a polynomial to its roots? Or are you perhaps suggesting there is a function that sends an arbitrary number to the set of all polynomials such that this number is a root for some polynomial?

    Neither seems true to me

    Or am I having a very bad day?
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    Quote Originally Posted by Guitarist View Post
    Quote Originally Posted by Strange View Post
    They {the roots of an arbitrary polynomial} are normally plotted on the complex plane.
    I am totally confused! What are you "plotting" against what?
    Sorry, we have slipped back into engineering. When designing a filter, for example, the roots of the transfer function [1] are plotted on the complex plane (rather like an Argand diagram) as this tells you important things about the behaviour of the system.

    I was just responding to local's apparent belief that the "normal" graph of x vs f(x) has some special significance.


    [1] the delta function and z-transforms may come into this; but I haven't had to think about any of this stuff for over 30 years...
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    Quote Originally Posted by logic View Post
    Please consider the depressed cubic equation when b is positive i.e.:
    x + 3x = +/-149,036 . The equation has only one root [x = +/-53].
    Is there a method to find quickly this solution with only a few operations?
    I will be extremely grateful if a great scientist/writer/linguist would show me where is the ambiguity or the poor wording in the OP,(which intelligent readers understood). If someone shows me that, I'll make amends and humbly apologize.
    Conceding that the OP was really badly written, anyway, I explained clearly in post #13 (I naively regarded it a 'dumbproof explanation'!) what I was looking for.
    It is not my style to reply directly to rude/snide/silly posts or members

    Quote Originally Posted by logic View Post
    Quote Originally Posted by Markus Hanke View Post
    Mathematically these are two distinct polynomials, so in general they will not have the same roots,....
    4) is this a reliable site ? :The Geometry of the Cubic Formula, and this :Cubic function - Wikipedia, the free encyclopedia
    If it is as you say, depress a cubic?
    The roots of this are all real, there are no imaginary roots ...
    5) see link quoted above.
    (If you are still interested in the main topic,) in all my previous posts I was referring to these links, and to Cubic curve and graph display - Math Open Reference . If you know a site where you can plot (is there anything wrong with my wording?) two equations at the same time, that would be even better.
    If you do not care to say what you would do with pencil and paper to solve OP equation (or x+12x=63), we can discuss the problem in general. Thanks, again for your attention!


    Quote Originally Posted by Guitarist View Post
    , I would close this thread ...
    Quote Originally Posted by logic View Post
    I'll avoid further sterile... , If you are interested start a separate thread ...
    Quote Originally Posted by Markus Hanke View Post
    .... - it is really quite embarrassing that you are even suggesting such a thing. BTP applies only to certain geometric objects, not to numbers.
    I would never be embarassed with you, Markus, as you have granted me the pleasure of being your "friend",
    Neither would I ever try to embarass you when I'll be able to reply.
    I said I do not wish my thread be closed before the OP is answered, please do not think I'm not replying for lack of arguments: I was fully aware I was stirring a hornet's nest. I will reply to you all, (and to your insults) at the end of the discussion, or in another thread. Please spare me your sarcasm before then. Thank you!

    P.S. Nobody has explained the necessity or even the usefulness of the imaginary roots of x=8
    Last edited by logic; April 19th, 2012 at 05:54 AM.
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  74. #73  
    Moderator Moderator Markus Hanke's Avatar
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    Quote Originally Posted by logic View Post
    I would never be embarassed with you, Markus, as you have granted me the pleasure of being your "friend",
    Neither would I ever try to embarass you when I'll be able to reply.
    Three things :

    1. This is not a social networking site, it's a science from - therefore :
    2. I will always point out statements which are plain wrong, regardless of whether that is from a "friend" or not, and
    3. Since you don't seem to think there is anything wrong with your claim, I shall be looking forward to your formal, rigorous proof using the BTP that 2=1.
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  75. #74  
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by logic View Post
    I would never be embarassed with you, Markus, as you have granted me the pleasure of being your "friend",
    Neither would I ever try to embarass you when I'll be able to reply.
    1. This is not a social networking site, it's a science from - therefore : I will always point out statements which are plain wrong, .
    It is not, Markus, but even in a science forum you can say that something is wrong, without saying that one should be ashamed, expecially when you are a forumprofessor or a moderator. I pretended not to see many things you said (like 'mathematical fact' etc...)
    Moreover, whenever or wherever one says something is wrong, should say why it is wrong.
    In post #10 Guitarist not ony qualified my post (me, of course) as arrogant, but dismissed my claim without challenging it. Is that fair or downright arrogance? If you thing a post is stupid you just do not answer it, do not pooh-pooh it.

    Edit: Lastly, you probably read that post too hastily (I hope), because that was a note that explained an adjective you had already overlooked or misunderstood once, and had nothing to do with the text, the text provided a numeric example.
    Your rebuff was, moreover, twice ungrounded: yes, it has been revived in topology, but it is the original Zermelo's axiom of ZFC.

    I am waiting for an answer to my question 2 in post 39. A question I asked a dozen times.
    Last edited by logic; April 19th, 2012 at 07:07 AM.
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  76. #75  
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    Quote Originally Posted by logic View Post
    Moreover, whenever or wherever one says something is wrong, should say why it is wrong.
    I think I did say to you that the BTP only applies to certain geometrical objects, not to numbers. That is the reason why your statement that the BTP implies 2=1 was wrong.

    In post #10 Guitarist not ony qualified my post (me, of course) as arrogant, but dismissed my claim without challenging it.
    Is that fair or downright arrogance?
    I cannot speak for Guitarist; it is up to himself to answer this.
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    We cross-posted, I was editing post #74 in order to prevent this objection: reason you gave is inappropriate and wrong. You should read carefully also the original post. The meaning of the note was : your system is arbitrary, you may have ZF or ZFC , choosing the right axioms, you can prove whatever you want. I had already had told you this to imply that your formal proofs (you, Guitarist etc..think have absolute value of Truth) have value only in your (flawed?) system and are worthless in any other system.
    I was sure a specialist would understand!
    Anyway, even if that note was meant as a numerical example, your objections are ungrounded anyway, to be punctilious, what is the difference between two spheres and two bananas? why a paradox?

    I hope you'll continue the general discussion, thanks for your attention.
    Last edited by logic; April 19th, 2012 at 08:34 AM.
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    Well if logic thinks I made no attempt to answer his query, and in particular explain how one extracts the roots of aan arbitrary polynomial, then either doesn't read my posts or, more likely, doesn't understand them.

    I will give one last try then close this thread if it fails.

    First we need to distinguish between a polynomial form where the unknown can be anything - it's an indeterminate - and a polynomial function where we need to have a very clear idea what is.

    In general, when the function is defined we MUST say what is the domain {X} and what is the codomain or range (Y)

    So if the polynomial function and if, for all that , then obviously the zero of this function can only be

    Or suppose (this a field since 2 is prime). This amounts to the vacuous statement that since

    Or try . Here so the solution is the same as for the reals

    If, on the other hand this same polynomial is a polynomial form we under no obligation to restrict ourselves to roots in any particular field, and must consider ALL possible zeros or roots. It easy to see, in this example we must consider ALL objects whose cube is . I have shown one, and there is no doubt in my mind that there are 2 others that "live" in .

    These seem to be (unless I made an error) then the FTA is satisfied and we have our 3 roots of this polynomial form
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  79. #78  
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    Logic,

    There's a neat little book titled, "An Imaginary Tale: The Story of the Sq. Rt. of -1" by Paul J. Nahin.

    It's cheap and worth reading. You might want to check it out.
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    I recommend all of Nahin's work actually. Brilliant recreational mathematics books.
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  81. #80  
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    Quote Originally Posted by logic View Post
    Anyway, even if that note was meant as a numerical example, your objections are ungrounded anyway, to be punctilious, what is the difference between two spheres and two bananas? why a paradox?
    I think you still don't understand. The BTP is concerned with how to disassemble a solid 3-dimensional ball into non-overlapping pieces, and then re-assemble the same pieces in a different manner to obtain two new balls which are identical to the original one. It is purely a geometrical theorem, and it's labeled a paradox because it is counter-intuitive that such a thing should be possible, which doesn't make it any less true.
    It does not imply that the number 2 is equal to the number 1, or any such thing. It has nothing to do with number theory.
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    Quote Originally Posted by Guitarist View Post
    ... an alternative must be offered..... Personally I dispute this can be done using logic's "logics"
    Quote Originally Posted by Guitarist View Post
    To these questions, I have not an effing clue, and as I am busy drinking beer on this Sat night, have no intention of even thinking about them
    Quote Originally Posted by Guitarist View Post
    This, if you never encountered it before, is called "proof by absurdity"
    Quote Originally Posted by Guitarist View Post
    ...logic's insistence that the Reals are "real" and the complex numbers are a mathematical "fiction" is inconsistent with his later assertion that , since ...=1[/tex])
    Quote Originally Posted by Guitarist View Post
    Well if logic thinks I made no attempt to answer his query, and in particular explain how one extracts the roots of aan arbitrary polynomial,These seem to be (unless I made an error) then the FTA is satisfied and we have our 3 roots of this polynomial form
    Thanks, Guitarists , for your contributions. My query has been thoroughly answered.
    Last edited by logic; April 22nd, 2012 at 06:36 AM.
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  83. #82  
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    Quote Originally Posted by logic View Post
    At physicsforums they prove that 0.9999999...... = 1
    Just came across this joke:

    An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer.

    The bartender says "You're all idiots", and pours two beers.
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