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Thread: 1=0 (no cancelling of zeros)

  1. #1 1=0 (no cancelling of zeros) 
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    Here i = iota = square root(-1)

    - square root(a^2 + b^2) <= asinx + bcosx <= square root(a^2 + b^2)
    now , lets take (sinx + icosx)
    - square root(1^2 + i^2) <=(sinx + icosx)<= square root(1^2 + i^2)
    = - square root(1-1) <=(sinx + icosx)<= square root(1-1)
    So,
    sinx + icosx = 0

    sinx = (-1)icosx

    squaring both sides,

    [sinx]^2 = [(-1)icosx]^2
    now, we know [sinx]^2 = 1 - [cosx]^2
    putting this in [sinx]^2 = [(-1)icosx]^2
    1 - [cosx]^2 = [i]^2[cosx]^2
    1 - [cosx]^2 = -[cosx]^2
    1- [cosx]^2 + [cosx]^2 = 0
    so,
    1 = 0


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  3. #2  
    Universal Mind John Galt's Avatar
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    Do you believe your result? If so, why do you think hundreds of thousands of mathematicians have missed it?

    If not, why post it?

    Anyway, welcome to the forum.


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  4. #3  
    Moderator Moderator Cogito Ergo Sum's Avatar
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    Quote Originally Posted by John Galt View Post
    Do you believe your result? If so, why do you think hundreds of thousands of mathematicians have missed it?

    If not, why post it?

    Anyway, welcome to the forum.

    I think he meant it as a mathematical exercise, rather than as an attempt to invalidate arithmetic.

    In essence, (s)he arrives at this part:

    or in which

    However, the last equation has no solutions.

    If one were to plot the two equations and ,
    one would notice that the two lines never intersect due to the fact that they share the same value for in a linear equation of the form .


    PS: I am certain that members whose knowledge of mathematics will be able to provide a more elegant and correct answer, but I hope that my answer suffices.
    Last edited by Cogito Ergo Sum; June 10th, 2014 at 03:38 PM.
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    ~ Arthur Schopenhauer, The Art of Being Right: 38 Ways to Win an Argument (1831), Stratagem XXXVIII.
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  5. #4  
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    "- square root(1^2 + i^2) <=(sinx + icosx)<= square root(1^2 + i^2)"

    Above is nonsense. You have a complex number in the middle of inequalities.

    The first line assumes a and b real.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    I don't think those are meant to be inequalities as the same thing appears on the left and right sides in every case. I'm not quite sure what they are meant to be though.
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  7. #6  
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    Quote Originally Posted by Magi-Master View Post
    I don't think those are meant to be inequalities as the same thing appears on the left and right sides in every case. I'm not quite sure what they are meant to be though.
    The above ineqality exists and here is the proof
    Let the eq. be (acosx + bsinx) ,
    Now,
    let a = rcosy and b = rsiny
    a^2 + b^2 = [rcosy]^2 + [rsiny]^2
    a^2 + b^2 = r^2{[cosy]^2 + [siny]^2} = r^2
    Putting a = rcosy and b = rsiny in (acosx + bsinx)
    = rcosycosx + rsinysinx = rcos(x-y) as cosycosx + sinysinx = cos(x-y)
    now,
    (-1) <= cosm <= 1
    so,
    (-1) <= cos(x-y) <= 1
    and thus ,
    (-r) <= rcos(x-y) <= r
    but r^2 = a^2 + b^2
    so,
    r = square root(a^2 + b^2)
    putting this in (-r) <= rcos(x-y) <= r
    = [-square root(a^2 + b^2)] <= rcos(x-y) <= [square root(a^2 + b^2)]
    but rcos(x-y) = rcosycosx + rsinysinx = acosx + bsinx
    So,
    [-square root(a^2 + b^2)] <= acosx + bsinx <= [square root(a^2 + b^2)]
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  8. #7  
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    Quote Originally Posted by mathman View Post
    "- square root(1^2 + i^2) <=(sinx + icosx)<= square root(1^2 + i^2)"

    Above is nonsense. You have a complex number in the middle of inequalities.

    The first line assumes a and b real.
    The above ineqality is not nonsense as (sinx + icosx) could be converted into a real expression by using r = square root( a^2 + b^2)and (sinx + icosx) = rcos(x-y) and thus (sinx + icosx) equates to a real expression and the above inequality thus becomes real and valid.
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  9. #8  
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    Just to start with, where did the and come from in , or if is just a generic definition, then why should ? (The latter isn't true for all x, y and r.)
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  10. #9  
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    Quote Originally Posted by shivanshverma View Post
    The above ineqality exists and here is the proof ...
    Quote Originally Posted by shivanshverma View Post
    The above ineqality is not nonsense ...
    I'm curious about what you are saying here. Are you just arguing about where the error is? Or do you not understand where the error is?
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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  11. #10  
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    note:
    the derivation of range of asinx + bcosx is as follows
    first we divide and multiply it by the square root of a^2 + B^2
    we get:
    sqrt{a^2 + b^2}[a/sqrt{a^2 + b^2} * sinx + b/sqrt{a^2 + b^2} * cosx]
    =
    sqrt{a^2 + b^2}[cosR * sinx + sinR * cosx]
    =
    sqrt{a^2 + b^2}[sin(x+r)]
    this shows that this should lie between -
    sqrt{a^2 + b^2} and +sqrt{a^2 + b^2}



    In the above question
    a=1
    b=i

    now
    a^2 + b^2= 1^2 +i^2
    =1-1
    =0

    we all know that we cannot divide nor multiply a number by 0

    therefore above question is statement is wrong and hence the answer we are getting is wrong
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  12. #11  
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    The answer to the posted question is as follows : -


    For sqaring all the terms should be on one side
    This is because :-

    Let consider [ a + b ] = 0
    now we can take b on R.H.S and then square or take [ a + b ] on one side and then square
    But if we take b on the other side
    Then,
    a= [-b]
    [a]^2 = [b]^2
    so,
    [a]^2 - [b]^2 = 0
    [a + b ][a-b] = 0
    thus either a + b = 0 or a-b = 0
    In the above post I took a-b = 0 and thus got a=b but actually a=[-b]
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  13. #12  
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    shivansh,

    the fault i told is truly viable and i hope that you do not think that it is not

    it is really a very important derivation in class 11th and is used in many other topics other than trigonometry



    also plz tell me the step in which you have done the mistake you were talking about in your above comment

    it will be really nice of you
    thnx
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  14. #13  
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    Quote Originally Posted by aman View Post
    shivansh,

    plz tell me the step in which you have done the mistake you were talking about in your above comment

    it will be really nice of you
    thnx
    The fault was that I took icosx on the other side thatis
    sinx = [-icosx]
    here as i took icosx on the other side on squaring I removed 2isinxcox which had to be there if i were to take all terms on one side and thus I got 1=0
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  15. #14  
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    I would be happy to tell you that there is no real derivation . All derivations which allow to get same results are viable. The scope of mathematics is to get all possible solutions to the equations and what you did made the inequality non valid . Recollect that even in limits we try to cancel 0/0 form for which mathematicians struggled for years.
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  16. #15  
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    Quote Originally Posted by Strange View Post
    Quote Originally Posted by shivanshverma View Post
    The above ineqality exists and here is the proof ...
    Quote Originally Posted by shivanshverma View Post
    The above ineqality is not nonsense ...
    I'm curious about what you are saying here. Are you just arguing about where the error is? Or do you not understand where the error is?
    No I am not arguing but just trying to tell that what you wrote was not the coorect answer to the post.
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  17. #16  
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    are you having tiff with strange
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  18. #17  
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    Just a small note, using the [ tex ] tag to format your equations makes them much more readable and isn't that much different than what yall have been typing so far: TeX tutorial
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  19. #18  
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    And umm ..... Dude this clearly means dat that the equation iz wrong, 0 is when u have nothing(ie. 1/infinite)
    so what ur equation basically means iz that if u hv 1 stick then that stick iz actually not stick(or it iz as good as no sticks)
    ie. Your stick iz imaginary(like this equation)
    try representing the equation in words before doing it mathematically.
    no hard feelings.
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  20. #19  
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    You have GOT to start spelling properly if you want people to take you seriously.
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