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Thread: Challenge: can you find the square root of n using only one of the 4 operations?

  1. #1 Challenge: can you find the square root of n using only one of the 4 operations? 
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    Here is a nice challenge for math lovers:

    can you work out an algorithm that uses only one operation [sign] + - : x , (excluding log, and sqrt) to find the square root of any number ?
    You are allowed to use your mind for preliminary 2-digit basic operations (times table , 8 +/- 3....and so on).

    It is not tough!


    Last edited by logic; April 17th, 2012 at 02:08 AM.
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  3. #2  
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    Suppose

    I'll express this as which implies that is . So all I used was multiplication

    That's my "algorithm". LoL.

    (can you elaborate on what you're asking?)


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  4. #3  
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    Quote Originally Posted by brody View Post
    (can you elaborate on what you're asking?)
    You can find on wiki the known algorithms to take a sqrt
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    b-b=a?
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    Quote Originally Posted by incorrect View Post
    b-b=a?
    That is incorrect, incorrect You have to define what "a" and "b" are to represent otherwise your proof/algorithm makes no sense.
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    Quote Originally Posted by logic View Post
    Quote Originally Posted by brody View Post
    (can you elaborate on what you're asking?)
    You can find on wiki the known algorithms to take a sqrt
    I'm assuming it's this "Babylonian method". (???)

    Quote Originally Posted by Wikipedia: Square root
    The most common iterative method of square root calculation by hand is known as the "Babylonian method" or "Heron's method" after the first-century Greek philosopher Heron of Alexandria...
    I think I understand it. See if this is right. I'll use as an example.

    . I know that . Therefore .

    So the average of and should yield a closer value to .

    So now I'll take this closer value, here being which I know is greater than , divide it from 2 and average that value to get an even closer approximation to . However, I don't think you can get the exact value without infinite iteration. And I'm sure the math and the numbers will get really nasty as the process goes deeper! (after all, it's approaching an irrational value)

    Interesting!
    Last edited by brody; May 1st, 2012 at 08:41 PM. Reason: Elaboration on iterative part
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    b squared + a squared =c squared . find the square root of c

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  9. #8  
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    Quote Originally Posted by brody View Post
    I'm assuming it's this "Babylonian method". (???)
    To my mind, that doesn't meet the requirements as it needs two operations (division and subtraction). Unless you "cheat" and say that comparison isn't "really" an operation. But that means you never write a program to do it.
    ei incumbit probatio qui dicit, non qui negat
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    You can approximate it to any degree by doing this:

    To find √n

    Starting at 1, try all the natural numbers in order until you reach the greatest such that
    k is then the whole number part of the answer.

    To find the first decimal digit try squaring k.0, k.1, k.2, ... k.9 in order until you find the greatest such that

    To find the second decimal digit repeat the same process with instead of

    Etc.

    I hope that kind of makes sense.

    I don't think you can find a non approximation algorithm though.
    Last edited by clonus; May 9th, 2012 at 01:13 AM. Reason: accidental subtraction
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  11. #10  
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    Quote Originally Posted by brody
    , I don't think you can get the exact value without infinite iteration. And I'm sure the math and the numbers will get really nasty as the process goes deeper! (after all, it's approaching an irrational value)....Interesting!
    You cannot get that with any method, brody, of course you must take a finite number x, say to a 10-digit accuracy. (Wiki chooses 6-digit precision).

    Now, only one operation is surely possible and is +, or - : addition or subtraction of the gnomons , odd numbers
    the problem is that it is too expensive: for n = x² it requires x operations
    The challenge is more sophisticate, asks for a solution less expensive than Newton's (20 operations) and Babylonian's (15)
    I assure you it is possible, not too tough!
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    Finding Square Roots arallel-method. Worteltrekken via parallel berekening. - YouTube
    I think this guy found a solution for the roots, he write that the red numbers are not a accidental, maybe right or not?
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  13. #12  
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    If you are saying 'any' number i.e choose our own number then yes fairly easy, if by any you really mean every then that's harder. Some guidance on what you mean would be helpful regarding this.
    “The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts.”

    Bertrand Russell
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