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Math Help - Inverse of 2 mod 108 and the quadratic equation

  1. #1
    Forum Admin topsquark's Avatar
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    Inverse of 2 mod 108 and the quadratic equation

    This is actually related to a problem in this thread, but is so much more basic I didn't want to interrupt any discussion that the original thread might generate.

    The problem is to solve
    x^2 + 3x - 22 \equiv 0~\text{mod(108)}

    Since the details of solving the problem are irrelevent here I will simply state a pair of solutions to the quadratic:
    x \equiv -\frac{3}{2} \pm \frac{1}{2} \cdot 23 \equiv \frac{1}{2} ( -3 \pm 23 ) ~\text{(mod 108)}

    So the solutions mod 108 are 10 and 95, which both work in the original equation.

    My question about all this. 1/2 has no multiplicative inverse mod 108. So even though,say, (1/2)(-3 + 23) = (1/2)(20) = (1/2)(2*10) = (1/2)*2*10 is apparently equal to 10, how can we say (1/2)*2 = 1?

    Thanks!

    -Dan
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  2. #2
    MHF Contributor alexmahone's Avatar
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    \frac{1}{2}\times 2 \equiv 1\ (mod\ 108)

    The multiplicative inverse of 1/2 (mod 108) is 2.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alexmahone View Post
    \frac{1}{2}\times 2 \equiv 1\ (mod\ 108)

    The multiplicative inverse of 1/2 (mod 108) is 2.
    I really don't want to dispute this because it obviously works, but both WolframAlpha and I agree that the equation 2x = 1 (mod 108) has no solution!

    -Dan

    Edit: I suppose you could say that this question is more Philosophically based than it is practical.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by topsquark View Post
    I really don't want to dispute this because it obviously works, but both WolframAlpha and I agree that the equation 2x = 1 (mod 108) has no solution!

    -Dan
    It really depends on whether you're restricting yourself to integral solutions.

    For example, 2x=1 (mod 108) does have rational solutions given by x=1/2 (mod 108).
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alexmahone View Post
    It really depends on whether you're restricting yourself to integral solutions.

    For example, 2x=1 (mod 108) does have rational solutions given by x=1/2 (mod 108).
    Well that would work. My apologies...I have never heard of allowing rational solutions in a modular system. \mathbb{Z} _{108} contains only integers so I thought all solutions had to be in \mathbb{Z} _{108}?

    -Dan
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by topsquark View Post
    This is actually related to a problem in this thread, but is so much more basic I didn't want to interrupt any discussion that the original thread might generate.

    The problem is to solve
    x^2 + 3x - 22 \equiv 0~\text{mod(108)}

    Since the details of solving the problem are irrelevent here I will simply state a pair of solutions to the quadratic:
    x \equiv -\frac{3}{2} \pm \frac{1}{2} \cdot 23 \equiv \frac{1}{2} ( -3 \pm 23 ) ~\text{(mod 108)}

    So the solutions mod 108 are 10 and 95, which both work in the original equation.

    My question about all this. 1/2 has no multiplicative inverse mod 108. So even though,say, (1/2)(-3 + 23) = (1/2)(20) = (1/2)(2*10) = (1/2)*2*10 is apparently equal to 10, how can we say (1/2)*2 = 1?

    Thanks!

    -Dan
    The 'division' in the field of integers mod 108 is performed as 'ordinary division' and that means that n must be an even number. The equation You have written is solvable only if is a 'perfect square' mod 108 and is even. In Your case is...



    ... so that it is 'all right'...

    Kind regards

    \chi \sigma
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