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Math Help - Modulo Arithmetic, set of all integers div 7

  1. #1
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    Modulo Arithmetic, set of all integers div 7

    Hi there.

    I've looked over my notes for my last lecture only to realise I don't understand what's going on with them at all and would like to know if anyone else can explain how the following works:

    in \mathbb{Z}/11:
    8+7 = 10 mod 11

    We all know that 8+7 does not equal 10 mod 11, so what's going on?

    It then goes on to prove that
    7+8 = ?
    7+8 = 15, too big so /11, remainder 4
    7+8 = 4 mod 11
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  2. #2
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    Okay, the problem was that I had typo'd my notes -.- It should be 7+3 = 10 mod 11 which is much more logical.

    My problem now is that I need to use the answer to "Find a number in \mathbb{Z}/7 which gives 1 mod 7 when multiplied by 3" (the answe of which is 5)
    to solve the equation "3n = 4 in \mathbb{Z}/7" and don't know where to start.

    Any ideas/insights?
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  3. #3
    Member HappyJoe's Avatar
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    It certainly looks like a mistake!

    If you quote it in its context, then we can probably tell for sure whether it's a mistake or not.
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  4. #4
    Member HappyJoe's Avatar
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    Ah, never mind my other post. I posted it before you found the typo.

    Anyway, you know that 3*5 = 1 (mod 7). Thus 4*(3*5) = 4 (mod 7), or rearranging the left hand side 3*(4*5) = 4 (mod 7). But 4*5 = 20 = 6 (mod 7), hence 3*6 = 4 (mod 7).
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Timsworth View Post
    Okay, the problem was that I had typo'd my notes -.- It should be 7+3 = 10 mod 11 which is much more logical.

    My problem now is that I need to use the answer to "Find a number in \mathbb{Z}/7 which gives 1 mod 7 when multiplied by 3" (the answe of which is 5)
    to solve the equation "3n = 4 in \mathbb{Z}/7" and don't know where to start.

    Any ideas/insights?
    what do you mean by  \mathbb{Z}_{/7}?? do you mean \mathbb{Z}_7.
    if that's the case then you need to solve 3n=1 \, in\, \mathbb{Z}_7, not the one you typed. for this you need to find the inverse of 3 \, in \, \mathbb{Z}_7 using euclidean algorithm.
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  6. #6
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    Cheers Joe. I was hoping that was the answer! It's the notation that threw me off, I think. There was no \equiv_7 or \pmod{7} etc that I'm used to seeing.
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  7. #7
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    Quote Originally Posted by abhishekkgp View Post
    what do you mean by  \mathbb{Z}_{/7}?? do you mean \mathbb{Z}_7.
    if that's the case then you need to solve 3n=1 \, in\, \mathbb{Z}_7, not the one you typed. for this you need to find the inverse of 3 \, in \, \mathbb{Z}_7 using euclidean algorithm.
    I'm just getting used to using TeX, what I meant was \mathbb{Z}/7 rather than \mathbb{Z}_/_7
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  8. #8
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    What's the difference between \mathbb{Z}/7 and \mathbb{Z}/7^*? Are there any online resources I'm missing about it? I can't find anything good.
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  9. #9
    Member HappyJoe's Avatar
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    The difference between \mathbb{Z}/7 and (\mathbb{Z}/7)^* is that the former consists of the numbers \{0,1,2,3,4,5,6\}, while the latter consists of \{1,2,3,4,5,6\}.

    In general, if n is some integer, then \mathbb{Z}/n consists of the numbers \{0,1,2,\ldots,n-1\}, while (\mathbb{Z}/n)^* consists of all numbers that are relatively prime to n (that is, all numbers a with gcd(a,n)=1, so that the greatest common divisor of a and n is 1).

    So for instance, we have that

    (\mathbb{Z}/8)^* = \{1,3,5,7\},

    and

    (\mathbb{Z}/15)^* = \{1,2,4,7,8,11,13,14\}.

    Importantly, whenever p is a prime, we have that

    (\mathbb{Z}/p)^* = \{1,2,3,\ldots,p-1\},

    like it happened in the case p=7 above.
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