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Thread: Modulo question

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    Modulo question

    Hi all. I'm having some trouble with Ex 6 of Pure Mathematics 1 by Parsonson (Scribd link below) on page 4:
    "Show that arithmetic modulo 11 does not suffer from the same defect as that exhibited by modulo 12 in example 2. Suggest why this is so." Example 2 is on page 4 also.

    The question is not clear to me, which doesn't help! Any guidance would be much appreciated!

    https://www.scribd.com/doc/310965239...cs-Volumes-1-2



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    Re: Modulo question

    Quote Originally Posted by flashylightsmeow View Post
    Hi all. I'm having some trouble with Ex 6 of Pure Mathematics 1 by Parsonson (Scribd link below) on page 4:
    "Show that arithmetic modulo 11 does not suffer from the same defect as that exhibited by modulo 12 in example 2. Suggest why this is so." Example 2 is on page 4 also.

    The question is not clear to me, which doesn't help! Any guidance would be much appreciated!

    https://www.scribd.com/doc/310965239...cs-Volumes-1-2



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    Well, I can't reference problem 12 because I don't wish to buy the membership. However I have a suspicion. Note that in mod 11 there aren't "zero divisors": non-zero numbers when multiplied together give 0. We do have those in mod 12: $\displaystyle 6 \times 2 \equiv 0 ~ \text{(mod 12)}$. 11 being a prime number we don't have the zero divisors.

    -Dan
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    Re: Modulo question

    Quote Originally Posted by flashylightsmeow View Post
    Hi all. I'm having some trouble with Ex 6 of Pure Mathematics 1 by Parsonson (Scribd link below) on page 4:
    "Show that arithmetic modulo 11 does not suffer from the same defect as that exhibited by modulo 12 in example 2. Suggest why this is so." Example 2 is on page 4 also.
    The question is not clear to me, which doesn't help! Any guidance would be much appreciated!
    I assume that you have shown that $4x \equiv 4(\bmod ~12)$ does not have a unique solution?
    That is: $x=1,~4,~7~\&~10$ are all solutions. YES?
    Does that also occur with $ax\equiv a(\bmod 11)~?$ HINT $11$ is prime.
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    Re: Modulo question

    Quote Originally Posted by topsquark View Post
    Well, I can't reference problem 12 because I don't wish to buy the membership. However I have a suspicion. Note that in mod 11 there aren't "zero divisors": non-zero numbers when multiplied together give 0. We do have those in mod 12: $\displaystyle 6 \times 2 \equiv 0 ~ \text{(mod 12)}$. 11 being a prime number we don't have the zero divisors.

    -Dan


    Hi Dan, Plato

    Thanks both for your responses. I have photographed the relevant page here.

    @Plato the Example 2 here shows the 4 solutions for 4x mod 12 = 4.

    This is my first time being introduced to modular arithmetic, I'm trying to think of it intuitively but can't figure out how to prove 4x mod 11 = 4 only has one solution. (I was trying to think of it algebraically. E.g. in the same way you could show a quadratic has one unique solution by factorising to show it is a perfect square)


    Sorry...some hand holding is needed on this one



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    Re: Modulo question

    Quote Originally Posted by flashylightsmeow View Post
    the Example 2 here shows the 4 solutions for 4x mod 12 = 4.
    This is my first time being introduced to modular arithmetic, I'm trying to think of it intuitively but can't figure out how to prove 4x mod 11 = 4 only has one solution. (I was trying to think of it algebraically. E.g. in the same way you could show a quadratic has one unique solution by factorising to show it is a perfect square)
    Here is a rough explanation. $a \equiv b(\bmod m)$ means $b$ is the remainder when $a$ is divided by $m$. that means $0\le b<m$

    EX: $45 \equiv 0(\bmod 5)$_______$4 \equiv 4(\bmod 6)$_______$33 \equiv 9(\bmod 12)$

    BUT $33 \equiv 0(\bmod 11)$

    So to answer $5x \equiv 9(\bmod 12)$ find integers $T$ such that $5\times T+9$ is a multiple of $12$.
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    Re: Modulo question

    Quote Originally Posted by Plato View Post
    Here is a rough explanation. $a \equiv b(\bmod m)$ means $b$ is the remainder when $a$ is divided by $m$. that means $0\le b<m$

    EX: $45 \equiv 0(\bmod 5)$_______$4 \equiv 4(\bmod 6)$_______$33 \equiv 9(\bmod 12)$

    BUT $33 \equiv 0(\bmod 11)$

    So to answer $5x \equiv 9(\bmod 12)$ find integers $T$ such that $5\times T+9$ is a multiple of $12$.
    Thanks Plato! This makes sense.

    So am I right that, for $4x\equiv4(\bmod 11)$, we have to find integers $T$ such that $4x = 11T + 4$ or $x = 11T/4 +1$. Given that $11T/4$ is an integer, T could be any multiple of $4$.

    This implies that x can take infinitely many values ($12,23,34$ etc).

    Doesn't that mean, in the words of the author, that mod $11$ DOES suffer from the same defect as mod $12$ in this case?

    I get very nervous saying the author is wrong though. So I'm sure I've misunderstood something key to answering this.
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    Re: Modulo question

    Quote Originally Posted by flashylightsmeow View Post
    Thanks Plato! This makes sense.
    So am I right that, for $4x\equiv4(\bmod 11)$, we have to find integers $T$ such that $4x = 11T + 4$ or $x = 11T/4 +1$. Given that $11T/4$ is an integer, T could be any multiple of $4$.
    This implies that x can take infinitely many values ($12,23,34$ etc).
    Doesn't that mean, in the words of the author, that mod $11$ DOES suffer from the same defect as mod $12$ in this case?
    I get very nervous saying the author is wrong though. So I'm sure I've misunderstood something key to answering this.
    The author is correct. He meant that in the set $\{0,1,2,3,4,5,6,7,8,9,10\}$ (the so-called equivalent classes) there is only one solution to $4x\equiv 4 \mod 11$.
    LOOK AT THIS and see only one solution.

    Then LOOK AT THIS to see three solutions.

    Using that site you can change the variables to explore.
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    Re: Modulo question

    Quote Originally Posted by flashylightsmeow View Post
    Thanks Plato! This makes sense.

    So am I right that, for $4x\equiv4(\bmod 11)$, we have to find integers $T$ such that $4x = 11T + 4$ or $x = 11T/4 +1$. Given that $11T/4$ is an integer, T could be any multiple of $4$.

    This implies that x can take infinitely many values ($12,23,34$ etc).

    Doesn't that mean, in the words of the author, that mod $11$ DOES suffer from the same defect as mod $12$ in this case?

    I get very nervous saying the author is wrong though. So I'm sure I've misunderstood something key to answering this.
    mod 11 we have cancellation so $4x=4$ implies $x=1$, a unique solution

    mod 11, $1=12=23=34=...$
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    Re: Modulo question

    Quote Originally Posted by Idea View Post
    mod 11 we have cancellation so $4x=4$ implies $x=1$, a unique solution

    mod 11, $1=12=23=34=...$
    @Plato @Idea thank you again both.

    I think I've got it.

    $4x \equiv 4(\mod 12) \Rightarrow 4x-4 \equiv 0 (\mod 12)$ therefore $12|4(x-1)$. In this case you have solutions {1,4,7,10}
    Whereas for $\mod 11$ we have, mutatis mutandis:

    $11|4(x-1)$ in this case, because 11 is prime, $4(x-1)$ has no solutions where $x\in$ Z (coudn't find the Z symbol in latex). So the only solution is 1.

    Is this right, and a satisfactory answer to the question!?
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    Re: Modulo question

    Looks good.

    note that $12$ divides $4(x-1)$ implies $\displaystyle 3 $ divides $\displaystyle (x-1) $ therefore $x=1+3 k , k=0,1,2,3$ giving all solutions mod $\displaystyle 12$

    on the other hand, $11$ divides $4(x-1)$ implies $\displaystyle 11$ divides $\displaystyle (x-1) $ since $\displaystyle 11 $ and $\displaystyle 4 $ are co-prime therefore $x=1+11k , k=0$ giving the unique solution mod $\displaystyle 11$

    you might also want to consider this example

    $\displaystyle 5x=4 \pmod{12}$ which has a unique solution namely $\displaystyle x=8$
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    Re: Modulo question

    Quote Originally Posted by Idea View Post
    Looks good.

    note that $12$ divides $4(x-1)$ implies $\displaystyle 3 $ divides $\displaystyle (x-1) $ therefore $x=1+3 k , k=0,1,2,3$ giving all solutions mod $\displaystyle 12$

    on the other hand, $11$ divides $4(x-1)$ implies $\displaystyle 11$ divides $\displaystyle (x-1) $ since $\displaystyle 11 $ and $\displaystyle 4 $ are co-prime therefore $x=1+11k , k=0$ giving the unique solution mod $\displaystyle 11$

    you might also want to consider this example

    $\displaystyle 5x=4 \pmod{12}$ which has a unique solution namely $\displaystyle x=8$
    Super! Thank you again!


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