1. ## 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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2. ## Re: Modulo question

Originally Posted by flashylightsmeow
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

Sent from my iPhone using Tapatalk
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

3. ## Re: Modulo question

Originally Posted by flashylightsmeow
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.

4. ## Re: Modulo question

Originally Posted by topsquark
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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5. ## Re: Modulo question

Originally Posted by flashylightsmeow
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 $12$.

6. ## Re: Modulo question

Originally Posted by Plato
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.

7. ## Re: Modulo question

Originally Posted by flashylightsmeow
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.

8. ## Re: Modulo question

Originally Posted by flashylightsmeow
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, $1=12=23=34=...$

9. ## Re: Modulo question

Originally Posted by Idea
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!?

10. ## 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 $\displaystyle x=8$

11. ## Re: Modulo question

Originally Posted by Idea
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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