0
$\begingroup$

My question reads:

a. Give 3 examples of the form ax=b in (mod 12) that has no nonzero solutions

b) In each equations, does equation ax=0 have a nonzero solution?

My proposed answer:

a. 2x=6, 4x=8 and 6x=6.

I am not too sure if this works or if I am misreading what no nonzero solutions means.

  • 0
    $6x=6$ has the solution $x=1$. – 2017-01-30
  • 0
    @lulu then does this not work? I am not too sure what is meant by having no nonzero solution – 2017-01-30
  • 0
    $2x=7$ is a good example. If $12$ divides $2x-7$ then $2$ would divide $7$, which it doesn't. Similarly $2x=1$, $3x=1$ are impossible $\pmod {12}$. – 2017-01-30
  • 0
    Please format your questions using Latex. – 2017-01-30

2 Answers 2

1

For section (a), if you pick a value for $a$ such that $d=\gcd(a,12)\ne 1$ - that is to say, $a$ and $12$ have a common factor $d$ - and then pick $b$ that is not a multiple of $d$, you will have an equation $ax\equiv b \bmod 12$ that does not have a solution. For example, $8x\equiv 10$, which has $d=4$, fulfills the requirements.

Then for section (b), for any equation $ax\equiv 0 \bmod 12$, where $a$ is chosen as above, then choosing $x=12/d$ will solve the equation.

In fact if you want to avoid non-zero solutions to $cx= 0$ in $\Bbb Z_{12}$, you need to have $\gcd(c,12)=1$. However then for any choice of $b$ there will be a solution for $cx=b$.

  • 0
    That's my second paragraph there. If we take an equation $ax\equiv b$ which has **no** solutions, then consider the equation $ax\equiv 0$, that one must have a non-zero solution. Of course any equation $ax\equiv 0$ has the solution $x\equiv 0$ – 2017-01-30
  • 0
    okay thank you! but then for part b I would set these to 0 and find an x that makes this possible? – 2017-01-31
  • 0
    You set the $b$ to zero, and there will be (at least one) non-zero $x$ to satisfy that, as I describe above – 2017-01-31
  • 0
    got it, thanks! – 2017-01-31
  • 0
    @Sam Um, I think those were the ones you used, but anyway they are fine. – 2017-01-31
0

The reason for the existence of such equations is because $12$ is not a prime. Due to this fact the ring $\Bbb Z_{12}$ has zero-divisors, e.g. $3$ and $4$ are zero divisors since $3 \cdot 4 = 0 \mod(12)$. Elements that are no zero-divisors have no common divisors with $12$ and are invertible (have a multiplicative inverse), like $5$ which is its own inverse : $5 \cdot 5 = 1 \mod(12)$. $a$ is invertible $\iff$ the equation $ax = b$ has a solution $\forall b \neq 0$. This is because $a^{-1}b$ is such a solution.