0

I'm currently working on the following problem:

Evaluate the following when they exist in $\mathbb{Z}_{9}$:

In $\mathbb{Z}_{9}$ we have $2 \times 5 = 4 \times 7 = 1$, and $5 \times 6 = 3$, so that $\frac{1}{2} = 5, \frac{1}{4} = 7$ and $\frac{3}{5} = 6$. However neither $\frac{1}{3}$ nor $\frac{5}{6}$ can exist in $\mathbb{Z}_{9}$, for otherwise it would follow quickly that one of the numerators would be a multiple of $3$, a contradiction.

I can't seem to first of all understand why $\frac{1}{3}$ and $\frac{5}{6}$ don't exist in $\mathbb{Z}_{9}$ as a result of one the numerators becoming a multiple of $3$, and also why this leads to a contradiction. I was hoping that someone could further elaborate.

Bill Dubuque
  • 282,220
Skm
  • 2,392
  • 2
    I think I see why you didn't find their explanation clear, but perhaps let's put it this way: Can you find $x$ such that $3x = 1$? Or such that $6x = 5$? (In $\mathbb{Z}_9$, that is.) – Brian Tung Oct 24 '24 at 00:01
  • You cannot deduce in $\mathbb Z_9$ that $5\times6=3$ implies $5=\dfrac12$ (Why?) – Ataulfo Oct 24 '24 at 00:04
  • @BrianTung: No, I'm not able to find such an $x$. – Skm Oct 24 '24 at 00:23
  • @Piquito: I'm not exactly sure why. – Skm Oct 24 '24 at 00:23
  • 2
    If $x = \frac{1}{3}$ exists it should satisfy $3x=1$ in $\Bbb Z_9$, equivalently $3x\equiv 1 \pmod 9$, which means $9 \mid 3x-1$ which is absurd because $3x-1$ is obviously not a multiple of $3$, let alone a multiple of $9$. – jjagmath Oct 24 '24 at 00:31

0 Answers0