0
$\begingroup$

I am studying modular arithmetic and I don't know how to prove or disprove the following :

If $x \equiv c \mod n $ then $x \equiv -c \mod n$

By trying different numbers it seems as this is true ,some attempt I've made so far is :

$x \equiv c \mod n \implies x \equiv -(-c) \mod n $

Since $-x \equiv -c \mod n $ I have that $$ x \equiv -(-x) \mod n \implies 0\equiv 0 \mod n $$

What does the last statement now mean ?

$\endgroup$
8
  • $\begingroup$ Did you try some specific numbers? It should be fairly clear, either from the definition of $\equiv$ or the final step you have that this is false. $\endgroup$ – user296602 Jan 23 '16 at 17:48
  • $\begingroup$ I've picked up the wrong numbers then ... $\endgroup$ – Nameless Jan 23 '16 at 17:49
  • 1
    $\begingroup$ You must have only picked $n=2$ or $x=nm$ for it to be true. $\endgroup$ – user208649 Jan 23 '16 at 17:50
  • $\begingroup$ Try doing this with $n=3$; what does that give you? $\endgroup$ – Arnaud D. Jan 23 '16 at 17:50
  • $\begingroup$ Before the edit, you had come to the conclusion that $2x \equiv 0 \pmod n$. Write down any number for which this is false. $\endgroup$ – user296602 Jan 23 '16 at 17:50
2
$\begingroup$

Mod 2, $1\equiv-1$ and the $\iff$ is true. When the mod $n$ is $>2$, $1\not\equiv-1$ and the $\iff$ is false.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.