1

Let $m \ge 2$ be a composite integer. Prove that there are elements $[x]_m$ and $[y]_m$ of $\mathbb Z$/m $\mathbb Z$ with $[x]_m,[y]_m$ $\neq$ $[0]_m$, such that $[x]_m\cdot[y]_m=[0]_m$

So, I understand the question but I am unsure of the approach to writing the proof. For me, it makes it easier to visualize a computational problem first, such as, let $x=4$,$y=5$ in $\mathbb Z$/4 $\mathbb Z$.

I suppose the proof I want to write is similar to the computational approach, let $x=m$ and show that, for any value $y$ and any composite value $m \ge 2$, $[x]_m\cdot[y]_m=[0]_m$

Any thoughts or suggestions?

Benji_Bombadill
  • 341

2 Answers2

3

If I understood your question correctly, your "computational approach" is wrong because $[4]_m=[0]_m$.

Anyway, first recall that

  1. $[x]_m\cdot [y]_m=[x\cdot y]_m$
  2. $[a]_m=[0]_m$ if and only if $a=km$ for some $k\in\mathbb{Z}$.

So you should look for two numbers $x$ and $y$ (that we can assume smaller than $m$ without losing generality) such that when you multiply them, you obtain something divisible by $m$.

There is a natural choice for them, can you guess it without reading the next few lines?

Suppose that the prime decomposition of $m$ is $$m=p_1^{e_1}\cdot p_2^{e_2}\cdots p_r^{e_r}.$$ Then you can pick $x=p_1$ and $y=p_1^{e_1-1}p_2^{e_2}\cdots p_r^{e_r}$.

I left to you the proof that they verify the hypothesis.

dadexix86
  • 970
  • 5
  • 16
  • The proof is incorrect if $,m = p_1^{e_1},$ has only one prime factor. But it is easy to fix. – Bill Dubuque Feb 21 '14 at 17:24
  • Thanks! I corrected the answer, now it should work :) – dadexix86 Feb 21 '14 at 17:55
  • 2
    That works. One needn't appeal to prime factorizations since, by definition, if $,m,$ is composite then it has a factorization $,m = xy,$ that is nontrivial, i.e. $,1< x,y< m,,$ which yields the sought $,x,y.\ \ $ – Bill Dubuque Feb 21 '14 at 18:29
0

$[m]_m=[0]_m$ so u cant use that.

$m$ is given to be composite so take a non-trivial factor $p$ and take $q=m/p$. Then $[p]_m.[q]_m=[0]_m$ .e.g. In $\Bbb Z/4\Bbb Z$ $[2]_4.[2]_4=[0]_4$.

viplov_jain
  • 913