# If both $a$ and $b$ $\not \equiv 0 \pmod{p}$ then $ab \not\equiv 0 \pmod{p}$

Any help with this proof would be great. Not even sure where to begin. I'm pretty much a total newbie.

If $a$ is not congruent to $0 \pmod{p}$ and $b$ is not congruent to $0 \pmod{p},$ where $p$ is a prime number, then $a*b$ is not congruent to $0 \pmod{p}.$

Also, not sure why it is necessary to assume that p is a prime number...?

-
Here is an example for $n$ composite. $3 \equiv 3 \pmod{12}$ and $4 \equiv 4 \pmod{12}$ but $3*4 \equiv 0 \pmod{12}.$ –  user2468 Jul 28 '12 at 2:53
Please don't use the proof-theory tag for question like this. That denotes a subfield of mathematical logic devoted to proof-theoretical matters. –  Bill Dubuque Jul 28 '12 at 3:08
noted. Sorry, won't happen again. –  rookie_02 Jul 28 '12 at 7:33

Let $Q=\{q_1,\ldots,q_m\}$ be the primes that occur in the prime factorisation of $a$, and $R=\{r_1,\ldots,r_n\}$ be the primes that occur in the prime factorisation of $b$. Then $Q\cup R$ is the set of primes that occur in the prime factorisation of $ab$. Your premise is that neither $a$ nor $b$ is a multiple of $p$. So $p\notin Q$ and $p\notin R$. Therefore $p\notin Q\cup R$ and $ab$ is also not a multiple of $p$.

-
1. Euclid's lemma: $p | a b \implies p | a \lor p | b.$

2. Contrapositive: $x \implies y$ is equivalent to $\lnot y \implies \lnot x.$

3. De Morgan $\lnot (x \lor y)$ is equivalent to $\lnot x \land \lnot y$

So $$p \not| a \land p \not| b \implies p \not| a b.$$

-
But I have to admit it's a terrible proof! –  user2468 Jul 28 '12 at 2:51
But it's still useful –  rookie_02 Jul 28 '12 at 7:34

For the last part of your question, let $m=6$ and $a=2$, $b=3$. Then $ab\equiv 0\pmod{m}$ but $a\not\equiv 0\pmod m$ and $b\not\equiv 0\pmod{m}$. One can also pick $a=4$, $b=3$.

One can produce a similar example for any composite integer $m$: just pick $a$ and $b$, neither equal to $\pm 1$, such that $ab=m$.

Thus if $m \gt 1$, then $m$ has the property of the problem iff $m$ is prime.

-

The condition is the contrapositive of Euclid's Lemma from your previous question.

-

Hint $\$ The contrapositive of your statement is $\rm\:p\:|\:ab\:\Rightarrow\:p\:|\:a\ \ or\ \ p\:|\:b,\:$ which I proved in your prior question. If $\rm\:p>1\:$ is not prime then $\rm\:p = ab\:$ for $\rm\:a,b>1\:$ so $\rm\:p = ab\:|\:ab,\,$ but $\rm\:ab\nmid a,b$.

-

When $p$ is a prime number $\Bbb{Z}/p\Bbb{Z}$ is a field and hence in particular an integral domain. So therefore for any $a,b \in \Bbb{Z}/p\Bbb{Z}$, if $$ab = 0$$

by definition of an integral domain we must have $a=0$ or $b = 0$. This is equivalent (by the contrapositive) to saying that for any $a,b \in \Bbb{Z}/p\Bbb{Z}$ both not equal to zero, we have

$$ab \neq 0.$$

-

Since this proof deals with congruence modulo $p$ (for some prime $p$), the first step is to ask what it means for an integer to be "congruent to $0\ \mod p$"?

Answer: It means the integer is a multiple of $p$.

The next step is to replace "congruence mod $p$" with its meaning:

If $a$ and $b$ are not multiples of $p$, then $ab$ is not a multiple of $p$.

Can you see how to prove the proposition now? Do you see why we have to assume that $p$ is prime?

-
I see it now. Thanks –  rookie_02 Jul 28 '12 at 9:16

$a$ not congruent to $0 \mod p$ and $b$ not congruent to $0 \mod p$ just mean that $a$ and $b$ are not divisible by $p$. Suppose $p$ divides $ab$. Since $p$ is prime, this means that $p$ divides $a$ or $p$ divides $b$. Contradiction.

-

Since p does not divide a then there is a linear combination (using the extended division algorithm) ua+vp=1 where u, v are integers. Now multiply by b to get uab+vpb=b. If p divides ab then it divides the left side of this relation and so p|b.

This would also work if a and n are relatively prime. The statement is: if n is relatively prime to a, and n divides ab then n divides b.

-