Proving a simple IFF modulo proof

Question

$a,b\in Z$ $a\equiv b$(mod10) IFF $a\equiv b$(mod2) AND $a\equiv b$(mod5)

Part 1: $a,b\in Z$ $a\equiv b$(mod10) ==> $a\equiv b$(mod2) AND $a\equiv b$(mod5)

Since $a-b-10k$ I can rewrite as $a-b=5(2m)$ and $a-b=2(5n)$. Therefore $2|(a-b)$ and $5|(a-b)$. So by definition, $a\equiv b$(mod2) AND $a\equiv b$(mod5)

Part 2: $a\equiv b$(mod2) AND $a\equiv b$(mod5) ==> $a,b\in Z$ $a\equiv b$(mod10)

I tried to do this as a direct proof, but my only cribs were $a-b=2m$ and $a-b=5n$ which I couldn't reduce to anything useful.

So instead, I take the contrapositive. Instead of $P\rightarrow Q$, I say $\neg Q\rightarrow \neg P$

$a\not \equiv b$ (mod10) $\rightarrow$ ($a\not \equiv b$ (mod5) AND $a\not \equiv) b$ (mod2)

The proof: Because $a\not \equiv b$ (mod10) THEN $a-b=10k+r$

I can rewrite that as $a-b=2(5m)+r$ $\rightarrow$ $a\not \equiv b$ (mod2) (because there is a remainder) and $a-b=2(5n)+q$ $\rightarrow$ $a\not \equiv b$ (mod5) because there is a remainder.

Is there a more efficient way to write that?

I think I see a better way for the $Q \rightarrow P$

Since $a\equiv b$(mod2) I know $2|(a-b)$ thus $a-b=2m$. Therefore $a-b$ is even.

Since $a-b=5n$ it also seems $5n$ is even.

Thus $n=2q$

Finally $a-b=5(2q)=10q$

This shows $10|(a-b)$

Answer 1

It is useful to see another way of doing this.

Your first part is correct. For the second part, suppose that $a-b = 2m=5n$.

Then, $2m=5n$, so $5 | 2m$, but because $5$ is prime and $5 \nmid 2$, so $5 | m$ (if you need a proof of this, ask me, it is given by Bezout's theorem).

Let $m = 5k$, then $2m = 5n = 10k$, and $a-b= 10k$, so that $a \equiv b \mod 10$.

Alternately, the theorem is equivalent to the fact that $\mbox{lcm}(2,5) = 10$, which is equivalent to the fact that $\gcd(2,5) = 1$ (product of two numbers is the product of their lcm and gcd), which may be seen from Bezout's lemma as $5 - 2 \times 2 = 1$.