Prove if $a \equiv x\pmod m$ and $b \equiv x \pmod m$ for $x \in \{0,1,...,m-1\}$, then $a \equiv b$

Question

I just have trouble proving the proof. I wrote that for $A$ and $B$ there exist $a$ and $b$ that $A-x = a\cdot m \Rightarrow A = x+a\cdot m$ and with that $B = x+b\cdot m$. Anything else that would strengthen my case?

Answer 1

$$a\equiv x\mod m$$ means $$m\mid a-x$$ and $$b\equiv x\mod m$$ means $$m\mid b-x$$ Hence , we have $$m\mid (a-x)-(b-x)=a-x-b+x=a-b$$ which means $$a\equiv b\mod m$$

Answer 2

It's basic idea of what an equivalence relationship means in particular what it means for an equivalence relationship to be transitive.

$a \equiv x \pmod m$ means $m|(a-x)$.

And $b \equiv x \pmod m$ means $m|(b-x)$.

Lemma: If $m|k$ and $m|j$ then $m|k \pm j$.

Pf: $m|k$ means there exists an integer $h$ so that $k = mh$ and $m|j$ means that there is an integer $g$ so that $j = mg$. So $k\pm j = mh \pm mg = m(h\pm g)$. Now $h\pm g$ is an integer so $h\pm g$ is an integer so that $k \pm j = m(h\pm g)$ so $m|k \pm j$.

So if $m|(a-x)$ and $m|(b-x)$ then $m|(a-x) - (b-x) = (a-b)$. So $a\equiv b \pmod m$.

That's it.

====

Review: An equivalence relationship intuitively means for the purpose of a certain property two elements are equivalent if the can more or less be considered to be pretty much the same thing.

More practically, it means if we take all the elements and group them into categories so that all the elements in the universe will each belong into precisely one and only one group, then we can say all the elements in any one of the groups are all equivalent to each other and each one can stand in for any other but they are not equivalent to any element in another group.

More formally that means being an element of the group is to be

1) Reflexive. $r$ is always equivalent to itself. $r$ is always in the group which it is in.

Claim: $a \equiv a \pmod m$.

Pf: $a-a = 0 = 0*m$ so $m|a-a$ so $a \equiv a \pmod m$>

2) Symmetric: If $r$ is equivalent to $s$ then $s$ is equivalent to $r$. If $s$ is in the same group as $r$ then $r$ is in the same group as $s$.

Claim: If $a \equiv b \pmod m$ then $b \equiv a \pmod m$.

Pf: If $a\equiv b\pmod m$ then $m|a-b$ so there is an integer $k$ so that $a-b = mk$. Then $b-a = m(-k)$ and $-k$ is an integer. So $m|b-a$ and $b\equiv a \pmod m$.

3) Transitive: If $r$ is equivalent to $s$ and if $s$ is equivalent to $t$ then $r$ is equivalent to $t$. If $r$ and $s$ are in the same group; and $s$ and $t$ are in the same group; then it follows that $r$ and $t$ are in the same group.

Claim: If $a \equiv x\pmod m$ and $x \equiv b \pmod m$ then $a\equiv b \pmod m$.

Pf: I proved in above... except for $b\equiv x \pmod m$. But by 2) $x\equiv b \pmod m \iff b\equiv x \pmod m$ so that doesnt make any difference.