equivalence properties of $\equiv \pmod n$ proofs

Question

Prove the identity: $$a \equiv b \pmod n \wedge a\equiv c \pmod n\implies b\equiv c \pmod n$$

I need to prove this property of $\equiv \pmod{n}$ along with a few others can someone link me to a site that can help me work through these proofs all I am finding are proofs that it is an equivalence relation

Answer 1

The property you are describing is transitivity. It is one of the three properties of an equivalence relation (along with reflexivity and symmetry). To prove transitivity, you simply show that if $aRb$ and $bRc$, then $aRc$.

So if $a \equiv b \pmod{n}$, then $n|(b - a)$, or $nq = b-a$, for some $q \in \mathbb{Z}$. Similarly, if $b \equiv c \pmod{n}$, then $nk = b - c$ for some $k \in \mathbb{Z}$. Now you want $a \equiv c \pmod{n}$. So consider $nk - nq = b - c - b + a = a - c$. Factor out the $n$ and then apply the quotient remainder theorem. Then you're back in the form of modular arithmetic and you're done.

Answer 2

Hint $\ $ The set of multiples of $\,n\,$ is closed under subtraction: $\,jn-kn = (j-k)n.$

Therefore if $\,a-c\,$ and $\,a-b\,$ are multiples of $\,n\,$ then so too is $\,\ldots$