Proving congruent modulo

Question

In my book there are three things that I need to prove. First of: $$a \equiv a \pmod n , \forall a \in Z.$$ I tried to prove it this way: $$ a-a = 0$$ $$0 \pmod n = 0$$ Is this correct? Second one: $$ a \equiv b \pmod n \Rightarrow b \equiv a \pmod n$$ I tried with this: $$a-b=c$$ $$b-a=-c$$ $$c \pmod n=-c \pmod n$$ Is this correct? And the third one: $$a \equiv b \pmod n \land b \equiv c \pmod n \Rightarrow a \equiv c \pmod n$$

No clue how to prove this so I need full help on this one.

Answer 1

In proving any relation involving congruences or such properties of congruences, I'll advise you to use the definition of congruence and convert the congruence to the division form and carry out the necessary calculations, i.e. use $n|a-b$ which means $a-b$ is divisible by $n$ instead of $a \equiv b \pmod n $

As for examples, I am working out your three problems.

1. $$a \equiv a \pmod n , \forall a \in Z$$ So $n|a-a$ by definition and hence $$n|0$$ which is always true. Hence proved.
2. $$ a \equiv b \pmod n \Rightarrow b \equiv a \pmod n$$ So $n|a-b$ by definition and hence $$n|b-a$$ or, $$b \equiv a \pmod n$$ Hence proved.
3. $$a \equiv b \pmod n \land b \equiv c \pmod n \Rightarrow a \equiv c \pmod n$$ So $n|a-b$ and $n|b-c$ by definition and hence $$n|(a-b)+(b-c)$$ or, $$n|a-c$$ or, $$a \equiv c \pmod n$$ Hence proved.

Answer 2

Hint: $a-a=0$ and $0\cdot n=0$, so $n\mid a-a$ thus $a\equiv a\pmod{n}$.

Answer 3

$$\exists k\in\mathbb Z:a=b+kn$$ is indeed an equivalence relation.

$$a=a+0n,$$ $$a=b+kn\implies b=a-kn,$$ $$a=b+kn\land b=c+k'n\implies a=c+k'n+kn=c+k''n.$$