Show that $a\sim b \iff 5a+5b\equiv 0\bmod10$ is an equivalence relation

Question

Let $R$ be the relation on $\mathbb Z$ defined by $$ a\sim b\text{ if and only if } 5a+5b\equiv 0\bmod10 $$ Show this is an equivalence relation. Using that $10\mid 5(a+b)$ if and only if $2\mid a+b$, describe the equivalence classes of $R$.

Reflexive: Let $x\in \mathbb{Z}$. Then $5x+5x\equiv 0\bmod10$. This means that $10\mid 5x+5x$. This can be simplified, giving us $10\mid 10(x)$, which is clearly true. Thus, since $10\mid 5x+5x$, we know that $5x+5x\equiv 0\bmod10$. Hence, $(x,x)\in R$, so R is reflexive.

Symmetric: Let $(x,y)\in R$. Then we have $5x+5y\equiv 0\bmod10$. By the definition of congruence, this means $10\mid 5x+5y$. It is clear that $5x+5y=5y+5x$, so $10\mid 5y+5x$. Thus, $5y+5x\equiv 0\bmod10$. This means that $(y,x)\in R$, so $R$ is symmetric.

Transitive: Let $(x,y)\in R$ and let $(y,z)\in R$. Then we have $5x+5y\equiv 0\bmod10$ and $5y+5z\equiv 0\bmod10$. By the definition of congruence, we have $10\mid 5x+5y$ and $10\mid 5y+5z$. By the definition of divisibility, there exists integers $l,k$ such that $10l=5x+5y$ and $10k=5y+5z$ ...

To prove equivalence I know I have to prove reflexivity, symmetry, and transitivity. I got reflexivity and symmetry but I'm not sure how to go about transitivity. I'm not too sure about the equivalence classes. Any pointers would be very helpful. Thank you.

Answer 1

Note that \begin{align*}2\mid a+b &\iff \exists\, k \in \mathbb{Z} \text{ such that }a+b=2k\\ &\iff a-b=2(k-b)\\&\iff2\mid a-b\\&\iff a\equiv b \mod 2\end{align*}

Transitivity: So if $(x,y)$ and $(y,z)\in R,$ then $x \equiv y \mod 2$ and $y \equiv z \mod 2$ and hence $x \equiv z \mod 2$ which gives $(x,z) \in R.$

Equivalence class: By the above equivalence we get for any $n \in \mathbb{Z},$ its equivalence class is $[n]=n+2\mathbb{Z}$ and hence there are exactly $2$ equivalence classes, namely the odd and even integers.

Answer 2

You want to prove $5x+5z \equiv 0 \pmod{10}$. But $$5x+5z \equiv 5x+10y+5z \equiv (5x+5y)+(5y+5z) \equiv 0 \pmod{10}.$$

Answer 3

A word on your style:

Reflexive: Let $x\in\Bbb Z$. Then $5x+5x\equiv 0 \bmod 10$.

This is a very confusing way of laying out a proof. It implies that the first step in your proof is to observe that if $x\in\Bbb Z$, then $5x+5x\equiv 0\bmod 10$. But that is precisely what you are trying to prove! Instead, you should say something like this::

Reflexive: Let $x\in\Bbb Z$. Then we have to show that $5x+5x\equiv 0 \bmod 10$.

The same applies, mutatis mutandis, to your other two cases.