Determining if a relationship x ~ y is symmetric

Question

I am to determine which of these relationships are symmetric. The variables $x$ and $y$ represent integers.

1. $x$ ~ $y$ iff $(x + y)$ is even
2. $x$ ~ $y$ iff $(x - y)$ is even
3. $x$ ~ $y$ iff $(x + 2y)$ is positive
4. $x$ ~ $y$ iff $(x - y)$ is positive
5. $x$ ~ $y$ iff $(x + y)$ is positive

I understand the concept of symmetry in relations when it comes to sets, but I'm not sure I understand these problems and what they're asking.

For number one, for example, are they trying to say that if x + y is even, then y + x is even? In which case, it would be symmetric.

Any insight would be hugely appreciated. I'm finding myself lost with this.

Answer 1

You can take examples to solve them.

(1) Let $x = 2$ and $y = 4$

$x + y = 2 + 4 = 6$

Now $x = 4$ and $y = 2$

$x + y = 4 + 2 = 6$

Answer is even number.

So it is symmetric.

(3) Let $x = 2$ and $y = 4$

$x + 2y = 2 + 8 = 10$

Now $x = 4$ and $y = 2$

$x + 2y = 4 + 4 = 8$

Answer is positive.

But for one value positive and other negative case is not symmetric.

Try with other cases.

Answer 2

Below I have shown that the second relation is symmetric and that the third relation is not symmetric. Hopefully this will provide some insight and you will be able to determine whether the remain three relations are symmetric.

For 2. define $x \sim y$ iff (x-y) is even.

To show that $\sim$ is symmetric, we have to show that for all $x,y \in \mathbb{Z}$, if $x \sim y$, then $y \sim x$. Thus let $x,y \in \mathbb{Z}$ be integers such that $x \sim y$. This means that $(x-y)$ is even. $(x-y)$ being even means that there exists a $k \in \mathbb{Z}$ such that $(x-y) = 2k$.

Now note that $(y-x) = (-x+y) = -(x-y) = -2k = 2(-k)$. Thus $(y-x)$ is an even integer and $y \sim x$ holds.

Note that there was nothing special about $x,y$ (other than assuming $x \sim y$), thus we know that whenever $x \sim y$ is true, $y \sim x$ is true and hence $\sim$ is symmetric.

For 3. define $x \sim y$ iff (x+2y) is positive.

I claim that $\sim$ is not symmetric. To show that a relation is not symmetric, we must find two integers $x,y \in \mathbb{Z}$ such that $x \sim y$ is true but $y \sim x$ is false.

If we let $x = -1$ and $y = 1$, we have that $x+2y = -1+2 =1>0$, thus $x \sim y$ is true. However $y + 2x = 1 + 2(-1) = 1-2 = -1 < 0$, thus $y \sim x$ is false. This shows that $\sim$ is not symmetric.