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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.

BabaSvoloch
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    You are right with number one. So number two is asking : if $x-y$ is even, is $y-x$ even? Number three is asking : if $x+2y$ is positive, is $y+2x$ positive? So you essential have to check if the proposition holds even when you switch $x$ and $y$. – Sarvesh Ravichandran Iyer Apr 24 '17 at 04:47
  • @астонвіллаолофмэллбэрг Thank you so much! I have the same problem regarding transitivity... how would I do that one? It's the exact same set of problems with x and y. Since there's only x and y instead of a third element z, would it basically be the same thing as answering whether it's reflexive? – BabaSvoloch Apr 24 '17 at 05:09
  • If $x+y$ is even and $y+z$ is even, is $x+z$ even? If $x-y$ is even and $y-z$ is even, is $x-z$ even? If $x+2y$ is positive and $y+2z$ is positive, is $x+2z$ positive? This is the pattern. – Sarvesh Ravichandran Iyer Apr 24 '17 at 05:14

2 Answers2

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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.

Kanwaljit Singh
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  • I don't that what you have written is a fully formed answer. While the examples you have given are useful for seeing why the relations might be symmetric, they do not answer the question. To show that the relationship is symmetric you have to show that $x \sim y$ implies $y \sim x$ for all integers $x,y$, not just for certain integers. – michaelhowes Apr 24 '17 at 06:07
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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.

michaelhowes
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