# Question about relation on real numbers

Let $R$ be the relation on the set of real numbers such that, $$R = \{(x, y): y = x2\}$$ Is $R$ an equivalence relation?

Sorry I'm quite new to discrete maths. What does the $| \;\;|$ mean anyway?

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tiny, why did you erase a key part of your question? –  Zev Chonoles Feb 4 '13 at 5:48

Here it means the absolute value. $$|a| = \begin{cases}a,& a \geq 0\\ -a, & a < 0.\end{cases}$$

1. $|x-x| = 0 \neq 2$, so $R$ is not reflexive.
2. $|y-x| = |-(x-y)| = |x-y|$, so $R$ is symmetric.
3. Transitivity is (slightly) less straightforward. If $|x-y|=2$ and $|y-z|=2$, does it follow that $|x-z|=2$? Try a few examples and see what conclusions you can draw.
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Another way to think of $|x|$ is to imagine an axis (that is, one-dimensional space) and calculate the distance from point $x$ to 0 (zero). In fact $\delta(x,y) = |x-y|$ does nothing else, but measures the distance (an unsigned number) between $x$ and $y$. So you could ask:

$$R = \{ (x,y) : \text{ the distance between } x \text{ and } y \text{ is } 2\}.$$

Of course, the distance between any $x$ and itself is zero, so $R$ is not reflexive. On the other hand, the distance from $x$ to $y$ is the same as from $y$ to $x$, so we can conclude that it is symmetric. Finally, if we know that the distance between $x$ and $y$ is 2 and between $y$ and $z$ is 2 too, can you guess what is the distance between $x$ and $z$ (or more precisely, is it $2$)?

I hope this will give you some intuition ;-)

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