Why is $\{(x,y)\mid x-y\text{ is a rational number}\}$ an equivalence relation and and why are $\{(x,y)\mid x-y\text{ is a irrational number}\}$ and $\{(x,y)\mid x+y\text{ is an integer}\}$ not?
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2For all $;x;,;;x-x\in\Bbb Q\implies;$ the relation is not reflexive. – DonAntonio Mar 02 '14 at 21:47
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2And the second is also not reflexive: $2x$ is not necessarily an integer for arbitrary real $x$. – J.R. Mar 02 '14 at 21:51
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And what about the first one? Is $0$ irrational? – alex Mar 02 '14 at 22:03
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1I presume you are talking about a subset of $\mathbb{R}^2$? – copper.hat Mar 02 '14 at 22:05
2 Answers
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The second one is not reflexive since $x-x$ is not irrational. It's also not transitive since one can have $x-y$ and $y-z$ irrational while $x-z$ is rational. That happens if $y=0$, $x=\pi$, and $z=10$.
The third one is not reflexive since $x+x$ can fail to be an integer, as, for example, when $x=1/3$. It is also not transitive: consider what happens when $x=1/3$, $y=2/3$, and $z=1/3$. Then $x+y$ and $y+z$ are integers and $x+z$ is not.
The first one is reflexive because $x-x$ is rational, transitive because if $x-y$ and $y - z$ are rational then so is $x-z$ (because $x-z$ is just the sum of those other two, and the sum of two rational numbers is rational). It is symmetric since if $x-y$ is rational then so is $y-x$.
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For the first one:
(1) x - x = 0 $\in \mathbb{Q} $
(2) if $ x - y = p \in \mathbb{Q} \ then \ y - x = -p \in \mathbb{Q} $
(3) if $ x-y = p; y-z=q; p,q \in \mathbb{Q} \ then \ x - z = x -y +y -z = p+q \in \mathbb{Q}$
For the second one fail condition 1 and 3 (the sum of two irrationals may be rational); for the third also fail condition 1 (as TooOldForMath pointed out) and 3.
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