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Define a relation $∼$ on $ R$ by $x ∼ y$ if and only if $x + y \in \mathbb Q$. Is this an equivalance relation?

For a relation to be an equivalance relation, it must be reflexive, symmetric and transitive. And note that a rational number $\in \mathbb Q$ is $\in a/b$ where $a,b \in\mathbb Z$ (integers).

a) Reflexive: we need show $x\sim x$ for all values of $ x \in\mathbb R$. But let $x = \sqrt 3$. $\sqrt 3 + \sqrt 3 = 2 \sqrt 3$ which is not a rational number, Therefore $\sim$ is not reflexive.

b) Symmetric: we need to show if $x\sim y$ then $y\sim x$. This is true because $x+y=y+x $ so $y+x \in\mathbb Q$ and $x+y\in\mathbb Q$. So $\sim$ is symmetric.

c) Transitive: we need to show that if $x\sim y$ and $y\sim z$ then $x\sim z$. $x\sim z = (x+y) + (y+z) \in\mathbb Q$, so $x+z \in\mathbb Q$ because an rational number plus a rational number is a rational number. So $\sim$ is transitive.

Finally, relation $\sim$ is not an equivalance relation because it is not reflexive.

Am i correct?

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    $\begingroup$ With $x-y$ rather than the sum, it is an equivalence relation, though. And a nice one: picking an element from every equivalence class (possible by the axiom of choice), ans restricting it to the interval $[0,1]$, one can obtain a set that is not Lebesgue measurable. $\endgroup$ – A. Pongrácz Aug 19 '18 at 20:01
  • $\begingroup$ b) and c) can be omitted, since a) is enough to prove that it's not an equivalence relation $\endgroup$ – Ludvig Lindström Aug 19 '18 at 20:07
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You’re right about a and b but not c. Take $x=\sqrt{2}$,$y =-\sqrt{2}$, $z =\sqrt{2}$. Then $x \sim y$, $y\sim z$ but $x+z = 2\sqrt{2}$ is irrational.

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  • $\begingroup$ Thanks. I understand my mistake. Thank you all! $\endgroup$ – Randy Rogers Aug 19 '18 at 19:07
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For transitivity, choose $x=1-\sqrt{3}$, $y=\sqrt{3}$ and $z=2-\sqrt{3}$, then $x+y \text{ and } y+z \in \mathbb{Q}$ but $x+z=3-2\sqrt{3} \not\in \mathbb{Q}$. So NOT transitive.

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The relation is not transitive. Note that $\sqrt2\sim-\sqrt2$ and $-\sqrt2\sim\sqrt2$, but $\sqrt2\not\sim\sqrt2$.

On the other hand, in order to prove that it is not an equivalence relation, proving that it is not reflexive is enough. The rest was a waste of time.

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  • $\begingroup$ in the exam question they want you to show all 3 $\endgroup$ – Randy Rogers Aug 19 '18 at 19:02
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    $\begingroup$ @RandyRogers If it's not an equivalence relation it is sufficient to show just 1. $\endgroup$ – Ovi Aug 19 '18 at 19:02
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    $\begingroup$ @RandyRogers They want you to show all three when it is an equivalence relation. In this case, it is not. $\endgroup$ – José Carlos Santos Aug 19 '18 at 19:06
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This is an explanation of why your proof of transitivity is wrong; and it ties back to the relation not being reflexive.

You said that "$(x+y)+(y+z) \in \mathbb{Q}$, so $x+z \in \mathbb{Q}$". This would be true if $y+y$ were in $\mathbb {Q}$. However, if $y+y$ is irrational (like you showed it can be) then $x+z$ must be irrational too in forder for $x+y+y+z$ to be in $\mathbb{Q}$.

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