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I have a relation $R$ defined on the set of integers as follows:

$(a,b) \in R ⇔ a-b\text{ is even}$

I have to prove that $R$ is an an equivalence relation. What I have:

I first check whether $R$ is reflexive. If $a = b$, $a-b = 0$. Zero is an even number. Therefore reflexivity is OK.

Than I check for symmetry: If
$$a-b = 2Z \\ b-a = (-1)(a-b) ⇒ (b-a) = -2Z$$
$-2Z$ looks also an even number, so symmetry i OK.

Then I should check for transitivity but don't know how to do it. Can someone please explain? Also: are my tests for reflexivity and symmetry correct? I didn't do math for several years, so my formulation of the solution is probably also incorrect.

Edit: I think I have it. Let's say that

$a-b = 2k$ and $b-c =2l$, where $k$ and $l$ are some integers. Then I sum the equations:

$a + b - b - c = 2l +2k$

Which is:

$a - c = 2(k+l)$

The right side is an integer times 2. So, when $(a-b) \in R$ and $(b-c) \in R$, $(a-c)$ is also $\in R$. Is this correct?

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    $\begingroup$ Assume that $a-b$ is even and $b-c$ is even. Then $a-c$ is even. Try first an example for yourself with actual integers. Then try to understand what transitive means. $\endgroup$ – Dietrich Burde Nov 23 '19 at 14:10
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    $\begingroup$ Write down the definition of transitivity and proceed. Yes, your work up until now is fine. $\endgroup$ – Qi Zhu Nov 23 '19 at 14:11
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    $\begingroup$ Well, your typesetting is not fine. Look up MathJax. $\endgroup$ – Dietrich Burde Nov 23 '19 at 14:12
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    $\begingroup$ Do not say "$-2Z$ looks like an even number" instead say $-2Z = 2(-Z)$ and $-Z$ is an integer, so $-2Z$ is even. $\endgroup$ – GEdgar Nov 23 '19 at 14:12
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What you have looks great so far. Transitivity is usually the longest section of the proof, but it is very often the same basic idea as symmetry.

So here, you want to want to take $a,b,c$ to be arbitrary numbers and assume that $(a,b)\in R$ and $(b,c)\in R$. Your goal is to prove that $(a,c)\in R$.

Similar to what you did before, you can choose $x$ and $y$ to be integers such that $a-b=2x$ and $b-c=2y$. Now think about whether you can find an integer $z$ such that $a-c=2z$.

Hint:

$a-c=(a-b)-(b-c)$

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  • $\begingroup$ Thanks for reply. Do You mean the following? b - a = 2x ⇒ a = b-2x and c -b = 2y ⇒ c = 2y + a. I put it in the c-a = 2z, it leads to: 2x + 2y = 2z. When I put in the orginal equations, I have: c - b + b - a = c - a. Which is: 0 = 0. So I can say, that it is true? $\endgroup$ – Radek John Nov 23 '19 at 16:02
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In general, proving that an integer is even requires showing that this number can be written in the form : 2 times some integer.

So, carrying the proof of symmetry to its completion would require, normally, to show that -2.Z can also be written as 2 times some integer.

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