1. ## equivalence relation

The relation R is defined by aRb if ab=ba for all a,b in Z(set of integers).

How to show that R is not an equivalence relation?
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R is definitely reflexive and symmetric.
How to prove that it is not transitive

2. ## Re: equivalence relation

The relation R is defined by aRb if ab=ba for all a,b in Z(set of integers).
How to show that R is not an equivalence relation?
-------------
R is definitely reflexive and symmetric.
How to prove that it is not transitive
OH? How do you define $0^0~?$

3. ## Re: equivalence relation

This is a strange problem.
I would like to see the textbook solution.

The relation $R$ is defined by: . $aRb\text{ if }a^b =b^a\text{ for all }a,b \in Z.$

How to show that $R$ is not an equivalence relation?

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$R$ is definitely reflexive and symmetric.

How to prove that it is not transitive? .I don't know . . .

Note that: $a,b \ne0$

Does $aRb$ and $bRc$ imply $aRc\,?$ .Yes!

$\begin{array}{ccccccc}aRb & \Rightarrow & a^b \:=\:b^a & [1] \\ bRc & \Rightarrow & b^c \:=\:c^b & [2] \end{array}$

Raise [1] to the power $\tfrac{c}{b}\!:\;\left(a^b\right)^{\frac{c}{b}} \:=\:\left(b^a\right)^{\frac{c}{b}} \quad\Rightarrow\quad a^c \:=\:b^{\frac{ac}{b}}\;\;[3]$

Raise [2] to the power $\tfrac{a}{b}\!:\;\left(b^c\right)^{\frac{a}{b}} \:=\:\left(c^b\right)^{\frac{a}{b}} \quad\Rightarrow\quad b^{\frac{ac}{b}} \:=\:c^a\;\;[4]$

Equate [3] and [4]: . $a^c \:=\:c^a$

Therefore, $R$ is transitive. . $R$ is an equivalence relation.

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One explanation . . .

We can find one such relation: . $2R4 \quad\Rightarrow\quad 2^4 \:=\:4^2$

The difficulty is finding the second relation: . $4Rc \quad\Rightarrow\quad 4^c \:=\:c^4$
. . where $c \ne 2.$

4. ## Re: equivalence relation

Originally Posted by Plato
OH? How do you define $0^0~?$
R cannot be reflexive as aRa does not hold for a=0 ( since 0^0 is not defined ). Am I right?

5. ## Re: equivalence relation

For me $0^0$ is not defined.