The relation R is defined by aRb if a^{b}=b^{a} 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
Hello, Suvadip!
This is a strange problem.
I would like to see the textbook solution.
The relation $\displaystyle R$ is defined by: .$\displaystyle aRb\text{ if }a^b =b^a\text{ for all }a,b \in Z.$
How to show that $\displaystyle R$ is not an equivalence relation?
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$\displaystyle R$ is definitely reflexive and symmetric.
How to prove that it is not transitive? .I don't know . . .
Note that: $\displaystyle a,b \ne0$
Does $\displaystyle aRb$ and $\displaystyle bRc$ imply $\displaystyle aRc\,?$ .Yes!
$\displaystyle \begin{array}{ccccccc}aRb & \Rightarrow & a^b \:=\:b^a & [1] \\ bRc & \Rightarrow & b^c \:=\:c^b & [2] \end{array}$
Raise [1] to the power $\displaystyle \tfrac{c}{b}\!:\;\left(a^b\right)^{\frac{c}{b}} \:=\:\left(b^a\right)^{\frac{c}{b}} \quad\Rightarrow\quad a^c \:=\:b^{\frac{ac}{b}}\;\;[3]$
Raise [2] to the power $\displaystyle \tfrac{a}{b}\!:\;\left(b^c\right)^{\frac{a}{b}} \:=\:\left(c^b\right)^{\frac{a}{b}} \quad\Rightarrow\quad b^{\frac{ac}{b}} \:=\:c^a\;\;[4]$
Equate [3] and [4]: .$\displaystyle a^c \:=\:c^a$
Therefore, $\displaystyle R$ is transitive. .$\displaystyle R$ is an equivalence relation.
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One explanation . . .
We can find one such relation: .$\displaystyle 2R4 \quad\Rightarrow\quad 2^4 \:=\:4^2$
The difficulty is finding the second relation: .$\displaystyle 4Rc \quad\Rightarrow\quad 4^c \:=\:c^4$
. . where $\displaystyle c \ne 2.$
That strictly depends upon who you think is correct. Study this page.
There is no general agreement on that point. So you must go by whatever your textbook/lecturer says.
For me $\displaystyle 0^0$ is not defined.