# Homework Help: Is this relation equivalence relation ?

1. May 3, 2012

### jd12345

1. The problem statement, all variables and given/known data
Relation is x^y = y^x........x and y belong to integers

2. Relevant equations

3. The attempt at a solution
Well i have already proven that they are reflexive and symmetric. I have doubt with transitive
I did the follwoing way x^y = y^x.......(1) and y^z = z^y.........(2)
from(1) x^z = y^(zx/y)
from(2) z^x = y^(zx/y)
Therefore x^z = z^x proving it is transitive

But the test paper solution does it this way
from(1) y = x^(y/x) and from(2) y = z^(y/z)
Now x^(y/x) = z^(y/z) ..... so its not transitive
Which is correct? Did i do anything wrong?

2. May 3, 2012

### glb_lub

How did it follow from x^(y/x) = z^(y/z) that x and z are not equivalent?
Raise both sides to the power of (xz/y) , we will get x and z as equivalent.
I think both methods prove transitivity. I can't see how the test paper solution is giving different answer after it has shown this :- x^(y/x) = z^(y/z) .

Note :- we don't have to worry about y being zero here. As 0 won't be equivalent to any other integer. As for any non-zero integer x ,
0^x(=0) won't be same as x^0(=1) .

3. May 3, 2012

### jd12345

So it is an equivalent relation right?
When the book gives the wrong solution i get real angry - more when it wastes so much time

4. May 3, 2012

### glb_lub

Anyways can you find three numbers which are equivalent according to the above relation. I am unable to.
I could think of pairs of numbers which are equivalent. For ex:- 2^4 = 4^2 .
I am unable to find 3 such numbers which are equivalent.

5. May 3, 2012

### glb_lub

Ah , as it happens other than the equivalence in the obvious case where an integer is equivalent to itself i.e x~x , there are only two pairs of integers which are equivalent , i.e (-2)~(-4) and (2)~(4) .

Thus there is no equivalence class containing 3 members or more for this equivalence relation. Either we have equivalence class of one member each i.e ...,{-3},{-1},{0},{1},{3}... or the two equivalence classes both of which contain two members i.e {2,4} and {-2,-4} .

This maybe a reason why your solution book gave the wrong answer for transitivity. Because there is no equivalence class of the form {x,y,z} where x,y,z are integers.

But there is another very common equivalence relation for which such a thing happens. I.e the equality relation.
I.e x~y if x=y. For this relation , all the equivalence classes contain one member each.