- #1

fishturtle1

- 394

- 82

## Homework Statement

If a, b ##\epsilon \mathbb{C}##, we say that ##a## ~ ##b## if and only if ##a^k = b^k## for some positive integer ##k##. Prove that this is an equivalence relation.

## Homework Equations

## The Attempt at a Solution

Proof:

(Reflexivity): Suppose ##a \epsilon \mathbb{C}##. Then ##a = x + iy## for some ##x,y \epsilon \mathbb{R}##. We know ##(x+iy)^k = (x+iy)^k##. So a ~ a. So ~ is reflexive.

(Symmetric): Suppose a ~ b. Let ##a = x + iy## and ##b = s + it## for some ##x,y,s,t \epsilon \mathbb{R}##. Then ##(x+iy)^k = (s+it)^k, k \epsilon \mathbb{N}##. So

##(s+it)^k = (x+iy)^k## is true. So b ~ a. So ~ is symmetric.

(Transitivity): Suppose a ~ b and b ~ c. Let ##a = x + iy, b + s + it, ## and ##c = m + ni##. Then ##(x+iy)^k = (s+it)^k## for some ##k \epsilon \mathbb{N}## and

##(s + it)^g = (m + ni)^g## for some ##g \epsilon \mathbb{N}##.

So we have to show there exists ##h \epsilon \mathbb{N}## such that ##a^h = c^h##.

Consider 2 cases:

Case1: k > g.

We know ##a^k = b^k##

##a^k b^g = b^k c^g##

##a^k b^g b^{-k} = b^k b^{-k} c^g##

##a^k b^{g-k} = (1) c^g##

...

I think the transitivity will have to do with the roots and factoring out b's somehow, can someone point me in a direction please