- #1
fishturtle1
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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