Transitivity question

1. Oct 4, 2017

fishturtle1

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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

2. Oct 4, 2017

andrewkirk

Think in terms of ratios. The condition $a^k=b^k$ is the same as $\frac{a^k}{b^k}=1$, provided $b\neq 0$. Can you think of a way of chaining ratios together to get the result you want for transitivity? .

3. Oct 4, 2017

SammyS

Staff Emeritus
It seems to me that you have made this much more complicated than necessary.

No need to even mention the express complex numbers in the form, $\ x+yi\$.

The reflexive and symmetric behavior of the ~ relation follow directly from the reflexive and symmetric behavior of equality .

Transitivity is not terribly difficult either.

To start, consider that $\ a\text{~}b \$ implies that there exists some $\ k\,$∈ℤ+ such that $\ a^k=b^k\$.

It then follows that $\ \left(a^k\right)^p=\left(b^k\right)^p\$ for every $\ p\,$∈ℤ+

etc.

4. Oct 4, 2017

fishturtle1

Yea I see your point that the expansion into a + bi is unnecessary.. Here is the updated proof,

Proof:
(Reflexivity) Suppose $a \epsilon \mathbb{C}$. Then $a^k = a^k$, so a ~ a. So ~ is reflexive.

(Symmetry) Suppose a ~ b. Then $a^k = b^k$ so $b^k = a^k$. So b ~ a. So ~ is symmetric.

(Transitivity) Suppose a ~ b and b ~ c. Then $a^k = b^k$ and $b^n = c^n$ for some $k, n \epsilon \mathbb{Z^{+}}$
By raising the first equation by n, we have $a^{nk} = b^{nk}$ and by raising the second equation by k, we have $b^{nk} = c^{nk}$. So $a^{nk} = b^{nk} = c^{nk}$ where $nk \epsilon \mathbb{Z^{+}}$. Thus a ~ c, so ~ is transitive.

We conclude ~ is an equivalence relation.

5. Oct 4, 2017

fishturtle1

Thank you for the suggestion.
So using this and the next post, my transitivity step would look like..

(Transitivity) Suppose a ~ b and b ~ c. Then $\frac {a^k} {b^k} = 1$ and $\frac {b^n} {c^n} = 1$ for some $k, n \epsilon \mathbb{Z^{+}}$. Raising the first equation by n and the second equation by k, we have $\frac {a^{kn}} {b^{kn}} = 1$ and $\frac {b^{kn}} {c^{kn}} = 1$. We know $\frac {b^{kn}} {c^{kn}} = 1 = \frac {c^{kn}} {b^{kn}}$. Then $\frac {a^{kn}} {b^{kn}} = \frac {c^{kn}} {b^{kn}}$ which simplifies to $a^{kn} = c^{kn}$ where $kn \epsilon \mathbb{Z^{+}}$. Thus, a ~ c. So ~ is transitive.

Not sure if this is what you meant though...

6. Oct 4, 2017

andrewkirk

Yes that's it.