# Homework Help: 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.