1
$\begingroup$

Let $ a,b \in \Bbb{N} $

We define the equivalence relation

$a \sim b:\Leftrightarrow$ there exist $r,s \in$$\Bbb{N}_{>0}$ such that $a^r=b^s$

Show that it is indeed an equivalence relation and find equivalence class of $144$.

What i have done so far :

Reflexive because $a^1=a^1$

Symmetric Let $a^r=b^s$ then $b^s=a^r$

Transivitive: Let $a^r=b^s $and $ b^l=c^k$ then there exists an $s$ such that $ a^{r\ell}=b^{s\ell} $ and $b^{\ell s}= c^{ks} $ so $a^{r\ell} = c^{ks}$

My thoughts on equivalence class of 144 is 12 and the powers of 144 but im not quite sure.

  • 0
    Your proof that it's a equivalence relation is fine, but why wouldn't $12^3=1728$ be part of the equivalence class? – 2017-01-17
  • 0
    I'm guessing asddf may have meant $[144] = \{12, 12^2 = 144, 12^3, \ldots\}$ (144 and all other powers $12^n, \;n\in \mathbb N_{>0}$ – 2017-01-17
  • 0
    No henrik is right i should have added powers of 12 aswell I ll fix it – 2017-01-17
  • 0
    That's exactly what I suggested in my comment: You meant powers of 12, and not powers of 144, as you say in your question. – 2017-01-17

1 Answers 1

1

Hint:

If you write down the exponents of each prime as a vector (for example, for $144$ is $(4,2,0,0,\ldots,0,\ldots)$) then $a\sim b$ iff their respective vectors are linearly dependent (exclude the case that $a$ or $b$ is $1$).

  • 0
    what how did you come up with this ? this seems so hard to spot – 2017-01-17
  • 0
    Well, I have sometimes used this way of writing natural numbers to solve problems. – 2017-01-17
  • 0
    If you have time could you give me another problem that is solved similiarly ? – 2017-01-17