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.

  • $\begingroup$ Your proof that it's a equivalence relation is fine, but why wouldn't $12^3=1728$ be part of the equivalence class? $\endgroup$ – Henrik supports the community Jan 17 '17 at 21:57
  • $\begingroup$ I'm guessing asddf may have meant $[144] = \{12, 12^2 = 144, 12^3, \ldots\}$ (144 and all other powers $12^n, \;n\in \mathbb N_{>0}$ $\endgroup$ – Namaste Jan 17 '17 at 22:19
  • $\begingroup$ No henrik is right i should have added powers of 12 aswell I ll fix it $\endgroup$ – asddf Jan 17 '17 at 22:24
  • $\begingroup$ 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. $\endgroup$ – Namaste Jan 17 '17 at 22:33


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 $1$).

  • $\begingroup$ what how did you come up with this ? this seems so hard to spot $\endgroup$ – asddf Jan 17 '17 at 21:57
  • $\begingroup$ Well, I have sometimes used this way of writing natural numbers to solve problems. $\endgroup$ – ajotatxe Jan 17 '17 at 21:58
  • $\begingroup$ If you have time could you give me another problem that is solved similiarly ? $\endgroup$ – asddf Jan 17 '17 at 21:59

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