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I need to prove that {9^n: n∈ℚ)={3^n: n∈ℚ).

So far I have proven {9^n: n∈ℚ}⊆{3^n: n∈ℚ}.

a∈{9^n: n∈ℚ}. Meaning for some a=9^n for some rational number. Thus, a=9^n=3^2n, showing that a is a rational number for some power of 3. so a∈{9^n: n∈ℚ}. Which also means {9^n: n∈ℚ}⊆{3^n: n∈ℚ}

But i don't know how to prove {3^n: n∈ℚ}⊆{9^n: n∈ℚ}.

So that I can say that {9^n: n∈ℚ)={3^n: n∈ℚ).

M.Maric
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Hint:$$\mathbb{Q}=\frac{1}{2}\mathbb{Q}$$

So $9^\mathbb{Q}=9^{\frac{1}{2}\mathbb{Q}}=(9^\frac{1}{2})^\mathbb{Q}=3^\mathbb{Q}$.

  • I know that for {3^n: n∈ℚ}⊆{9^n: n∈ℚ}. we need to somehow show that 9^1/2(n)= 3^n, but i'm not quite sure how. – M.Maric Nov 26 '16 at 00:25
  • @M.Maric Sorry, I don't know how to make the boxes alert "Spoiler" :(. This answer prove what you need, you just need to read every element as a set; i.e. for example the equality $9^\mathbb{Q}=3^\mathbb{Q}$ is the set equality ${9^q:q\in\mathbb{Q}}={3^q:q\in\mathbb{Q}}$, which is what you want!! –  Nov 26 '16 at 00:33
  • @M.Maric Try to prove every equality in the order they appear in my answer, I'll be glad to help if you get stuck. =) –  Nov 26 '16 at 00:36
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    Thanks, I understand what you are saying. I actually did something similar for other problems involving proofs with modulo – M.Maric Nov 26 '16 at 00:41