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Prove that $ℕ^n$ is countable for every n∈ℕ

I'm stuck please help this quesion

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closed as off-topic by Brian Fitzpatrick, Dan Rust, MJD, user63181, user61527 Mar 17 '14 at 23:15

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Hint: Let $n\in\mathbf{N}$ and $p_1,\dots,p_n$ be distinct primes. Consider $f:\mathbf{N}^n\to\mathbf{N}$ given by $f(m_1,\dots,m_n)=p_1^{m_1}\cdots p_n^{m_n}$. What can you deduce from such a function?

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A typical element of the product set is something like (3,58,15), the case $(n=3)$. Consider base 11 number system where this represents a unique (positive) integer by treating comma as symbol (digit) for ten: So the example above represents $$ 5 + ( 1\times 11) + (10\times 11^2)+ (8\times 11^3) + (5\times 11^4) +(10\times 11^5) +(3\times11^6)$$ This gives an injective function to positive integers, so the product set is countable.

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  • $\begingroup$ Never seen that before! That's pretty clever :) $\endgroup$ – Stella Biderman Mar 17 '14 at 3:55
  • $\begingroup$ @ Stella Biderman. I don't deserve the credit. Not my original. Read it in American Math Monthly (don't know the author) 25 years ago. That proof was for rational numbers where $1/2$ is read as a base 11 integer treating the slash as digit 10. I reused for this case. $\endgroup$ – P Vanchinathan Mar 17 '14 at 6:54
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Hint: note that $\mathbb{N}^n =\mathbb{N}^{n-1} \times \mathbb{N}$. If you've proven that $\mathbb{N}^{n-1}$ is countable, then this must be the same as $\mathbb{N}\times \mathbb{N}$.

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