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I want to show that the set of all polynomials with rational coefficients is countable.

I know that to prove this I need to find a mapping $f$ from the natural numbers to this set that is a bijection. But I'm struggling conceptually of how to think of this. In fact, my intuition is that this isn't even true since can't we plug in an uncountable set of numbers into our polynomial? It's because of this thought that I can't even think of what a bijection would be like. If anyone could help untangle these conceptual issues I'm having that would be good -- I don't necessarily need a proof of the claim, though that's fine too.

Edit: not really a duplicate.

Wilson Brians
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  • Do you know already a proof that the set of rational numbers is countable? – J.-E. Pin Nov 01 '16 at 18:44
  • @J.-E.Pin Yeah I do. – Wilson Brians Nov 01 '16 at 18:52
  • @Charter Thanks sheriff but I found the answer a little inadequate (doesn't address the conceptual problem I'm having). – Wilson Brians Nov 01 '16 at 18:53
  • You don't need to find a mapping -- you can just do calculations with sets and cardinal numbers. –  Nov 01 '16 at 19:07
  • Can you explain why you think the set of numbers you can plug into a polynomial has relevance here? Also, you can plug uncountably numbers into $\sin$ -- but clearly the set ${ \sin }$ is not uncountable. –  Nov 01 '16 at 19:08
  • @WilsonBrians wow you called me "sheriff" xD. The title of your question suggest that you were looking for a proof. I recommend you next time to put a better title, e.g., "I having trouble with the proof of a theorem" or something like that. – Xam Nov 01 '16 at 20:25

2 Answers2

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It's important to distinguish the polynomial from the function it defines. A polynomial of degree $n$ with rational coefficients has the form $$ f(X)=a_nX^n+\dots+a_1X+a_0 $$ with $a_n\neq 0$, hence corresponds to an $(n+1)$-tuple of rational numbers (with the last entry being non-zero). Now what can you say about a finite product of countable sets? What about a countable union of countable sets?

carmichael561
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The set of all $(n+1)$-tuples $(a_0, a_1, ... a_n)$ , ($a_i \in \mathbb Q, 0 \le i \le n$) for a given polynomial of $n$th degree, $a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$ is the finite Cartesian product of $\mathbb Q$ with itself $n$ times, which is of course countable, since the finite Cartesian product of countable sets is countable, and $\mathbb Q$ is countable. Thus there are countably many such polynomials with degree $n$. For each $n$, denote the set of all $n$th degree polynomials with rational coefficients by $P_n$.

So now take $\bigcup_{n \in \mathbb N} P_n$, and we have the countable union of countable sets. Thus the set of all polynomials with rational coefficients is countable.

David Bowman
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