0
$\begingroup$

How to prove $\left| P_n \right| = \left|\mathbb{Q}^n\right|$ Where $P_n =\{p(x)=a_n x^n+ a_{n-1} x^{n-1}+...+ a_1 x+a_0 |a_i \in \Bbb Q \}$.

This is, the set of polynomial of degree $n$ and coefficient in $\mathbb{Q}$

I was thinking in this: For example, How many $4$ digits number are there? In a $4$ digit number, we have $4$ places to fill which can be filled by only $0,1,…,9$. So, number of $4$ digits number are 9*10*10*10 (since first place cannot be zero) Similarly, i have n+1 places (coefficients) and |Q| ways to fill them.

But, formally i cannot find a biyection, or give a formal proof of this. Can someone help me?

$\endgroup$
  • $\begingroup$ There is a bijection from $ \Bbb N$ to $\Bbb Q^n$ or to $\Bbb N^n$ for any positive integer $n.$ Search this site. I suggest searching "cardinal of $\Bbb Q^n\;$" and similar phrases. $\endgroup$ – DanielWainfleet Apr 9 '18 at 21:08
1
$\begingroup$

Hint: Consider $P_n \to \mathbb Q^n$ given by $p \mapsto (a_{n-1}, \dots a_1, a_0 )$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.