0
$\begingroup$

Let $f:\mathbb{Q} \rightarrow \mathbb{Q}$ be defined as ${\forall}x:f(x)=a_0x^0+a_1x^1+...+a_{n_1}x^{n-1}+a_nx^n$ and let $g:\mathbb{Q} \rightarrow \mathbb{Q}$ be defined as ${\forall}x:g(x)=b_0x^0+b_1x^1+...+b_{n-1}x^{n-1}+b_nx^n$, where ${\{a_0,...,a_n}\}$ and ${\{b_0,...,b_n}\}$ are indexed sets of rational numbers. Let $f*g$ be the function $h$ such that ${\forall}x:h(x)=f(x)g(x)$.

How to prove that for all such polynomials $f,g$, $h$ too is a polynomial $h:\mathbb{Q} \rightarrow \mathbb{Q}$ and therefore that multiplication of polynomials is closed?

$\endgroup$
1
  • 1
    $\begingroup$ Use the distributive law to derive the well-known formula for the product of two polynomials, and observe that it's also a polynomial. $\endgroup$
    – Nick
    Jan 15 '17 at 14:33
1
$\begingroup$

This is true for polynomials over any ring. We have:

$$h(x) = c_0 + c_1x^1 + \ldots c_nx^n$$

with

$$c_n = \sum_{i=0}^n a_ib_{n-i}$$

You can verify by comparing coefficients of $f(x)g(x)$. Hence $\mathbb{Q}[x]$ is a ring.

$\endgroup$
2
  • $\begingroup$ Taking $\displaystyle \sum_{i=0}^n a_ix^i{\big{(}}\displaystyle \sum_{i=0}^nb_ix_i{\big{)}}$ as $h(x)$, how do I get $c_n$ as coefficients? The "coefficient" here seems to be $a_i{\big{(}}\displaystyle \sum_{i=0}^n b_ix_i{\big{)}}$, which can't be a constant since its formula involves $x$. How to prove that it's equal to $c_n$? $\endgroup$
    – asdasdfsss
    Jan 15 '17 at 15:08
  • $\begingroup$ First, you have $(\sum a_ix^i)(\sum b_ix^i)$ as $h(x)$. Second, $a_i(\sum b_ix^i)$ can't be the coefficient, since $x^i$ is in the term. To get $c_n$, you have to use the distributive law to find all the summands in each factor, so that when they are multiplied, the $x$ has power $n$. $\endgroup$
    – ryanblack
    Jan 15 '17 at 15:17

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.