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?

  • 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

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

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


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

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

  • $\begingroup$ 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

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