Question

Prove $\mathit{f}\left(x\right)\mathbf{\equiv }\mathit{g}\left(x\right)\left(modp\left(x\right)\right)$that if and only if $\mathit{f}\left(x\right)$and $\mathit{g}\left(x\right)$leave the same remainder when divided by $\mathit{p}\left(x\right)$ .

Short answer

It is proved$\mathit{f}\left(x\right)\mathbf{\equiv }\mathit{g}\left(x\right)\left(modp\left(x\right)\right)$

Step-by-step solution

Write the theorem

Assume that F is the field,$a\left(x\right),b\left(x\right),c\left(x\right)$belongs to the field,$a\left(x\right)dividesc\left(x\right)$, if$a\left(x\right)/b\left(x\right)c\left(x\right)anda\left(x\right)b\left(x\right)$are relatively prime.

Proof

Assume that $q\left(x\right)$and $r\left(x\right)$are quotient and remainder, respectively, when $f\left(x\right)$is divided by $p\left(x\right)$. Also, $q\text{'}\left(x\right)$and $r\text{'}\left(x\right)$are quotient and remainder when $g\left(x\right)$is divided by $p\left(x\right)$.

Then, by division algorithm, we have:

$$\begin{gathered} f\left(x\right)=q\left(x\right)p\left(x\right)+r\left(x\right) \\ and, \\ g\left(x\right)=q\text{'}\left(x\right)p\left(x\right)+r\text{'}\left(x\right) \end{gathered}$$

Here, both the remainders being 0 or having a degree less than the degree of $p\left(x\right)$.

$ForF\left(x\right)\equiv g\left(x\right)\left(modp\left(x\right)\right)considerthedefinitionp\left(x\right)/\left(f\left(x\right)-g\left(x\right)\right),whichimplies$

$p\left(x\right)/\left(q\left(x\right)-q\text{'}\left(x\right)\right)p\left(x\right)+\left(r\left(x\right)-r\text{'}\left(x\right)\right)\dots \left(1\right)$

But, $\left(r\left(x\right)-r\text{'}\left(x\right)\right)$is either 0 or less than the degree of ; then we have

$$\begin{gathered} \left(r\left(x\right)-r\text{'}\left(x\right)\right)=0 \\ \Rightarrow r\left(x\right)=r\text{'}\left(x\right) \end{gathered}$$