When dividing polynomials, the remainder is uniquely defined.

Question

How can we prove that when dividing a polynomial $P(x)$ by polynomial $Q(x)$, the remainder is uniquely defined. Let :

$P(x) = S_1(x)Q(x)+R_1(x)$

$P(x) = S_2(x)Q(x)+R_2(x)$

$\deg(R_1(x)),\deg(R_2(x))<\deg(Q(x))$

Prove that $S_1(x)=S_2(x),R_1(x)=R_2(x)$

I tried playing with equation and using the euclidean algorithm and using degrees of functions but did not work.

Can you give me an idea for the solution where should I start?

Answer 1

Quoting from the document links supplied in comment,

If $a$ and $b$ are integers, with $a>0$, there exist unique integers $q$ and $r$ such that $b=qa+r$ $0≤r<a$ The integers $q$ and $r$ are called the quotient and remainder, respectively, of the division of $b$ by $a$.

As for your question, assume $q$ and $r$ to be functions of $x$.

To show that these values for q and r are unique.

Effectively, we need to show that if $b=q_1a+r_1$ and $b=q_2a+r_2$, then it can only be the case that $q_1=q_2$ and $r_1=r_2$.

Without loss of generality, let us suppose that $r_2≥r_1$. Substituting for b, we have $$q_1a+r_1=q_2a+r_2$$ But then $q_1a−q_2a=r_2−r_1$ and thus,$a(q1−q2)$ Consequently, $r_2−r_1$ is some multiple of a.

However, since $0≤r_1,r_2<a$ and $r_2>r_1$, it must be the case that $0≤r_2−r_1<a$. The only multiple of $a$ in this range is zero, so $r_2−r_1=0$, or equivalently, $r_1=r_2$.

Further, as seen before, $q_2=b−r_2a$ and $q_1=b−r_1a$ Since $r_1$ and $r_2$ agree in value, this forces $q_1$ and $q_2$ to agree in value as well.

Thus, having shown that if $b=q_1a+r_1$ and $b=q_2a+r_2$, then it can only be the case that both $q_1=q_2$ and $r_1=r_2$, we can conclude that if there exist integers q and r such that $b=qa+r$ with $0≤r<a$, then those values of q and r are unique.

Since the first part of our argument established the existence of these integers q and r, we are done. It is indeed the case that given integers a and b, with a>0, there exist unique integers q and r such that $b=qa+r$ and $0≤r<a$.

Answer 2

$0 = P(x)-P(x) = (S_1(x)-S_2(x))Q(x)+(R_1(x)-R_2(x)) = Z(x)$

If $S_1 \neq S_2$, $deg(Z(x)) \ge deg(Q(x))$. If $deg(Q(x)) > 0$, we have contradiction (for $deg(Q(x)) = 0$ we have $R_i(x)=0, S_i(x)=P(x)/Q$).

For $S_1 = S_2$, first line becomes $0 = R_1(x) - R_2(x)$ which completes the proof.

Answer 3

You start well. Suppose $$ P(x)=S_1(x)Q(x)+R_1(x)=S_2(x)Q(x)+R_2(x) $$ with $\deg R_1<\deg P$ and $\deg R_2<\deg P$. Then $$ (S_1-S_2)Q=R_2-R_1 $$ so $\deg (S_1-S_2)Q\le\max\{\deg R_1,\deg R_2\}<\deg Q$, if $(S_1-S_2)Q\ne0$.

Assuming the coefficients are in a field (such as the reals) or a domain, the degree formula implies that, if $S_1\ne S_2$, $$ \deg (S_1-S_2)Q\ge \deg Q $$ which would lead to a contradiction. The degree formula is: if $A$ and $B$ are nonzero polynomials, then $$ \deg (AB)=\deg A+\deg B $$