I am tasked with the following:
Proposition (Euclidean Division of Polynomials): Let $A,B \in \mathbb{R}[X]$ be nonzero. There exists a unique pair $(Q,R) \in \mathbb{R}[X]^2$ such that:
$$A=BQ+R,\;\;\;\;\;\;\;\;\;\deg R<\deg B.$$
I actually have a proof but I need someone to check it for me.
$\mathbf{Proof\;of\;the\;Uniqueness:}$
Let $A(x)$ and $B(x) \in \mathbb{R}$ be nonzero, let $(Q_1,R_1), (Q_2,R_2)$ such that $A=Q_1B+R_1=Q_2B+R_2$. we will prove this is not possible.$$$$ first we see that $B(Q_1-Q_2)=R_2-R_1$ $$$$We have $\deg(R_2-R_1)<\deg B$, because $\deg R_2<\deg B$ and $\deg R_1<\deg B$ so $\deg R_2-R_1=\max(\deg R_1,\deg R_2)<\deg B$.
$$$$but $B(Q_1-Q_2)=R_2-R_1$, then $\deg B(Q_1-Q_2)<\deg B$ which is false, except if $Q_1-Q_2=0$. But if $Q_1-Q_2=0$, then $R_2-R_1=0$.
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square$$$$$
$\mathbf{Proof\;of\;the\;Existence:}$
We will prove this by induction on the degree of $A$.
Let $P(X)$ be the statement "$A=BQ+R:\deg A=x$ and $\deg R<\deg B$'
$$$$$P(0)$ that is when $\deg A=0$, we have $\\$ $A=0.B+R$ where $R=A$ ,$\deg R=0<\deg B$ $$$$Assume P is true for $A$ with degree less than $N>0$, then we have $A=BQ+R, \deg R<\deg B,\;\deg A<N$ $$$$Consider $a$ such that $\deg A=N+1$ $\\$ we have the monomial $a_{N+1}X^{N+1}$, but there exists a $P \in \mathbb{R}[X]$ such that $\deg P=h$ and has the monomial $p_hX^h$, such that for the monomial $b_kX^k$ in $B$, $(p_hX^h)(b_kX^k)=p_hb_kX^{h+k}=a_{N+1}X^{N+1}$
$$$$Then we have $A-BP=a_{N+1}X^{N+1}+…+a_0$$\;$–$[p_hb_kX^{h+k}+…+b_0(p_hX^h+…+p_0)]$
(I have skipeed some steps because this is tiring), but $p_hb_kX^{h+k}=a_{N+1}X^{N+1}$, therefore $A-BP=a_NX^N+…+a_0+…+b_0(p_hX^h+…+P_0)$ $$$$We see that $\deg(A-BP)\le N$, Hence the division exists for $A-BP$. $\\$We have $A-BP=BP+(A-2BP)$ and $\deg(A-2BP)<\deg B$, but $\deg(A-2BP)=\deg(A-BP)<\deg B$
$$$$ Thus, we have $\\$ $A=BP+(A-BP)$ with $\deg A=N+1$ and $\deg(A-BP)<\deg B$.
We have proved $P(0)$ and if $P(N)$ is true then $P(N+1)$ is true, therefore by the principle of induction, $P$ is true for all $N \in \mathbb{N}$.$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ $\square$
$$$$Phew, writing that was harder than the proof. I have skipped some details because this is getting tiring and long, but please let me know what you think.
Best Answer
Minor Problems:
Since you are trying to prove the $n+1^{\text{th}}$ case, your inductive hypothesis should read "Assume $P$ is true for all $A$ with degree less than or equal to $n$" rather than just assuming it for degree less than $n$.
In your paragraph after the inductive hypothesis, you have not dealt with the case where the degree of $B$ is larger than the degree of $A$. In this case, it would not be possible to find such a $P$ that you have specified. This is easily rectified by observing that $A = B(0)+A$ since $\text{deg}(A)<\text{deg}(B)$.
Slightly Bigger Problem:
You have not shown $\text{deg}(A-2BP) < \text{deg}(B)$. In fact, this statement is not true in general. What you should do instead is to use your inductive hypothesis on $A-BP$ since $\text{deg}(A-BP) \leq n$. This will give you that $$A-BP = BQ+R$$ for some $Q,R \in \mathbb{R}[x]$ where $\text{deg}(R) < \text{deg}(B)$. Adding $BP$ to both sides will get what you want.