I'm an engineer, and while you get introduced to many concepts of mathematics, but only with a subset of the vocabulary, and none of the rigor and proofs. So while trying to read a mathematical book, the following exercise is posed:
Let $p \in \mathbb{P}_k$ with $k ≥ 1$. Assume that $p$ vanishes on the $\mathbb{R}^d$-hyperplane of equation $λ = 0$. Prove that there is $q \in \mathbb{P}_{k-1}$ such that $p = λq$.
I've attempted a prove by induction, but I have no idea if it is correct, and if it would be accepted by a mathematician as proof. So the question is, how do I get to the point that I feel comfortable with my rigor and notation, given the state of iqnorance I'm currently in?
My attempt at the proof goes like this: Base case: $k = 1$, $λ = \sum_{i=1}^d a_i x_i = 0$, $p = \sum_{i=1}^d b_i x_i + c$. For $p$ on $λ$, $p$ has to vanish: $$p(x_1 = -\frac{1}{a_1} \sum_{i=2}^d a_i x_i) = 0$$ $$\Leftrightarrow 0 = \sum_{i=2}^d (b_i - \frac{b_1}{a_1} a_i) x_i + c.$$ Because this has to hold for all $x_i$: $$c = 0,$$ $$b_i = \frac{b_1}{a_1} a_i,$$ which means $p = \frac{b_1}{a_1} \sum_{i=1}^d a_i x_i$. Inserting this into $p = λq$: $$\frac{b_1}{a_1} \sum_{i=1}^d a_i x_i = q \sum_{i=1}^d a_i x_i$$ $$\Leftrightarrow q = \frac{b_1}{a_1}.$$ This shows that the base case is true. It exists, and it is of degree zero.
Inductive step A polynomial of degree $(k+1)$ can be written as a product of two polynomials, one of degree $k$ and the other of degree 1. $$p_{k+1} = p_k p_1$$ Given the base case, and the induction hypothesis, we can write $$p_{k+1} = λ q_k λ q_1$$ $$\Rightarrow p_{k+1} = λ q_{k+1}, \text{ with } q_{k+1} = λ q_k q_1$$ $q_{k+1}$ exists, and is of order $k$, since it is a product of polynomials of degree one [$λ$], $(k-1)$ [$q_k$] and zero [$q_1$].
Q.E.D.