Expanding a Proof of Induction on $\Bbb N $ to $\Bbb Q $ (Linear Algebra)

Question

My problem is the following:

I have an $\Bbb R$ Vectorspace called $V$ and had to show via induction that $\langle nv, w \rangle=n \langle v, w \rangle$ for $v,w \in V$ and $ n\in \Bbb N$. (it's not already given that $\langle \rangle$ is an inner product.) I've already shown this, but now i have to expand the prove for $n\in \Bbb Q$ and after that, for $n \in \Bbb R$. I would do the last step with the knowledge that Q is dense in R and just use a series in Q that converges to the $n\in \Bbb R$. My problem is the second step. How can i get from $n\in \Bbb N$ to $n\in \Bbb Q$. Does someone have an idea?

Thanks in advance

Edit: The definition for $\langle \rangle$ ist given via $$ \langle v,w\rangle := \frac 1 2(||v+w||^2-||v||^2-||w||^2)$$ where $||. ||$ is a norm on $V$

Answer 1

First, extend the statement from the natural numbers to the integers by negation, simple enough. Now use the statement for integers to prove it directly for rationals. No further induction is necessary.

Proof: Let $v, w$ be vectors in $V$, let $p/q$ be a rational number. Then

$\langle{p\over q}v, w\rangle = {q\over q}\langle{p\over q}v, w\rangle = {1\over q}q\langle p({1\over q}v),w\rangle = {1\over q}qp\langle {1\over q}v,w\rangle = {1\over q}p\langle q({1\over q}v),w\rangle = {p\over q}\langle v,w\rangle$.

The third equality follows from the integer case for $n = p$, the fourth follows from the integer case for $n = q$.

Answer 2

If $v'=\frac pq v$, then $qv'=pv$ and therefore $$q \langle v',w\rangle = p\langle v,w\rangle $$ This shows your goal for positive rational multipliers.

You need negative ones too, but it will be enough if you can can show $\langle -v,w\rangle = -\langle v,w\rangle$.