Proof that the Arf invariant is independent of choice of basis

I'm confused about the proof of the following claim:

Set $Z_2 = \mathbb{Z}/2\mathbb{Z} = \mathbb{F}_2$. Let $V$ be a $Z_2$-vector space of dimension $2n$ and let $e_i, f_i$ be a symplectic basis. Let $q: V \to Z_2$ be a non-degenerate quadratic form. Define $c(q) = \sum_i q(e_i) q(f_i)$. Then $c$ is independent of the choice of basis for $V$.

We can show this by showing that any $q(v) = q(x_1 e_1 + y_1 f_1 + \dots + x_n e_n + y_n f_n)$ can be written in the form $x_1 y_1 + \dots x_n y_n + c(q) (x_n^2 + y_n^2)$.

Here's an outline of the proof(as given for example here on page 318 or here p. 98):

First, we show it for $\mathrm{dim}(V) = 2$. In dimension $2$ there are only two non-equivalent non-degenerate quadratic forms, $q_0, q_1$ and they are such that $q_0(v) = x_1 y_1$ and $q_1 (v) = x_1 y_1 + x_1^2 + y_1^2$ and $c(q_0)= 0$ and $c(q_1) = 1$.

Next we note that $q_0 + q_0$ and $q_1 + q_1$ are equivalent quadratic forms.

Then if $q$ is any quadratic form on $V$ with $\mathrm{dim}(V) = 2n$ we can either write $q = q_0 + \dots + q_0 = n q_0$ or $q = (n-1)q_0 + q_1$. The respective forms written in coordinates are $x_1 y_1 + \dots + x_n y_n$ and $x_1 y_1 + \dots + x_n y_n + x_n^2 + y_n^2$ and we have $c(nq_0 ) = 0$ and $c((n-1)q_0 + q_1) = 1$.

To conclude the proof one needs to show that $n q_0$ and $(n-1)q_0 + q_1$ are not equivalent.

Why is this last step necessary? At that time we already have that we can write any $q$ in a coordinate independent way and that there are only two non-equivalent non-degenerate quadratic forms on any symplectic vector space $V$. Thanks for your help.

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I went to wikipedia and found the sentence "William Browder has called the Arf invariant the democratic invariant[4] because it is the value which is assumed most often by the quadratic form." So one could show that the two quadratic forms are not isomorphic by counting how often they take the value $0$ and $1$. – HenrikRueping Aug 30 '12 at 19:35
@HenrikRueping Yes, that's what the sentence "To conclude the proof one needs to show that nq0 and (n−1)q0+q1 are not equivalent." is saying. But my question is: why do we want to show that they are not equivalent? – Rudy the Reindeer Aug 30 '12 at 20:29
But my question is a basic logic / proof question and has not much to do with the actual statement, I think. – Rudy the Reindeer Aug 30 '12 at 20:35
Hmm. Is it not more accurate to say (before that last step): At that time ... there are at most two non-equivalent non-degenerate quadratic forms... In other words, the Arf invariant would not be an invariant, if it turned out that $nq_0$ and $(n-1)q_0+q_1$ are equivalent? – Jyrki Lahtonen Aug 30 '12 at 20:48
@Matt: I think we already know that $c(nq_0)\neq c((n-1)q_0+q_1)$, so it is not an invariant, if those two forms are equivalent. – Jyrki Lahtonen Aug 31 '12 at 11:59