Proving this additive function is continuous.

[daon]

I needed to prove that a function f:R^m->Rⁿ with the property that f(x+y)=f(x)+f(y), for all x,y in R^m has the following properties:

1) for all rationals q, f(q) = a*q where a=f(1)
2) if f is continuous and r a real number, then f(r) = a*r.

This was relatively easy for me to do with the integers, but I'm clueless on the rationals and reals. For the integers I could write f(z) as f(1)+f(1)+...+f(1) z times, and for negative integers it was easy to show f(-x)=-f(x) for all reals x so that f(-z) = -f(z) = -z*f(1).

Any ideas?
Thanks

edit: Also, in a related question, I am to show that if f is monotone then (f is additive[i.e. f(x+y) = f(x)+f(y)] implies f must be continuous.)I figure this is the same as "If f is monotone and additive,then f is continuous."

I think I am close on this but I can't figure out where the additive part comes in. I am attempting this by contra-positive, but if this is the wrong path please redirect me. I want to show that: "If f is not continuous, then f is not additive or f is not monotone." I am assuming it is continuous and additive and trying to show it must not be monotone.

 

[pka]

All that is need is a sleight of hand.

If \(\displaystyle \alpha \in Q\), a rational number then \(\displaystyle \alpha = \frac{m}{n}\) the ratio of two integers. Thus \(\displaystyle n\alpha = m\) and

\(\displaystyle \begin{array}{l}
f(m) = f(n\alpha ) = f\left( {\underbrace {\alpha + \alpha \cdots \alpha }_n} \right) = nf(\alpha ) \\
nf(\alpha ) = f(m) = mf(1) \\
f(\alpha ) = \frac{m}{n}f(1) = \alpha f(1) \\
\end{array}.\)

For #2, note that any real is the limit of a sequence of rational numbers.
Now one characterization of a continuous function is in terms of preserving converging sequences.