Show that $f(x)=x$ for every $x$ 
I've done part a and part b. Need help with the last two parts. 
For part c, I know the value of $f(1/2), f(1/4)...$ but I don't know how to prove it for all rational numbers.
And for part d, can I say "suppose $g(x)=f(x)-x$ and $g(x)$ is continuous." But then I don't know how to proceed from that. Thanks!
 A: For part c): Let $\frac{p}{q} \in \mathbb Q$ with $p,q \in \mathbb N$. 
Note that according to b) $f(1) = qf(\frac{1}{q})$ and that, once you know $f(\frac{1}{q})$, you can write the following (again using b): $$f(\frac{p}{q})=pf(\frac{1}{q})$$
Finally, to take negative rationals into account, observe that by a) $0=f(0)=f(q - q)=f(q) + f(-q)$, i.e. $f(-q)=-f(q)$ for every $q \in \mathbb Q$.
For part d): The trick here is that for each real number $r$, there exists a sequence of rationals $(q_n)$ such that $\lim_{n\rightarrow \infty}q_n = r$.
By definition of continuity, what's $\lim_{n\rightarrow \infty}f(x_n)$ when $\lim_{\rightarrow \infty}x_n = x$?
A: c) Let $r = n/m \in \mathbb Q+$ so that $n, m \in \mathbb Z; \gcd(m,n) = 1;m >0; n > 0$.
$f(n/m) = f(n/m) = f(1/m + .... + 1/m) = f(1/m)+....+f(1/m) = nf(1/m)$
Note: $m*f(1/m) = f(1/m) + ... + f(/1m) = f(m*1/m) = f(1)$ so $f(1/m) = f(1)/m$.
So $f(n/m) = f(1)*n/m$.
That's all positive rationals.
$f(0) = f(0 + 0) = f(0) + f(0) \implies f(0) = 0$.
$ f(r) + f(-r) = f(r +(-r) ) = f(0) = 0$ for $f(-r) = -f(r)$.
So for any $q \le 0, f(q) = -f(-q) = -(-q)f(1) = q  f(1)$.
So that's all rationals.  That's c).
d) Let {$a_n$}$\rightarrow x$ be a sequence of rational numbers that converge to real $x$.   As $f$ is continuous {$f(a_n)$}={$a_n f(1)$}$\rightarrow f(x)$.  But {$a_n f(1)$}$\rightarrow xf(1)$.  So $f(x) = xf(1)$.
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A function where $f(x+y) = f(x) + f(y)$ is called a Cauchy function.  An interesting thing is $f(x) = xf(1)$ for all real $x$ if one of the following conditions hold.
a) $f$ is continuous at one point (just one).
b) $f$ is bounded on an interval $(a, b)$ (any interval).
c) $f(x)$ is positive for all $0<x<\epsilon$ for some $\epsilon$.  
If any of the conditions are true, then $f(x) = xf(1)$.
If $f$ is cauchy but $f(x) \ne xf(1)$ for some real $x$ then.
0) $x$ is irrational
1) $f(x)$ is discontinuous everywhere.
2) $f(x)$ is unbounded on every interval.
and 
3) $f(x)$ will have positive and negative values in every interval.
