Necessity of linear map conditions I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.
I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.
 A: You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $n\in \mathbb{N}$. However, for $c=\pi$ (say) it isn't obvious how one would show  that it follows that $f(\pi x)=\pi f(x)$. 
Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(\lambda x)=\lambda f(x)$ for all $\lambda \in \mathbb{R}$.
A: A problem with the proposed way of defining $f(cx)$,
$f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), \; \text{and so forth}, \tag 1$
is that, unless $c \in \Bbb N$, the recursive process won't terminate.  What happens is $c = \sqrt 2$ or $c = \pi$, for example?  Or even if $0 > c \in \Bbb Z$?  Of course here one may take
$f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), \tag 2$
since we have $f(-x) = -f(x)$ from
$f(x) + f(-x) = f(x + (-x)) = f(0) = 0. \tag 3$
But with $c \notin Z$, we will never arrive at a result for $f(cx)$.
If $c \in \Bbb Q$, one can make some progress in this direction via the observation that with
$c = \dfrac{p}{q}, \; p, q \in \Bbb Z, \tag 4$
we can write
$pf(x) = f(px) = f \left (q \dfrac{p}{q}x \right ) = qf \left (\dfrac{p}{q}x \right ), \tag 5$
whence
$f \left (\dfrac{p}{q}x \right ) = \dfrac{p}{q}f(x). \tag 6$ 
We can handle $f(cx) = cf(x)$, $c \in \Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $\Bbb R^m$ and $\Bbb R^n$, and such homomorphisms extend in a natural way to $\Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . . 
Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$.  Then if $c_n \to c$ with $c_n \in \Bbb Q$, we have
$c_n f(x) = f(c_n x) \to f(cx) \tag 7$
by the continuity of $f(x)$.
