The rest of the question states the following: "If true, give an explanation as to why. If false, give a counter example."
Here is the following statement:
If $f$ is a linear function of the form $f(x)=mx+b$, then $f(u+v)=f(u)+f(v).$
I know that this statement is false because I arbitrarily chose a function of the form $f(x)=mx+b$ and I plugged in random numbers for $u$ and $v$.
My actual question is this: is it sufficient enough to just use variables for counter-examples or must actual numbers be used?
Here is my work:
False. Counter-example:
Give $m$ and $b$ the values of $1/2$ and $3$, respectively. Therefore, $f(x)=mx+b$ becomes $f(x)=x/2+3$.
Give $u$ and $v$ the values of $2$ and $4$, respectively. Now we have:
$f(2)+f(4)=(1/2)(2)+3+(1/2)(4)+3=9.$
$f(2+4)=f(6)=(1/2)(6)+3=6.$
$9\neq6,$ so therefore, this statement is false.
Yes, your counterexample suffices. However, there is an easier one: Let $b \neq 0$. Then $$ f(0 + 0 ) = b \neq 2b = f(0) + f(0), $$
hence $f(u+v) = f(u) + f(v)$ does not hold for arbitrary $u,v$. On the other hand, if $b = 0$, it's easy to see that $f(u+v) = f(u) + f(v)$ for all $u,v$.