# “If $f$ is a linear function of the form $f(x)=mx+b$, then $f(u+v)=f(u)+f(v).$” True or false?

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.

• Well, why bother plugging in values? You can see that $f(u)+f(v) = mu +mv + 2b$. Now you can actually find your answer if you set this equal to $f(u+v)$: namely, you'll find that $b$ must equal zero in order for equality to hold, and you're done. However, you are correct. All you need is to exhibit a single counterexample to show that the statement is false, so your method is completely correct. – Rellek Aug 13 '16 at 23:47
• @Rellek Sorry. I only read your comment after typing my answer. Should you decide to post it as an answer yourself and want me to delete mine, just let me know. – Stefan Mesken Aug 13 '16 at 23:52
• @Stefan Don't worry about it! I will keep it as a comment. – Rellek Aug 13 '16 at 23:58
• @C.Guan If you have a statement that asserts that a condition holds for ALL possible situations, then it is sufficient to provide a single counterexample to prove the statement false. However, if you are trying to prove that a statement is false in general, then a single counterexample is NOT sufficient. – Rellek Aug 13 '16 at 23:59
• This is basically the difference between an affine function and a linear function. – bjb568 Aug 14 '16 at 1:41

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 $u,v$.