# "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. Aug 13, 2016 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. Aug 13, 2016 at 23:52
• @Stefan Don't worry about it! I will keep it as a comment. Aug 13, 2016 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. Aug 13, 2016 at 23:59
• This is basically the difference between an affine function and a linear function. Aug 14, 2016 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 $u,v$.