# If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$?

If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$. True or False.

Verification of work: I found a similar problem where $f(x)=y-mx+b$ and test values for $m$ and $b$ were used. $f(x)=y-mx+b$ My problem has $m$ and $x$ as the values so I worked it as such and came to the conclusion that the question is True.

Give $m$ and $x$ the values of $3$ and $1$, respectively. So that, $f(x)=mx$ becomes $f(x)=(3)(1)$.

Give $a$ and $b$ the values of $2$ and $4$. Now we have:

$$f(2)+f(4)=(3)(2)+(3)(4)=18$$

$$f(2+4)=f(6)=(3)(6)=18$$

$18=18,$ so that $f(x) =mx$, then $f(a + b) = f(a) + f(b)$

• Okay, so it works with those numbers, but in order to say True you have to be certain that it works for all numbers, even the ones you didn't try explicitly. Try a similar argument with variables instead of specific numbers. – Nate Eldredge Jun 4 '17 at 2:56
• You have to show it more general: $f(a)+f(b)=ma+mb=m(a+b)=f(a+b)$ with $f(x)=mx$ – callculus Jun 4 '17 at 2:58
• This question doesn't deserve the hate. It's a prime opportunity to teach somebody to do the general case, not just one case. – Jacob Claassen Jun 4 '17 at 3:01

Your solution shows that $f(a+b)=f(a)+f(b)$ for the specific case where $a=2,b=4$, and $m=3$. To prove the statement in general, you need to show that it is true for all values of $a$ and $b$, no matter what $m$ is. My hint for you would be to repeat your steps above, but instead of 2,3 and 4, simply leave them as $a,b$ and $m$, and treat them like they were concrete numbers.

I encourage you to do this yourself, but if you are still confused the proper argument would go something like this:

$$f(a+b)=m(a+b)=ma+mb=f(a)+f(b)$$

• I see we had the same thought lol – Jacob Claassen Jun 4 '17 at 2:58
• Funny how that works! – Shaun_the_Post Jun 4 '17 at 2:59
• @Shaun_the_Post If you want to hide it so that OP doesn't see that equation immediately (to better encourage him/her to do it him/herself), then you can add >! to the beginning of that line -- not that it'll really help in this case since there are already 2 other answers with that same work -- but just something to keep in mind in case you weren't aware how to hide text. :) – user137731 Jun 4 '17 at 3:01
• Thanks, that's really helpful to know. – Shaun_the_Post Jun 4 '17 at 3:02
• @Carefulle where did the x value go There is no $x$ value. Think of $\,x\,$ as a "placeholder" (or "free variable") in the definition of the function $\,f(\color{red}{x})=m\color{red}{x}\,$. When you write $\,f(\color{red}{a+b})\,$ you just replace $\,x\,$ with $\,a+b\,$, so you get $\,f(\color{red}{a+b})=m(\color{red}{a+b})\,$ – dxiv Jun 4 '17 at 5:05

$$f(x)=mx$$ $$f(a+b)=m\cdot(a+b)$$ $$f(a+b)=ma+mb$$ $$f(a)=ma,f(b)=mb$$ $$\therefore f(a+b)=f(a)+f(b)$$

So if $f(x) = mx$, $f(a+b) = m(a+b) = ma + mb = f(a) + f(b)$