1
$\begingroup$

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 $f(a + b) = f(a) + f(b)$

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jun 4 '17 at 2:56
  • 2
    $\begingroup$ You have to show it more general: $f(a)+f(b)=ma+mb=m(a+b)=f(a+b)$ with $f(x)=mx$ $\endgroup$ – callculus Jun 4 '17 at 2:58
  • 4
    $\begingroup$ This question doesn't deserve the hate. It's a prime opportunity to teach somebody to do the general case, not just one case. $\endgroup$ – Jacob Claassen Jun 4 '17 at 3:01
2
$\begingroup$

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:

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

$\endgroup$
6
  • $\begingroup$ I see we had the same thought lol $\endgroup$ – Jacob Claassen Jun 4 '17 at 2:58
  • $\begingroup$ Funny how that works! $\endgroup$ – Shaun_the_Post Jun 4 '17 at 2:59
  • 2
    $\begingroup$ @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. :) $\endgroup$ – user137731 Jun 4 '17 at 3:01
  • $\begingroup$ Thanks, that's really helpful to know. $\endgroup$ – Shaun_the_Post Jun 4 '17 at 3:02
  • 2
    $\begingroup$ @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 $\,f(\color{red}{a+b})=m(\color{red}{a+b})\,$ $\endgroup$ – dxiv Jun 4 '17 at 5:05
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.