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

Linus Rastegar
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Carefulle
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    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 04 '17 at 02:56
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    You have to show it more general: $f(a)+f(b)=ma+mb=m(a+b)=f(a+b)$ with $f(x)=mx$ – callculus42 Jun 04 '17 at 02:58
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    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 04 '17 at 03:01

3 Answers3

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

Jacob Claassen
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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:

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

Shaun_the_Post
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  • I see we had the same thought lol – Jacob Claassen Jun 04 '17 at 02:58
  • Funny how that works! – Shaun_the_Post Jun 04 '17 at 02:59
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    @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. :) –  Jun 04 '17 at 03:01
  • Thanks, that's really helpful to know. – Shaun_the_Post Jun 04 '17 at 03:02
  • one last question, I see what you mean by using the general case approach- where did the x value go. Would this also be correct: f(a +b) = (a + b)x = xa +xb = f(a) + f(b). granted you distribute from L to R. – Carefulle Jun 04 '17 at 03:48
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    @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 04 '17 at 05:05
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So if $f(x) = mx$, $f(a+b) = m(a+b) = ma + mb = f(a) + f(b)$

Araske
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