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:



$9\neq6,$ so therefore, this statement is false.

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    $\begingroup$ 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. $\endgroup$ – Rellek Aug 13 '16 at 23:47
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    $\begingroup$ @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. $\endgroup$ – Stefan Mesken Aug 13 '16 at 23:52
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    $\begingroup$ @Stefan Don't worry about it! I will keep it as a comment. $\endgroup$ – Rellek Aug 13 '16 at 23:58
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    $\begingroup$ @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. $\endgroup$ – Rellek Aug 13 '16 at 23:59
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    $\begingroup$ This is basically the difference between an affine function and a linear function. $\endgroup$ – 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 $u,v$.


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