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Quick question about validity, just to make sure.

When I have a fraction in a form: $$\frac{3a + 3b}{a+b}$$ and I extract the common factor 3 out to get: $$\frac{3(a+b)}{a+b} \;=\; 3\frac{a+b}{a+b}$$ is it valid now to reduce the fraction to 1 to end up with just 3?

Could it change anything if $b = -a$ ? Because then I would have: $$3\frac{a-a}{a-a} \;=\; 3 \frac{0}{0}$$ and mathematicians say that $0/0$ is an indeterminate symbol. Does it make the whole product indeterminate too?

And what if the common factor is not 3 but some unknown symbol, say $c$?

What can we say about the divisibility of the numerator by $a+b$ in that particular situation? Is it safe to say that the numerator is divisible by $a+b$ even if $a+b$ can be 0 sometimes? Or should I first enforce a condition on $a+b$ that it is never 0?

Edit: I know about the general case when I have $a+b$ in the denominator, that then it cannot be 0. But I wonder if it changes anything if $a+b$ is also a factor of the numerator. Because then, we have somewhat different situation: division by 0 is undefined, but $0/0$ is just undetermined, that is, it can be anything (that's what mathematicians usually say to me).

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3 Answers 3

up vote 2 down vote accepted

The original fraction $\dfrac{3a+3b}{a+b}$ is defined if and only if $a+b\ne 0$. When that’s the case, then you have

$$\frac{3a+3b}{a+b}=\frac{3a+3b}{a+b}\cdot1=\frac{3a+3b}{a+b}\cdot\frac{\frac1{a+b}}{\frac1{a+b}}=\frac31=3\;.$$

What you can say, then, is that if $\dfrac{3a+3b}{a+b}$ is defined in the first place, it’s equal to $3$.

If, for instance, you have a function $f$ of two variables defined by

$$f(a,b)=\frac{3a+3b}{a+b}\;,$$

this function is equal to the constant function $g(a,b)=3$ wherever it is defined, but while $g$ is defined on the whole plane, $f$ is defined only where $a+b\ne 0$, i.e., where $a\ne -b$.

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You've made a very important point here, and that's what I waited for. $g(a,b) = 3$ is defined everywhere. But $f(a,b) = \frac{3a+3b}{a+b}$ is defined everywhere except the line $a=-b$. So we can put the equal sign between these two functions only if we enforce additional condition upon $g(a,b)$. In other words, $g(a,b)$ is something more than $f(a,b)$, so we have to limit $g(a,b)$ to make them both equal. Is my stream of thought correct? –  SasQ Nov 8 '13 at 19:12
    
@SasQ: Yes, that’s exactly right. –  Brian M. Scott Nov 8 '13 at 19:15

The latter of your statements is correct. You should first enforce a condition on $a+b$ so that it is never $0$. In general, keep track of domains from the start, and then this factoring is legitimately $3$. If $b=-a$, then the fraction is not even defined from the start.

Regarding your edit. Consider assigning a domain to the fraction. For example, we are considering the fraction to be a member of the real numbers. In terms of real numbers, $\frac{0}{0}$ has no more meaning that $\frac{1}{0}$ as neither are actually real numbers. So for practical purposes, it changes nothing.

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Thanks for your answer. I added an update to my question to address that in somewhat more detail. –  SasQ Nov 8 '13 at 18:37

If $a+b$ is on the downside of the fraction, so it's need to be $\neq 0 $, else the fraction is meaningless. And if have a $c$ factor you can do the same what you done with the $3$.

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