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In my maths lesson today we were simplifying fractions by factorising. One question was something like this: $\frac{x-2}{3x-6}$, which I simplified as $\frac{x-2}{3x-6}=\frac{x-2}{3(x-2)}=\frac{1}{3}$. It got me wondering however, whether these expressions are really equal, specifically in the case $x=2$, where the former expression is undefined but the latter takes the value $\frac{1}{3}$.

Since the expressions only differ at a single point are they for all intents and purposes equal, or are they theoretically different? If I wanted to be entirely correct would I have to write $\frac{x-2}{3x-6}=\frac{1}{3}$ where $x \neq 2$?

My maths teacher explained that at $x=2$ the expression evaluates to $\frac{0}{3 \times0}$ and the zeros effectively cancel out. I wasn't altogether satisfied with this explanation because as far as I know $\frac{0}{0}$ is undefined.

Thanks in advance!

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    $\begingroup$ Your teacher is wrong.You are right. The expression is undefined if $x=2$. $\endgroup$ – MathematicsStudent1122 Jun 16 '16 at 18:02
  • $\begingroup$ Maybe you could provide a link? @Ahmedhussein $\endgroup$ – Olba12 Jun 16 '16 at 18:04
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    $\begingroup$ For related discussion and a few subtleties, see Is $\frac{x^2+x}{x+1}$ a polynomial?. $\endgroup$ – Henning Makholm Jun 16 '16 at 18:04
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    $\begingroup$ @Olba12 math.stackexchange.com/questions/477765/… $\endgroup$ – user258700 Jun 16 '16 at 18:07
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    $\begingroup$ @acernine this is technically what is referred to as a removable singularity. $\frac{1}{3}$ in this case is the unique function which matches the one given which is analytic. $\endgroup$ – JMoravitz Jun 16 '16 at 18:15
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I think it may help to frame it as this:

One one hand, $\frac{x-2}{3x-6}$ is a function that is defined on $\mathbb{R}$ except at $x=2$, and at every point it is defined it is equal to $\frac{1}{3}$.

On the other hand, $\frac{1}{3}$ can be seen as a function that must take in ANY input and give you $\frac{1}{3}$ back. In particular, the domain restriction at $x=2$ prevents $\frac{x-2}{3x-6}$ to be the same thing as $\frac{1}{3}$.

Thus, It would not be correct to simplify $\frac{x-2}{3x-6}$ to $\frac{1}{3}$ unless we are working in a domain where $x \neq 2$.

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  • $\begingroup$ Thanks for an excellent answer, as well as to all the people who commented. I'll have a read about removable singularities now! $\endgroup$ – acernine Jun 16 '16 at 18:24
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Yes... and no.

Math notation is ambiguous; it is impractical to fully spell out all of the fine details of what one means. We've had centuries of wisdom in developing notations where the ambiguities usually don't matter, but they often do in the fine detail.

There are a few different things one might mean by $\frac{x-2}{3x-6}$; the most significant disagreement between the alternatives is the status of "evaluation at $x=2$".

The most basic interpretation is that $\frac{x-2}{3x-6}$ is a recipe for performing a sequence of arithmetic operations upon a given input; in this interpretation, evaluation at $x=2$ is, in fact, undefined.

Many other interpretations can be described as "take the continuous extension": roughly speaking, you take the graph of the function and fill in all of the holes; here you would add in $(2,1/3)$. Also, if you were using the extended real numbers you would add in $(+\infty, 1/3)$ and $(-\infty, 1/3)$, so that evaluation at $\pm \infty$ is defined. (similarly if you were using the projective real numbers)

Your teachers description is nonsense when taken literally; however, the likely intention is that he is using "$0$" as a stand-in for some sort of 'witness' of vanishing; e.g. we factor out $(x-2)$ from both the numerator and denominator to get

$$ \frac{x-2}{3x-6} = \frac{1}{3} \cdot \frac{x-2}{x-2} = \frac{1}{3} \cdot 1$$

If we use one of these "continuous extension" interpretations, the witnesses do 'cancel' to leave behind $1/3$.

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