# Ask Uncle Colin; A variable fraction

Dear Uncle Colin,

I have a fraction, $\frac{x^2-x}{x-1}$, and I want to cancel it down to $x$ - but I’m not sure those are the same. Are they?

- Got A Lot Of Interesting Sums

Hi, GALOIS, and thanks for your message!

The short answer is, yes and no.

Everywhere *except* $x=1$, the two things are the same – if you factorise the top, you get $x(x-1)$, and $\frac{x(x-1)}{x-1}$ is clearly $x$ as long as the bottom isn’t zero.

However, when $x=1$, we have a problem: the function evaluates to $\frac{1 \times 0}{0}$, which is not defined. We’re not **physicists**; we take great care not to divide by zero.

It’s worth pointing out that as $x$ gets closer to 1, your fraction also gets closer to one - and *in the limit* as $x$ goes to 1, the fraction becomes equal to 1; unfortunately, “goes to in the limit” isn’t quite the same thing as “is equal to” and treating them as the same can lead to some fairly nasty paradoxes and errors.

In summary: so long as you know $x$ isn’t 1, the fraction is equal to $x$. If $x$ happens to be 1, the fraction is undefined.

Hope that helps!

-- Uncle Colin