*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

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

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## fred terracina⚡

Here is a probability question encountered in pubs.

The pub has a game that requires players to ante up one dollar (1£ in GB).

The game begins with a cup that has six dies inside it. The cup has a predetermined number x on the outside where x is any integer from 1-6 So the predetermined integer is either 1 or 2 or 3 or 4 or 5 or 6. Each die has six faces and on each face is either of any of the integers 1-6 or 1-6 dots representing those six integers. So the player has three rolls of the dies to get a total of Six of the predetermined integer written on the cup. Before rolling the cup with the dies is shaken angd then the dies are splayed onto a flat surface. In each roll, if the predetermined integer appears it is removed from the cup. So that the next roll has fewer dies. For instance assume two x’s appear in the first roll. Then two dies are removed and the second roll is done with four dies. If in the second roll three x’s appear then they are removed. So the third roll would be with one die. So getting six dies with the predetermined integer in three rolls is the the object. The winner gets the pot. Another sometimes added scenario is for the house to give a free pint to any player who gets no appearances of the predetermined integer. So what are the odds ie probability of getting eithe six of the predetermined integer in three rolls given the rules? What are the odds of getting no dies exhibiting the predetermined integer in three rolls? I have solved this using traditional maths and using a computer. But then again I was more fxacile at maths then.

If you like I will send you my solution.

## fred terracina⚡

Colin The comment I submitted has nothing to do with fractions. Perhaps with time I will learn how to submit any comments in a more appropriate format.