Dear Uncle Colin,

I’m ok with my basic power laws, but I don’t understand why $x^0$ is always 1, and I get mixed up when it’s a fraction or a negative power. Can you help?

Running Out Of Time

Hi, ROOT, and thanks for your message!

If it’s any consolation, you’re not alone – this is something that I struggled with when I was learning, and it’s a really common problem with my GCSE and A-level students.

### First and biggest bit of advice

My first and biggest bit of advice, when you see a power you’re unfamiliar with: slow down and think. If I had a list of Important Mathematical Skills, somewhere near the top would be “recognising where something is tricky” or “knowing your weaknesses”. This doesn’t just apply to awkward powers – every mathematician has different strengths and weaknesses – but the advice is good every time. If you think “this is something I’ve messed up frequently in the past, I’d better take extra care”, you will save yourself a lot of headache in the end.

### Descending powers

Right, let’s get on to zero and negative powers. While you’re learning these, I recommend writing out a list of powers to remind yourself of the pattern, which turns out to be completely logical – once you look at it in exactly the right way.

Let’s work with 9 as our base – although there’s nothing special about 9 ((Well, I say that. There’s plenty that’s special about 9, but none of its special things are relevant to this discussion.)), you could pick any number of variable you wanted here.

Start by writing $9^3 = 9 \times 9 \times 9 = 729$ on one line.

On the next, write $9^2 = 9 \times 9 = 81$ and notice that you’ve divided by the base to drop the power by 1.

On the next, write $9^1 = 9$ – and again, you’ve divided by the base to drop the power.

So, logically, what must $9^0$ be? Well, you divide by the base again and get $9^0=1$ on the next line.

And how about $9^{-1}$? Guess what, you divide by the base again. $9^{-1} = \frac{1}{9}$, and $9^{-2} = \frac{1}{81}$, and so on.

You can also see from here, using the power laws you’re happy with, that $9^{-2} \times 9^{2} = 9^0$, which is exactly the same as $\frac{1}{81} \times 81 = 1$.

Fractions are a bit harder to see, but they do still make sense: it’s pretty clear from your list of powers that $9^{\frac{1}{2}}$ is between 1 and 9, but where exactly? Again, your power laws can help here: you know that $9^{\frac{1}{2}} \times 9^{\frac{1}{2}} = 9^1$ because $\frac{1}{2} + \frac{1}{2} = 1$.
This means that $9^{\frac{1}{2}}$ is the number you multiply by itself to get 9 - or the square root of 9, which is 3.