Written by Colin+ in basic maths skills.

"There's more than one way to do it."

- Perl programming motto

One of my few regrets about writing maths books is that I usually only get to show one way of doing things - and that gives the impression that I believe Method X is absolutely, no questions, the best method of doing things.

Let me put that straight right now: I don't think there is a *best* method of doing anything in maths. Any method that reliably gets *you* to the correct answer is an excellent method.

Just because your kid is being taught a different method to the one you learned, doesn't mean either of you is doing it wrong - it just means you're going to have to do a bit of work to understand the new method.

This is really important, and it might sound harsh: your attitude here makes a big difference to your kid's learning - if your reaction is "I don't get it, ask your dad/mum/auntie Carol," you're teaching them that the correct response to being confused is to give up. If your reaction is "oo! I've not seen that way before, let's look at it together," you're teaching them that learning is a positive thing.1

End rant.

Of course, I want to help you develop the skills and confidence to help like that. So, I'm starting a little series explaining some of the common variations of basic maths ideas, and why they work; the first episode is all about subtraction.

The classical method for taking numbers away involves putting the small number below the big number like this:

$\begin{array}{cccc}

1 & 6 & 7 & 4 \\

& 8 & 9 & 2 \\ \hline

& & & \end{array}$

Working from the right, you take the bottom number from the top if you can, and write it directly below; the first step here would look like this, because 4-2=2:

$\begin{array}{cccc}

1 & 6 & 7 & 4 \\

& 8 & 9 & 2 \\ \hline

& & & 2 \end{array}$

But then we have a problem: you can't really take 9 away from 7 in this context. Instead, you 'borrow 10' from the next column to the left. You'd knock the six down by one (to 5), and add 10 to the 7 to get 17 - so it looks like this:

$\begin{array}{cccc}

1 & ^5\cancel{6} & ^17 & 4 \\

& 8 & 9 & 2 \\ \hline

& & 8 & 2 \end{array}$

... and, of course, you can do 17-9 to get 8. I don't really like the word 'borrow' here - I think of it as 'making change', as if you were working with banknotes. You've started with £1,674 (made up of one £1,000 note, six £100 notes, seven £10 notes and four £1 notes2. If you spend £892, you might deal with the pound notes first (they're easy) and hit a problem with the tenners: you've got 7, but need to spend 9. Luckily, you can change one of the £100 notes into ten £10 notes with no problem at all - meaning you'd have 5 £100s left, and now 17 £10 notes, of which you can spend nine.

Lastly, you'd do the same thing: You've only got five £100 notes, but need to spend eight, so you change your £1000 into ten £100s like this:

$\begin{array}{cccc}

^0\cancel{1} & ^{15}\cancel{6} & ^17 & 4 \\

& 8 & 9 & 2 \\ \hline

& 7 & 8 & 2 \end{array}$

... and your answer is 782.

A close relative of the borrow 10 method is, I think, a bit neater: instead of making change from the number immediately to the left, you increase the number just below it. Faced with the same problem as before:

$\begin{array}{cccc}

1 & 6 & 7 & 4 \\

& 8 & 9 & 2 \\ \hline

& & & 2 \end{array}$

... you'd make the 7 into 17, but instead of knocking the 6 down to 5, you'd knock the 8 up to 9 like this:

$\begin{array}{cccc}

1 & 6 & ^17 & 4 \\

& ^9 \cancel{8} & 9 & 2 \\ \hline

& & 8 & 2 \end{array}$

It's not obvious why this should work unless you think about it: here, you're borrowing £100 from the person you're paying, and then paying them back in the next step. Although the reasoning isn't as straightforward, the mechanics are much easier (you never have to knock down a 10, for instance, or adjust a number more than once. Just to finish it off (and throwing in a 0 to show we weren't planning to spend any thousands):

$\begin{array}{cccc}

1 & ^16 & ^17 & 4 \\

^1\cancel{0} & ^9\cancel{8} & 9 & 2 \\ \hline

& 7 & 8 & 2 \end{array}$

Which is what we had before. Phew!

This one is *my* favourite, but doesn't mean it has to be yours.

Imagine, for a moment, a football match that finished 1674-892. A bit of a thrashing, to be sure; the defending was terrible. Just roll with it, ok? How much did the first team win by?

Well, the thing is: you could let both sides score extra goals and the margin of victory would stay the same (as long as they both score the same number of goals). Alternatively, you could disallow the same number of goals from each side, and that would be fair, too. So that's what you do: you try to make the second number of goals something that's easy to subtract from the first.

For example, if both teams scored eight goals, the score would become 1682-900. If they then scored 100 more each, it'd be 1782-1000; you can then disallow 1000 goals each to get 782 as your answer.

This is particularly useful if you quite like the column methods, apart from all that borrowing malarkey. (I don't think anyone teaches this, but maybe they should!). The trick is to use the football method to make the first score full of 9s, then use the column method. It's only really practical for first numbers that end with a lot of 0s in, like - say - 1000 - 475.

All you'd do is disallow one goal on either side to make it 999-474, which you can do using the column method without a carry. It's 525.

I like to think of this as a living review. Do you know of any other subtraction methods taught in schools or elsewhere? I'd love to hear.