- Learn a Tudor Algorithm to do hard times tables
- See a simple (and useful) example of what algorithms are.
- Understand why algebra is really useful – for proving algorithms ALWAYS work
Algorithms are magic. Even simple algorithms can sometimes look like witchcraft. My favourite is one of the ways the Tudors did multiplication. It is all down to Welsh Tudor polymath Robert Recorde. As well as being a mathematician, who wrote the first arithmetic text book, he was also the physician to two Tudor Monarchs: Edward VI and Mary. Not only that, but along the way he also invented the equal sign using it for the first time as a shorthand for the phrase “ is equalle to” when doing maths (he would have liked the C programming language!). He was particularly good at algorithms and his book contained a cunning way to make the times tables easier.
Learning times tables is a rite of passage of primary school students. Those up to 5 are not too difficult but the 6, 7, 8 and 9 times table are particularly hard to remember (though of course the 9 times table does have a cunning pattern that makes it easier). Robert Recorde had a solution. In his book he gave an algorithm that made those larger times tables easy. If you know your tables up to the 5 times table (the easy ones), then the rest become simple too. It is just a matter of knowing the algorithm and doing some simple subtractions combined with the multiplications you do know. That the algorithm works is barely believable though, it is so odd.
The Algorithm
Here is how it goes. Suppose you have two of those nasty higher numbers to multiply like 8×7.
- Draw 7 boxes labelled A, B, C, D, E, F and G. Write one of the numbers you want to multiply in box A and the other in box B.
- eg if you want to calculate 8 x 7, put 8 in box A and 7 in box B.
- Subtract the number in box A from 10 and put the answer in box C.
- 10-8 = 2, so put 2 in box C.
- Subtract the number in box B from 10 and put the answer in box D.
- 10-7 = 3 , so put 3 in box D.
- Multiply the two numbers in box C and D together. Put the answer in box E.
- 2 x 3 = 6 so we write 6 in box E
- Because your original two numbers in A and B were above 5, the new ones you must multiply instead are both below 5 so its a nice easy multiplication.
- Now subtract the number in box C from Box B (ie the diagonal numbers) and multiply the answer by 10 (stick a 0 on the end) writing the answer in box F.
- 7-2 = 5, so multiplying by 10 means you put 50 in box F.
- Construct the final answer in box G by adding the number in box F to the number in box E.
- E holds 50 and F holds 6 so 50+6 = 56 is the answer to the original multiplication of 8×7.
Try the algorithm yourself on some more examples, like 6×9, 7×6 and 8×9. It seems a bizarre way to do multiplication – were the Tudors slightly bonkers? Well possibly, but Robert Recorde was just very clever. At a stroke his clever algorithm gives a way to do all those hard multiplications using only a few much simpler calculations.
Does it always work?
This little computational ‘spell’ seems like witchcraft! Surely it doesn’t really always work? Actually it does. The point of an algorithm is that it always guarantees you get the right answer. We can prove this one does with a bit of algebra. Of course, as long as you trust Recorde, you don’t have to understand the proof for the algorithm to work for you. It is good to do the proof though (or at least have a rigorous argument) to be sure the algorithm really is fool proof.
The Proof
STEP 1: Let’s call the number in box A, a, the number in box B, b and so on. So we are trying to work out the answer to:
ab (thats just another way to write: a x b)
Now (see the diagram below) instead of doing the multiplication directly we work out
STEP 2: c = 10-a and
STEP 3: d = 10-b
We multiply these together e = c x d so to get the number in box E, replacing c and d by the terms they are equal to, we do
STEP 4: e = (10-a)(10-b).
We also subtract 10-a from b, and multiply it by 10 to get the number in box F, so
STEP 5: f = 10(b-(10-a))
The final answer is then just g = e + f. This means the calculation we ultimately do is
STEP 6: g = 10(b-(10-a)) + (10-a)(10-b)
We need to show this value g is the same as a x b. It looks a bit unlikely but let’s simplify it all.
g = 10(b-(10-a)) + (10-a)(10-b)
Expanding the brackets for the part that came from f gives:
= 10(b -10 + a) + (10-a)(10-b)
= 10b -100 + 10a + (10-a)(10-b)
Similarly, multiplying out the brackets from e’s part gives:
= 10b -100 + 10a + 100 -10a -10b + ab
Now having expanded everything, we simplify: the terms +100 and -100 cancel out
= 10b + 10a -10a -10b + ab
The +10b and -10b cancel out too.
= 10a -10a + ab
The +10a and -10a also cancel out, leaving just a single term.
= ab
So what this says is that following the algorithm leaves the answer of calculating a x b in box G which is just what we wanted.
Algorithms
Algorithms are just sequences of steps to follow that guarantee a result (whether you understand why it works or not). Here the result is to multiply two numbers doing only simpler calculations. Of course, computers don’t know what they are doing. They can only follow rules blindly. While perhaps not using exactly this algorithm, they do use lots of clever algorithms to allow them to do arithmetic quickly.
This bit of witchcraft also shows why algebra is such a useful thing. It is a great way of proving useful algorithms really do always work.
The Tudors may still mistakenly have believed in witchcraft, and may not have had computers, but their computational witchcraft was still a really useful thing.
This one is for Peter, the only person I imagine ever to get spontaneous applause from a class of teenagers for algebra… and a wonderful computational magician.
He also loved hiding Easter Eggs in our work.
