# Secrets Of The Mathematical Ninja: Adjusting using errors.

Your average ninja (if there is such a thing) doesn’t need to do the swirly-sword manoeuvre. If he wanted, he could just run you through and you’d be dead before you even noticed he was there. But that’s rather rude and unsophisticated, the kind of thing a pirate might do. The true ninja is always striving to be a little bit more intimidating, a little bit more on the ball.

So it is with the mathematical ninja. You don’t need to adjust your answers to account for errors in your estimate, but wouldn’t you rather say “About 19.6” rather than “About 20”?

This technique only works when you’re calculating a power, a root, a multiplication or a division sum - and when the error is fairly small.

Here’s what you do:

- If you’re multiplying two numbers together,
*add*their errors. - If you’re dividing one number by another,
*take away*the bottom error from the top. - If you’re taking a power of a number,
*multiply*the error by the power. - If you’re taking a root of a number,
*divide*the error by the root (e.g., if you’re taking a cube root, divide the error by 3)

(If those laws look suspiciously like the rules for working with powers… well, see if you can figure out why!)

Once you’ve worked out this new error, it’s all plain sailing: you work out the sum using your rough answers and then correct by the combined error at the end!

So, for instance, to work out $\frac{g}{\sqrt{2}}$, you’d work out $10 \div \frac{10}{7}$, which is 7. The errors are (-2%) on top and (-1%) underneath, and you take them away, making (-1%) - so the answer is 1% less than 7, or 6.93. (It’s 6.9296).

If you wanted to work out $\left(\frac{g}{\sqrt{2}}\right)^3$, it would simply be seven cubed less 3%. That’s 343 less 3%, which is about 10 - or 333. The actual answer is 332.76.