Monthly Archives: March 2019

182 – approximating 9 as 10

A few nights ago, I asked myself

Is 10 a workable substitution for 9?

I don’t quite remember the context, but I think it stemmed for wondering how far I overshot products when multiplying several things rounded up from 9 to 10.

I thought about it and set an arbitrary bar for goodness: if we stay within one order of magnitude of the true answer, then I consider 10 a good approximation for 9. Fortunately, this equation is easily expressed and easily solved:

10^{x} = (9^{x})(10)

for which we obtain x = 21.854. I believe this means that in any given product, you can up-round 9 (to 10) for 21 terms and still stay within an order of magnitude of the correct answer. So by this arbitrarily loose metric, it’s perfectly acceptable to round 9 up to 10 to perform your calculations. I rarely do products of 21 terms anyway.

The situation sours quickly if we tighten the goal, though. What if we want to come within 10% of the true answer?

(9^{x})(1.10) = 10^{x}

Unfortunately, solving for this yields x = 0.905. That is, you can’t even perform a single 10-for-9 substitution in any product without inflating your result in excess of 10%. You will always overshoot unacceptably. This should have been trivially obvious to me, but ah well. What if we accept a 30% error?

(9^{x})(1.30) = 10^{x}

We find that x = 2.490. We can make 2 10-for-9 substitutions in any product and come within 30% of the true answer – that’s not bad.