How precise should an absolute error be?

Question

While adding approximate numbers, our result must be as precise as the least precise number that was given to us. For instance, $101+1.001+1.0≈103$ because the number with the least precision viz., $101$, is correct to the nearest unit.

But what about when evaluating errors? Suppose $2/3$ is approximated to $0.6667$. Then the absolute error would be $|2/3-0.6667|$. But this value is $0.00003333333...$, which, according to our rules, we must approximate to $0.0000$ (as precise as $0.6667$).

This is certainly not right.

Answer 1

By precision rule $0.0000 \approx 0$ up to the 4th decimal place. However $0.0000 \ne 0$ since we don't know (or can't measure) the following digit

The best you can say is

$$ 0 < |2/3 - 0.6667| < 0.0001 $$

Or the error is somewhere on the order of $10^{-5}$

Answer 2

Mathematically, $0.666667=0.6666670000$. Your confusion probably comes from the fact that you implicitly think that $0.6666670000$ is more accurate, presumably because of the use of significant figures in physics.

Answer 3

Thats right, with an absolute error of $0.0001$,

$$\left|\frac23-0.6667\right|=0.0000$$