IEEE Floating Point Number Rounding

Question

For the IEEE floating point number, why do these equal epsilon m?

(1 - 1e-16) - 1 ans = -1.1102e-16

1 + (1e-16 - 1) ans = 1.1102e-16

While, on the other hand, below equals 0

(1 + 1e-16) - 1 ans = 0

Can someone explain to me why? For the last one, I understand that fl(x) = 1 for 1 ≤ x ≤ 1 + M and fl(x) = 1 + 2M for 1 + M < x ≤ 1 + 2M

Shouldn't the first two equal 0 since fl(x) = 1 for 1 - M < x ≤ 1 and fl(x) = 1 - 2M for 1 - 2M ≤ x ≤ 1 - M ?

• Mark Dickinson: What value are you using for M here? 2^{-53} or 2^{-52}?

Answer 1

Machine epsilon is not really a consistently well-defined term, as different sources and languages define it in different ways.

If we take M to be 2^-53 ≈ 1.11 × 10^-16 (i.e. the gap between 1 and the previous floating point number, or half the gap between 1 and the next floating point number) then the floating point number line near 1 looks something like:

--|----|----|----|---------|---------|------
 1-2M 1-M 1 1+2M 1+4M

So, under standard round to nearest, ties to even, we have

• fl(x) = 1-M for 1 - 3M/2 < x < 1 - M/2
• fl(x) = 1 for 1 - M/2 ≤ x ≤ 1 + M
• fl(x) = 1+2M for 1 + M < x < 1 + 3M

So in the first case, 1 - 1e-16 would round to 1 - M, and in the second case 1 + 1e-16 would round to 1

• Kevin Lee: So, what you are saying is that previous floating point number is rounded by every M, while the next floating point number is rounded by every 2M. This is because the previous fpn is 1/2 while the next fpn is 2. Is that right?
• Simon Byrne: Yes: the gap between floating point numbers doubles for every binade: en.wikipedia.org/wiki/Binade
• Kevin Lee: Awesome! Thanks for the help!
• Mark Dickinson: @KevinLee: Please accept the answer if you found it helpful!
• Kevin Lee: Do I do that by clicking the green check? I am not really used to this website.