Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, as the question asks, is we have only normalised floating-point values and normalised results, could you please explain how x + y = x?

I know it all relates to precision, but how can I explain that?

share|cite|improve this question
If y/x is tinier in magnitude than machine epsilon, well... – J. M. May 1 '11 at 22:54
Can you please continue your explanation? – Sorin Cioban May 1 '11 at 23:03
I would suggest that you look up what machine epsilon is first, and then we can talk. – J. M. May 1 '11 at 23:05
up vote 5 down vote accepted

Suppose you have a floating-point system which keeps track of, say, two decimal digits. If $x=1.0 \times 10^0$ and $y=3.4 \times 10^{-10}$, then the exact value of $x+y$ is $1.00000000034$, which would be represented in your system by the closest floating-point number, which is... Well, what is it?

share|cite|improve this answer
1.0 :) Thanks a lot! – Sorin Cioban May 2 '11 at 23:26
You're welcome! :) – Hans Lundmark May 3 '11 at 7:11
please see link. How to prove general problem that mentioned in this link? – MathMan Apr 18 '14 at 7:00
@MathMan: I already saw that question, but unfortunately I must say that it was too poorly formatted for me even to bother to figure out exactly what it was about... – Hans Lundmark Apr 18 '14 at 10:53
how to prove it in general for any $x,y$ with $|y|<eps\times \beta^{-1}\times |x|$? – MathMan Apr 18 '14 at 11:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.