# How does x+y = x for non-zero value of y in floating-point arithmetic?

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?

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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

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?

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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