2

I am reading a numerical analysis book which talks about nearest numbers.

They quickly state that you can easily find nearest machine numbers, but never go on to show how exactly to do so.

Can someone please explain how you can find the two closest numbers $y'$ and $y''$ of a given number $y$, where $y'$ is the number below and $y''$ is the number above for double precision?

That is, to be precise, for $\mathbb{D}$ the set of floating point (double) values, $y' = \max\{x \in \mathbb{D} | x \leq y \} $ and $y'' = \min \{ x \in \mathbb{D} | x > y \}$.

Let say the number is $y = 0.8$.

How do you number the closest number above and below that number in double precision?

MichaelChirico
  • 4,131
JabberJabber
  • 135
  • 5

1 Answers1

3

You would have to determine the 53 binary digits of the mantissa in $$ x=2^e·(1.m_1…m_{53})_2 $$ The exponent is obtained from the condition $2^e\le x <2^{e+1}$. Then for the mantissa compute the integer part of $x·2^{53-e}$.

As for $x=0.8$ you get $e=-1$, you need to compute the integer part of $0.8·2^{54}$, $14411518807585587$, and its successor, $14411518807585588$, (and additionally the binary representation of them if you want to construct the floating point format).

Lutz Lehmann
  • 126,666