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In Numerical Mathematics (2e) by Quarteroni, Sacco, and Saleri; they describe this set $$\mathbb F(\beta, t, L, U) = \{0\} \bigcup \biggr\{x\in\mathbb R: x = (-1)^s\beta^e\sum_{i=1}^{t}a_i\beta^{-i}\biggr\}$$ as the set of floating point numbers with $t$ significant digits, base $\beta \geq 2$, $0\leq a_i\leq \beta -1$, and range $(L, U)$ with $L\leq e \leq U$.

I can see why the set would be defined this way, but I am having a difficult time trying to actually apply it with case examples. Every resource I look at either skips intermediate steps or does not explain those steps. I can find the minimum and maximum values, but after that I'm stuck.

Can anyone demonstrate how to construct the set based on a particular case of $(\beta, t, L, U)$? Any help is appreciated.

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  • $\begingroup$ What do you mean by "construct"? Do you mean to list all the elements one after the other? That could be a very long list unless $\beta$, $t$, and $U-L$ are all very small! $\endgroup$
    – David K
    Sep 23 at 1:46
  • $\begingroup$ For base 2, with 3 digits (I am not sure why they are called "significant") the possible numbers are $\endgroup$ Sep 23 at 12:02
  • $\begingroup$ For base 2, with 3 digits (I am not sure why they are called "significant") the possible numbers are o, 1, 10, 11, 100, 101, 110, 111. 2^3- 1= 7 numbers. I would be inclined to conjecture that with base $\beta$ and n digits, there are $\beta^n-1$ possible numbers. $\endgroup$ Sep 23 at 12:12
  • $\begingroup$ @GeorgeIvey There's also a notion of "normalization" of a floating-point representation, which essentially means that $a_1$ is not allowed to be zero. Normalization is so common that I would usually just assume it, but on second thought it might not be intended in this case. $\endgroup$
    – David K
    Sep 23 at 21:40
  • $\begingroup$ @Dee Are we allowed to set $a_1=0$? Are we allowed to set all the $a_i$ to zero simultaneously? If not, what are the restrictions? If we are allowed to set them all to zero, doing so gives us a representation of the number $0$; why then do we have to explicitly take a union with the set $\{ 0 \}$? $\endgroup$
    – David K
    Sep 26 at 3:09

1 Answer 1

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Taking $\beta = 10, t = 1, L=-1, U=1$, we have that \begin{align*} \mathbb{F} =& \{0.1\times 10^{-1}, 0.2 \times 10^{-1},\cdots, 0.9 \times 10^{-1}, \\ &0.1 \times 10^0, 0.2\times 10^0, \cdots 0.9 \times 10^0, \\ & 0.1 \times 10^1, 0.2 \times 10^1, \cdots, 0.9 \times 10^1, \\ &-0.1\times 10^{-1}, -0.2 \times 10^{-1},\cdots, -0.9 \times 10^{-1}, \\ &-0.1 \times 10^0, -0.2\times 10^0, \cdots -0.9 \times 10^0, \\ & -0.1 \times 10^1, -0.2 \times 10^1, \cdots, -0.9 \times 10^1\} \cup \{0\} \end{align*}

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