Prove that in a floating point system with truncation the number of significant digits is $n$.

Question

I was requested to prove that in a floating point system $\text{F}(\beta, n, m, M)$ with truncation the number of significant digits is $n$. (where $n$ is the number of digits and $m < \text{exponent} < M$.

I tried using the definition that the number of significant digits is the maximum natural $d$ for which the relative error of $x$ is $\leq\beta^{1-d}$, and so:

$$ x = b_1b_2\ldots b_n\cdot\beta^e,$$

$$ fl(x) = b_1b_2\ldots b_{n-1}\cdot\beta^e.$$

Therefore:

$$\dfrac{|x-fl(x)|}{|x|} = \dfrac{|b_1b_2\ldots b_n\cdot\beta^e - b_1b_2\ldots b_{n-1}\cdot\beta^e}{|b_1b_2\ldots b_n\cdot\beta^e|} = \dfrac{b_n}{b_1b_2\ldots b_n} \leq \beta^{1-d}.$$

And I'm stuck here because I need to prove $d = n$ but I have no idea how to advance. Any suggestions?

Thanks.

Answer 1

Figured out the solution:

$b_n <= \beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $\beta$

so we get: $\dfrac{b_n}{b_1b_2…b_n} <= \beta^{1-n} =>$ the number of significant digits is $n$.