# t digits are used to represent the mantissa in floating-point system, but rounding unit is calculated for doubles with 53 bits

I'm reading the book "A First Course in Numerical Methods" by U. Ascher and C. Greif, and in the 2nd chapter it's written that

... we associate $x$ a floating point representation $fl(x)$ of the form similar to that of $x$ but with only t digits, so

$$fl(x) = sing(x) \times (1.d_1d_2d_3 \cdots d_{t-1}d_t) \times 2^e$$

Then it says that

the relative error $$\frac{fl(x) - x}{|x|}$$ is bounded by $n = \frac{1}{2} \times 2^{-t}$.

So for a double-precision floating pointer number (represented using IEEE), which uses $52$ for the mantissa, $n = 2^{-53} \times 1.1 \times 10^{-16}$

Why $-53$ if the mantissa of a double floating point number is $t = 52$?

• Well I guess you'll get your answer here: stackoverflow.com/questions/4930269/… It's related to what is called the implicit bit. – G.Clavier Apr 6 '16 at 10:35
• @G.Clavier I thought it could be because of this, since I'm aware of this implicit digit 1, but I was not sure, so I decided the same to ask the question. – nbro Apr 6 '16 at 10:42