The required number of digits (in base t) to represent the double in base t?

Question

Theorem:
The required number of digits (in base t) to represent the positive integer S in base t is ⟦log_tS⟧+1 (⟦.⟧: floor function).

I wondered, what is the required number of digits (in base 2) to represent the maximum positive double (floating point) number in computer. I have 64-bit OS and 32-bit R on it. Hence, I did:

.Machine$double.xmax # 1.797693e+308
typeof(.Machine$double.xmax) # double
floor(log(.Machine$double.xmax, 2))+1  # 1025
.Machine$integer.max   # 2147483647
class(.Machine$integer.max) # integer
floor(log(.Machine$integer.max, 2))+1  # 31; (1 bit for sign bit)

So, the theory is OK for integers.

(1) But what about the double equivalent of the theorem? I.e., what is the required number of digits (in base t) to represent the double in base t?

(2) This may be difficult with real numbers with decimals. So, perhaps, one may know the equivalent of the theorem for decimalless reals (that is ">2147483647").

In particular, where does the 1025 above come from?

(3) Would I get 63 if I used 64-bit OS and 64-bit R for the following?

floor(log(.Machine$integer.max, 2))+1  # 63??; (1 bit for sign bit??)

Answer 1

Ad 3) I don't know about doubles but the integer internal representation is still 32 bits even on 64 bit systems. If you want to go bigger you need to use some sort of library for that for example 'bit64'

You will get more detailed information with help(double) and help(integer)