Binary Floats Represented as Decimal Numbers - binary

Question

Not all decimal numbers can be represented exactly using binary floats.
http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
There are two reasons why a real number might not be exactly
representable as a floating-point number. The most common situation is
illustrated by the decimal number 0.1. Although it has a finite
decimal representation, in binary it has an infinite repeating
representation.
What about the other way around? Can every single IEEE 754 float be represented exactly using a decimal number, if enough digits are used?

Answer 1

Yes, every finite IEEE 754 float be represented exactly using a decimal number, if enough digits are used.
Each additional binary digit of precision requires at most one additional decimal digit of precision to represent exactly.
For instance:
0.1b -> 0.5
0.01b -> 0.25
0.11b -> 0.75
0.001b -> 0.125
A double-precision (binary64) number between 1 and 2 requires only 52 decimal digits after the dot to be represented exactly:
#include <stdio.h>
int main(void) {
printf("%.55f\n", 1.1);
}
Result:
1.1000000000000000888178419700125232338905334472656250000
It's all zeroes after the four displayed at the end of the representation above. 1.100000000000000088817841970012523233890533447265625 is the exact value of the double nearest to 11/10.
As pointed out in the comments below, each additional unit of magnitude for a negative exponent also requires one additional decimal digit to represent exactly. But negative exponents of a large magnitude have leading zeroes in their decimal representations. The smallest subnormal number would have 1022 + 52 decimal digits after the dot, but the first nearly 1022*log10(2) of these digits would be zeroes.