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.
