What is the relationship between digits of significance and precision loss in floating point numbers?

Question

So I have been trying to wrap by head around the relation between the number of significant digits in a floating point number and the relative loss of precision, but I just can't seem to make sense of it. I was reading an article earlier that said to do the following:

1. Set a float to a value of 2147483647. You will see that its value is actually 2147483648
2. Subtract 64 from the float and you will see that the operation is correct
3. Subtract 65 from the float and you will see that you actually now have 2147483520, meaning that it actually subtracted 128.

So why is this 128 when there are 10 significant digits? I understand how floats are stored (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) and understand how you will lose precision if you assume that all integers will automatically find exact homes in a float data structure, but I don't understand where the 128 comes from. My intuition tells me that I'm on the right track, but I'm hoping that someone may be able to clear this up for me.

I initially thought that the distance between possible floats was 2 ^ (n-1) where n was the number of significant digits, but this did not hold true.

Thank you!

Answer 1

The "distance" between two adjacent floating point numbers is 2^(1-n+e), where e is the true exponent and n the number of bits in the mantissa (AKA significand). The exponent stored is not the true exponent, it has a bias. For IEEE-754 floats, this is 127 (for normalized numbers). So, as Peter O said, the distance depends on the exponent.