Why number are (not) representable in double precision IEEE754?

问题

I am confused on IEEE754 double precision, I consider two questions:
1. Why each number from interval -2⁵⁴, -2⁵⁴+2, -2⁵⁴+4...2⁵⁴ is representable ?

2. Why 2⁵⁴+2 is not representable ?

Can you help me ? I understand way of working IEEE754 - however, I have a problem with seeing it.

回答1:

There are 53 bits in the significand (or mantissa) of an IEEE 754 double. −2⁵⁴ can be exactly represented, as

mantissa: 1.00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00 (bin)
exponent: 54
sign:     1

Now let's forget the sign bit for a moment. It is irrelevant for this explanation. So assume we have +2⁵⁴.

With this exponent, the lowest -- rightmost -- bit of the significand has the value 2^-52 * 2⁵⁴ = 4. So 2⁵⁴ + 4 is encoded as:

mantissa: 1.00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 01 (bin)
exponent: 54                                                             ^
                                                                lowest bit

But there is no value inbetween. So you cannot encode 2⁵⁴ + 2.

Why is this not a problem for −2⁵⁴ + 2? Because that is the same as −(2⁵⁴ − 2), and that is represented as:

mantissa: 1.11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11
exponent: 53 !!
sign:     1

And the exponent 53 means you have steps of 2^-52 * 2⁵³ = 2. The next value toward 0 is then:

mantissa: 1.11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 10
exponent: 53
sign:     1

which is −2⁵⁴ + 4, or actually −(2⁵⁴ − 4). And you can go on like that until you reach −2⁵³.

来源：https://stackoverflow.com/questions/39291410/why-number-are-not-representable-in-double-precision-ieee754