Floating Point Representation

Suppose we are in a floating-point regime where the mantissa had 3 digits and the exponent has one digit.

Then, 0 = .000e-9 is relatively isolated from next two positive numbers .100e-9 and .101e-9. Which are 10e-12 apart.

This is from Hamming's book on numerical methods, but how is .100e-9 the next positive number, why not .001e-9?

[Walter Roberson]

Hidden bit or no hidden bit?

Normalization required or no normalization required?

[James Tursa]

@Walter: OP has stated a decimal digit system, so hidden bit & normalization is not applicable.

[Walter Roberson]

Okay on the hidden bit, but I think normalization is still a factor.

You seldom want a system that has a distinct zero for every different exponent value (e.g., 000E-9 000E-10).

You seldom want a system that has three different representations for some values, such as 100E-12 010E-11 001E-10

IEEE 754, referenced in the tags, *requires* normalization of binary values, except for the extremely small denormalized numbers, which it has special rules for.

IEEE 754 did adopt an optional decimal representation; my recollection at the moment is that it uses normalization as well, but I would need to cross-check that.

Answer 1

[James Tursa]

1) What does this have to do with MATLAB?

2) How can you have numbers in the e-9 and e-12 range with only one exponent digit? Is this some type of decimal digit floating point system you are talking about? If so, what is the exponent bias? How can you get negative exponents in e-9 range with only one digit?

In short, please explain the floating point format you are talking about in detail.

[Stuart]

Decimal digits, rest isn't relevant to answer.

[James Tursa]

Obviously you don't think so, but I do. Why isn't 0.001e-10 or 0.001e-11 or 0.001e-12 the next positive number? The exponent bias *does* make a difference to me, and I can't answer your question without knowing this.