CODEWITHSUNDEEP
floating-pointieee-754
k314159
k314159

Reputation: 11062

How come all floating-point numbers can be represented in decimal?

It's well known that not all decimal numbers can be exactly represented in binary floating point.

However, it seems that all binary floating-point numbers can be exactly represented using decimal notation.

How come there aren't any floating-point numbers that can't be represented in decimal, when the reverse is not true? It seems somewhat asymmetric.

Upvotes: 2

Views: 821

Answers (2)

chux
chux

Reputation: 153303

Just another way to look at it

A binary floating point number has individual binary digits with values like:

...
24 = 16.0
23 = 8.0
22 = 4.0
21 = 2.0
20 = 1.0
2-1 = 0.5
2-2 = 0.25
2-3 = 0.125
2-4 = 0.0625
...

It then is easy to see that large + powers of 2 are exactly representable values in decimal as well as large - powers of 2.

As all binary floating point numbers are the sum of their digits, the decimal value of all binary FP is simple the sum of the above representable decimal values.

Base 10 numbers, (the ones we write like "123.456") are not always representable in floating point binary as the individual digits like 0.1, 0.01, 0.001,m etc. lack an exact finite binary FP representation.

10-1 = 0.00011001100110011...2

Upvotes: 3

Eric Postpischil
Eric Postpischil

Reputation: 222274

A binary floating-point format represents a number as ±M•2e, where M is an integer within specified bounds, and e is an integer exponent within specified bounds. (Representations may also be defined where M is a fixed-point number instead of an integer. These are mathematically equivalent with suitable adjustments to the bounds.)

A decimal numeral is equivalent to ±M•10e (generally for some other M and e, of course). For example, 3.4 is equivalent to +34•10−1.

For any decimal numeral in the form ±M•10e, we can rewrite it: ±M•10e = ±M•(5•2)e = ±M•(5e•2e) = ±(M•5e)•2e. Now, if M•5e is an integer within bounds, this is a binary floating-point representation. But if M•5e is not an integer, as with 34•5−1, there is no way to make it an integer. This number cannot be represented in the binary floating-point format.

Conversely, consider a binary floating-point number ±M•2e. If e is nonnegative, it is an integer, so this is already in the decimal numeral form; it is an integer times 100. If e is negative, we can rewrite it: ±M•2e = ±M•2e•5e•5−e = ±M•10e•5−e = ±(M•5−e)•10e. Then, since e is negative, M•5−e is an integer, so ±(M•5−e)•10e is in the decimal form of an integer multiplied by a power of ten.

In other words, we can make any binary floating-point number that is not already an integer into an integer by multiplying it by 10 until we have an integer. This is because 2 is a factor of 10, so each multiplication by 10 cancels one of the negative powers of 2 in the floating-point representation.

Conversely, given a decimal numeral that is not an integer, such as .1, we cannot always make it into an integer because multiply by 2 will not cancel negative powers of 5 that are in it.

Upvotes: 7

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