Is integer multiplication implemented using double precision floating point exact up until 2^53?

Question

I ask because I am computing matrix multiplications where all the matrix values are integers.

I'd like to use LAPACK so that I get fast code that is correct. Will two large integers (whose product is less than 2^53), stored as doubles, when multiplied, yield a double containing the exact integer result?

Housewarming

Answer 1

Your analysis is correct:

• All integers between -2⁵³ and 2⁵³ are exactly representable in double precision.
• The IEEE754 standard requires calculations to be performed exactly, and then rounded to the nearest representable number.

Hence a product of two values that equals an integer in that range will therefore be represented exactly.

Reference: What every computer scientist should know about floating-point arithmetic. The key section is the discussion of the IEEE standard as pertaining to operations. That contains the statement of the second bullet point above. You already knew the first bullet point and it's the second point that completes the argument.

Constrictor

Answer 2

Yes! A double's data is split into its sign, exponent, and fraction:

Wikipedia has an article explaining the ranges of representable numbers

All integers between -2^53 and 2^53 are representable in double precision.

Between 2^52=4,503,599,627,370,496 and 2^53=9,007,199,254,740,992 the representable numbers are exactly the integers. For the next range, from 2^53 to 2^54, everything is multiplied by 2, so the representable numbers are the even ones, etc. Conversely, for the previous range from 2^51 to 2^52, the spacing is 0.5, etc.

Ruscher

Please don't paste other peoples words and images without attribution. – Constrictor

Hey! I made that image! EDIT: I guess I attributed it in my comment, too. – Historic

Everything's from the wiki link. – Ruscher

Yes, we know. But you must make that totally clear in your answer. Note my edit to include quotes. – Constrictor