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The function for Oakley group 2, p = 2^1024 - 2^960 - 1 + 2^64 * ⌊2^894 * π + 129093⌋, produces a prime number. It contains pi, an irrational number. How can an equation containing an irrational number produce a rational product?

I looked on the internet and could not find an answer. I learned in my mathematical structures class that the product of a rational and irrational number is irrational so this makes me wonder how pi is used to produce a prime.

Cad
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    Look at the section with pi: "⌊2^894 * π + 129093⌋" Those ⌊...⌋ brackets are the floor function, which rounds a number down to the next lowest integer. – rossum May 07 '23 at 07:47

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Rossum already answered your question in the comments, but I'm going to elaborate a bit more on the structure of this prime.

As Rossum says, the ⌊⌋ in the ⌊2^894 * π + 129093⌋ portion denote the mathematical floor function, yielding the largest integer less than or equal to the argument. The 2^1024 - 2^960 produces a number where the first 64 bits are 1s. Multiplying the ⌊2^894 * π + 129093⌋ by 2^64 and subtracting 1 produces a number where the last 64 bits are 1s. Combining these, we have a number where the first and last 64 bits are ones, since there will be no carrying in the addition of the two.

Within the ⌊2^894 * π + 129093⌋, pi is multiplied by 2^894 to extract 896 bits (1024-64-64)bits of pi. The+ 129093` is just an offset to make the overall result prime.

Dillon Davis
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