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Asked: 2026-05-13T12:49:43+00:00

I am wondering if this is true: When I take the square root of

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I am wondering if this is true: When I take the square root of a squared integer, like in

f = Math.sqrt(123*123)

I will get a floating point number very close to 123. Due to floating point representation precision, this could be something like 122.99999999999999999999 or 123.000000000000000000001.

Since floor(122.999999999999999999) is 122, I should get 122 instead of 123. So I expect that floor(sqrt(i*i)) == i-1 in about 50% of the cases. Strangely, for all the numbers I have tested, floor(sqrt(i*i) == i. Here is a small ruby script to test the first 100 million numbers:

100_000_000.times do |i|
  puts i if Math.sqrt(i*i).floor != i
end

The above script never prints anything. Why is that so?

UPDATE: Thanks for the quick reply, this seems to be the solution: According to wikipedia

Any integer with absolute value less
than or equal to 2^24 can be exactly
represented in the single precision
format, and any integer with absolute
value less than or equal to 2^53 can
be exactly represented in the double
precision format.

Math.sqrt(i*i) starts to behave as I’ve expected it starting from i=9007199254740993, which is 2^53 + 1.

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1 Answer

  1. Editorial Team

    For “small” integers, there is usually an exact floating-point representation.

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