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    It's true, that if the sqrt function comes up with the exact integer solution, it is representable.

    But is it also true, that the sqrt calculation is sure to come up with the exact integer solution without round-off errors?

    Moreover, can we be sure, that no sqrt of a non-square double will yield an integer due to round-off errors?

  • Default User Avatar

    No, because every int is exactly representable as a double.