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0 pointssebzim45004mo ago0 comments

Right but there isn't a theorem saying `(sqrt x)^2 = x`, there's a theorem saying `x >= 0 -> (sqrt x)^2 = x`

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Ah, that makes sense. Thank you. As long as every use of sqrt has such a condition.

j / k navigate · click thread line to collapse

Ah, that makes sense. Thank you. As long as every use of sqrt has such a condition.

j / k navigate · click thread line to collapse