Closed asmeurer closed 8 years ago
The problem is that sqrt(x^2) is not equal to x in general (take x = -1). sqrt(x)^2 does always equal x, by definition.
Fixes #42
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Thanks!
The problem is that sqrt(x^2) is not equal to x in general (take x = -1). sqrt(x)^2 does always equal x, by definition.
Fixes #42