Talk:Richardson's theorem
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Use of sqrt(x^2)[edit]
The article states at the present moment that if sqrt(x^2) is in E, then the A(x) = 0 is unsolveable. But sqrt(x^2) = x. This seems to be a weird formulation. — Preceding unsigned comment added by 217.227.80.156 (talk) 16:43, 29 June 2011 (UTC)
- sqrt(x^2) is unequal to x for x < 0. —Mark Dominus (talk) 19:56, 29 June 2011 (UTC)