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 Quote by poutsos.A $$\forall a\forall b$$[( a>0 & b>0)------> (a$$\leq b$$ <------>$$a^{2}$$$$\leq b^{2}$$)]. or in words: for all a and for all b , if a>0 and b>0 then .a$$\leq b$$ iff $$a^{2}$$$$\leq b^{2}$$ is there a possibility for a proof within the predicate calculus??
yes thank you it is $$a^{2}$$$$\leq b^{2}$$,instead of $$a^{2}$$$$\leq a^{2}$$