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Oct23-08, 06:11 AM
P: 101
Quote Quote by poutsos.A View Post
[tex]\forall a\forall b[/tex][( a>0 & b>0)------> (a[tex]\leq b[/tex] <------>[tex]a^{2}[/tex][tex]\leq b^{2}[/tex])].

or in words: for all a and for all b , if a>0 and b>0 then .a[tex]\leq b[/tex] iff [tex]a^{2}[/tex][tex]\leq b^{2}[/tex]

is there a possibility for a proof within the predicate calculus??
yes thank you it is [tex]a^{2}[/tex][tex]\leq b^{2}[/tex],instead of [tex]a^{2}[/tex][tex]\leq a^{2}[/tex]