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StephenPrivitera
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Let F be the set of all functions whose domain is R. Considering the binary operations + and * on F, show that P10-P12 cannot hold.
P10 is essentially the trichotomy law. In my book, it is written, "For all a in R, a=0, xor a in P, xor -a in P," where P is the set of positive numbers. Um, how does this make ANY sense for functions?! I can't say that the function F is positive! P11 and P12 are closure under addition and multiplication respectively. Again these make no sense in this context. So my answer to this question presently is that P10-P12 make no sense for the operations on F.
Anyone else see some sense in this?
Edit: an afterthought
I can see sense in this if they mean f(x)=0, f(x)>0 xor -f(x)>0 for a given x, but f(x) is not a function - it's a number.
P10 is essentially the trichotomy law. In my book, it is written, "For all a in R, a=0, xor a in P, xor -a in P," where P is the set of positive numbers. Um, how does this make ANY sense for functions?! I can't say that the function F is positive! P11 and P12 are closure under addition and multiplication respectively. Again these make no sense in this context. So my answer to this question presently is that P10-P12 make no sense for the operations on F.
Anyone else see some sense in this?
Edit: an afterthought
I can see sense in this if they mean f(x)=0, f(x)>0 xor -f(x)>0 for a given x, but f(x) is not a function - it's a number.
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