I started today studying real numbers axiomatic and I'm a bit stuck showing a proof of a property of order in real numbers. This is the problem:
if $r<0,$ then $-r>0$.
I have to do it just using axiomatic. Thank you for any ideas.
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if $r<0$, then $0=-r+r<-r$ – J. W. Tanner Apr 01 '20 at 17:58
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you mean axioms? – J. W. Tanner Apr 01 '20 at 20:18
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Yes I do. Proving it by axioms – Richard Kingston Apr 01 '20 at 20:25
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I see you say r+(-r)<-r, its "obvius" from r<0 and adding (-r) to both sides. But you are using a property and not axioms. Do you get my point? – Richard Kingston Apr 01 '20 at 20:27
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What are the axioms you are using? You should edit them into you question – J. W. Tanner Apr 01 '20 at 20:37
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I have available to use two axioms; – Richard Kingston Apr 01 '20 at 20:38
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I)A number only can be >0, <0. If not then the number is 0. – Richard Kingston Apr 01 '20 at 20:39
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Don't you have any axioms regarding the interaction between $<$, $+$, and $\times$? – J. W. Tanner Apr 01 '20 at 20:40
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If a>0 and b>0, then a+b>0 and ab>0. I don't have the interaction if a,b<0. But I can use all axioms of real numbers equality (conmutativity, asociativity etc) – Richard Kingston Apr 01 '20 at 20:44