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  1. To prove $0 <1$

1Now if $a \leq b$ and $0 \leq c$ then $ac \leq bc$

Assume that $1 \leq 0$. Let $a=1$ and $b=0$ and $0 \leq c$. So we get $c \leq 0$ which is a contradiction. Is this correct?

  1. To prove if $0 < a$ , then $0 < a^{-1}$

so we have $aa^{-1}=1$. So $a a ^{-1} \geq 0$. Again using same axiom as above with $c=1$ we get $0 \leq aa^{-1}$ which implies $a^{-1} > 0$ or $a^{-1}=0$. But since $a \neq 0 $ so $a^{-1} > 0$

Thanks

ReadThyOwnBook
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2 Answers2

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It's not a contradiction yet, because $c$ could be $0$. Specifically pick $c\geq0$ such that $c\neq0$ (if you can guarantee that such a thing exists), and you will have your contradiction.

Arthur
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  • You mean to say pick c=1 or 2 like that. – ReadThyOwnBook May 12 '18 at 07:38
  • I write c=2 at beginning to arrive at contradiction. what would be complete proof – ReadThyOwnBook May 12 '18 at 07:42
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    @FrejaJessen Not $1$ or $2$, because we have assumed that $1<0$. On the other hand, under that assumption, there is $-1>0$. – Arthur May 12 '18 at 07:43
  • is my 2nd part correct? – ReadThyOwnBook May 12 '18 at 07:48
  • @FrejaJessen 1) You should only ask one question per post. 2) You shouldn't significantly change your question after an answer has been posted (or at any time, really, since people may be writing answers already). That being said, no, not really. Here you need to be careful not to get letter-confused. But we know that $a>0$, so that is our $c$. We assume $a^{-1}<0$, multiply on both sides by our $c$, which is $a$, and by your axiom get $1<0$ which contradicts the previous result. – Arthur May 12 '18 at 07:56
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  1. Suppose $1<0$. Since $x^{2}\ge0$ holds for all $x$, we also have $$(1)\cdot(1)=1^{2}=1\ge0,$$ hence a contradiction.$$\\ \\$$

  2. Your second part is correct, for suppose $a>0$ and $a^{-1}<0$. Multiplying through by $a$ doesn't change the order, so $$a^{-1}<0\implies a^{-1}\cdot a<0\cdot a\implies1<0,$$ which is a contradiction. Whence $a^{-1}>0$.

Alex D
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