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I've started studying calculus. As part of studies I've encountered a question. How does one prove that $1 > 0$?

I tried proving it by contradiction by saying that $1 < 0$, but I can't seem to contradict this hypothesis.

Any help will be welcomed.

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2 Answers 2

up vote 5 down vote accepted

If $1<0$ then $-1>0$, hence $1=(-1)\cdot(-1)>0$.

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Why do you assume that (−1)⋅(−1)>0 ? –  vondip Nov 6 '12 at 16:38
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He doesn't assume it. The product of positives is positive. That's part of the definition of an ordered field. –  kahen Nov 6 '12 at 16:41
    
I do not assume that. I assume that $1<0$. By adding $-1$ on both sides, I find $0<-1$. By multiplying with the positive(!) number $-1$ then $1>0$ follows. –  Hagen von Eitzen Nov 6 '12 at 16:41
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I don't know if it's called anything, but the definition goes like this: An ordered pair $(F,P)$ where $F$ is a field and $P \subset F$ satisfies (1) $0 \in P$ and (2) $x,y \in P \implies x+y \in P$ and $xy \in P$ is said to be an ordered field and $P$ is said to be the *positive cone of $F$*. $\quad$ It's an easy exercise to prove that if $(F,P)$ is an ordered field then $x \leq y \iff y-x \in P$ defines a total order on $F$. –  kahen Nov 6 '12 at 16:45
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Oh. I forgot part of the definition. You need (3) $F$ is the disjoint union of $P\setminus\{0\}$, $-P\setminus\{0\}$ and $\{0\}$. Otherwise $P$ is just a prepositive cone. –  kahen Nov 7 '12 at 6:34

You can use the trivial inequality $x^2 \geq 0$ for all $x$. Prove this fact and use it to prove $1 >0$.

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