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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.

7 Answers 7

0

It is not provable that $0$ unequals $1$. It is part of the field-axioms.

11

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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    If you use this, you also have to prove that $1\neq 0$. It's ok, but the prove is incomplete by a trifle.2018-06-11
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If $1<0$ then $-1>0$, hence $1=(-1)\cdot(-1)>0$.

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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.2012-11-07
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Let’s consider the question in the ordered field of real numbers. By the trichotomy axiom of inequality, only one of the following is true: $ 1=0 $ $ 1<0 $ $ 1>0 $ Now, by the nontriviality axiom of the real numbers, $ 1\ne 0 $, so we’ve ruled out the first possibility.

Now suppose $ 1<0 $. Then for $ a\in R $, $ a>0 $, by the multiplication axioms of an ordered field, $ a\cdot a^{-1}<0 $ $ a\cdot a^{-1}\cdot a<0\cdot a $ $ a<0 $ A contradiction, since by the trichotomy axiom, we cannot have $ a>0 $ and $ a<0 $. Thus we must have that $ 1>0 $.

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Previous answers are not complete.

Axioms of ordered set give us only $\forall x \le 0 \ y \le 0 \ \ \ \ xy \ge 0$. And we can get only $0 \le 1$.

Now we need to proof the $0 \ne 1$.

$x = x \cdot 1 = x \cdot (1 + 0) = x \cdot 1 + x \cdot 0 \rightarrow x \cdot 0 = 0$

Suppose that $0 = 1$. By the axiom $\forall x \in \mathbb{F} \ \ \ \ x \cdot 1 = x $. But $x \cdot 0 = 0$ by above. Contradiction.

$0 \ne 1 \\ 0 < 1$

QED

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Let x $\in$ R and $ x>0 $. Now, $x.1 = x >0 $ $\implies$ $ x^{-1}.x.1 >x^{-1}. 0 $ $\implies 1>0 $ [$\because a.0 = 0 $ $ \forall $a $\in $ R $]$