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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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8 Answers

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

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

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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,\begin{align} a\cdot a^{-1}&=1<0 \\ &\Downarrow\\ a\cdot a^{-1}\cdot a&<0\cdot a \\ &\Downarrow\\ a&<0 \end{align}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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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 $]$

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

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Using that$$ x \neq 0 \Rightarrow x^2 > 0 $$leads to$$ 1^2 > 0 $$and thereby$$ 1 > 0 $$

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Suppose $0 \ge 1$ and let $x > 0$. We have:$$0 \ge 1$$$$0 \cdot x \ge 1 \cdot x$$$$0 \ge x$$ A contradiction with our assumption that $x > 0$

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