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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.

Matthew Conroy
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vondip
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8 Answers8

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

Hagen von Eitzen
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  • Why do you assume that (−1)⋅(−1)>0 ? – vondip Nov 06 '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 06 '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 06 '12 at 16:41
  • product of positives is positive - What is the name of this characteristic of the ordered field? Thanks again! – vondip Nov 06 '12 at 16:42
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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 06 '12 at 16:45
  • @kahen thanks for bear with me. – vondip Nov 06 '12 at 16:57
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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 07 '12 at 06:34
  • Please tell me if I'm wrong but IMO the answer is a circumlocution: $1=1\cdot 1>0$ - done. – Filippo Sep 17 '21 at 06:10
  • @Filippo: yes square of a non-zero member of an ordered field is positive. – Paramanand Singh Sep 17 '21 at 18:15
  • @ParamanandSingh Thank you for the confirmation. – Filippo Sep 19 '21 at 12:05
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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 $1 >0$.

Filippo
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ILoveMath
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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 $.

Filippo
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Francis Pejril
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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 $]$

Satish
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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

kamensky
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  • This is not a true contradiction, in fact you showed that all terms in the field are zero since if we suppose $1=0$ then $ x = x \cdot 1 = x \cdot 0 = 0 $, which is the trivial field $\mathbb{F} = {0} $. To fix it we might say , suppose we are not in the trivial field, and thus there exists $x \neq 0 $, then your proof follows and we derive a contradiction. But the only way to show we are not in a trivial field is to assume $0 \neq 1 $, which is what we want to prove. This circularity cannot be avoided. Thus it is taken as a field axiom that $ 0 \neq 1 $ it is not necessary to prove it. – john Oct 07 '20 at 16:36
  • See (also) [https://math.stackexchange.com/questions/1437123/prove-axiomatically-that-1-does-not-equal-0/1663445] – john Oct 07 '20 at 16:44
  • @john "The trivial field" is not a field. By definition, a field contains more than one element, since $(F\setminus{0},\cdot)$ is a group (in particular, non-empty). – Filippo Sep 19 '21 at 12:16
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It is not provable that $0$ unequals $1$. It is part of the field-axioms.

Armando j18eos
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user565759
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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 $x > 0$

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

Ulfu
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