4
$\begingroup$

How can one show/prove that $0<1$?

$1$ is not actually defined to be greater than zero, but I think it can be proven. I already know that the real numbers are an ordered field and I am familiar with the field axioms and the definition of a total ordering. However I do not know how to start such a proof.

Any kind of help or advice will be really appreciated.

$\endgroup$
1
  • 5
    $\begingroup$ What is the definition you are using for "<" ? There are different equivalent definitions. In some of the $0< 1$ is actually part of the definition. $\endgroup$
    – Ramiro
    Sep 30, 2015 at 16:42

2 Answers 2

7
$\begingroup$

$1 = 1^2$ and $x^2>0$ if $x\ne 0$.

For the last claim, the main point is that by definition of positive $x,y>0$ implies $xy>0$. This is used as follows:

If $x>0$ then $x^2 = x \cdot x > 0$.

If $x<0$ then $-x > 0 $ and so $x^2=(-x)^2 > 0$.

In turn, this follows from $(-x)y=-(xy)$, which follows from distributivity: $0 = (-x + x)y = (-x)y+(xy)$.

$\endgroup$
4
  • $\begingroup$ For the details, see for instance Chapter 1 of Spivak's Calculus. $\endgroup$
    – lhf
    Sep 30, 2015 at 16:47
  • $\begingroup$ Nice. However, as you pointed out, your proof relies on the property "if $x,y>0$ then $x.y>0$". Since @Fullbright did not specify which definition of $<$ is being used, the property you used may not be "evident". In fact, it may be harder to prove than $0<1$. $\endgroup$
    – Ramiro
    Sep 30, 2015 at 16:52
  • 1
    $\begingroup$ @Ramiro, that's usually in the definition of ordered field. $\endgroup$
    – lhf
    Sep 30, 2015 at 16:54
  • 1
    $\begingroup$ I know. But, for instance, there are equivalent variants of the definition where $0<1$ is part of the definition and the property "if $x,y>0$ then $x.y>0$" is deduced using that $0<1$. $\endgroup$
    – Ramiro
    Sep 30, 2015 at 17:01
2
$\begingroup$

I'll add some details/more information to @lhf answer:

In an ordered field $K$, there is a distinguished subset $P$ such that:

(i) $\{ P,\{0\},-P\}$ is a partition of $K$,

(ii) $\forall x,y\in P$, $\; x+y\in P\;$ and $\;xy\in P$.

In particular, as $\;1=1\cdot 1=(-1)\cdot(-1)$, we see $1\in P$.

We then define a strict order relation on $K$ by setting $x>y$ if $x-y\in P$. In particular, $\;x\in P\iff x>0$.

$\endgroup$
2
  • $\begingroup$ The other approach to defining ordered fields is to say that there is total order that is compatible with addition and with multiplication by positive elements, where positive means > 0. $\endgroup$
    – lhf
    Sep 30, 2015 at 18:09
  • $\begingroup$ They're equivalent, of course, but I personally prefer this one, as I find it more natural. $\endgroup$
    – Bernard
    Sep 30, 2015 at 18:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .