$\forall x\in\mathbb{R}\enspace \enspace, \forall y\in\mathbb{R} : (x\ge y)$ or $(x\le y)$ proof? Well, it's obvious that it's true.
but our teacher just proved it by saying "for each $x$ and $y$ we will input, we will receive true element".
My question is - How can I prove it in the best way?
I did the contradiction proof and said
$\exists (x,y)\in\mathbb{R} : (x<y)$ AND $(x>y)$.
I input one time $x < y$ , $x=y$ and $x>y$ numbers $($like $2,3$ , $2,2$ , $2,1$$)$
and showed in truth table that its always false which means the main claim is true.
Is it ok to do so or there are better proofs?
Thank you very much
 A: It comes from definition of Real numbers - Cauchy sequences, Dedekind cuts or other. In any way aim is to have so called The complete ordered field.

*

*Dedekind cuts:

We take rational numbers $\mathbb{Q}$ and consider its $\alpha$ subsets with following properties:
$$\begin{array}{}
\alpha \neq \emptyset, \alpha \neq \mathbb{Q} \\
q \in \mathbb{Q} \land p \in \alpha \land q<p \Rightarrow q \in \alpha \\
p \in \alpha \Rightarrow \exists r \in \mathbb{Q}, p<r
\end{array}$$
Such subsets are called "cuts". We define for cuts "$\alpha < \beta$",  when $\alpha$ is proper subset of $\beta$. Now can be obtained all desired properties of order, including least upper bound property.


*Cauchy sequences:

Again we consider rational numbers $\mathbb{Q}$ with metric $|x-y|$ and look at all Cauchy sequences. We called 2 Cauchy sequence equivalent, when they have same limit. By this we divide $\mathbb{Q}$ into equivalence classes, which we called real numbers. Order for equivalence classes $(x_n) \leqslant (y_n)$ iif $(x_n)$ is equivalent to $(y_n)$ or $\exists N, n>N, x_n \leqslant y_n$. Now again is possible to have all order properties.
