Prove formally that $a \le x \le a \Rightarrow x = a$ for $x, a \in \mathbb R$ not using contradiction.

Question

Prove formally that $a \le x \le a \Rightarrow x = a$ for $x, a \in \mathbb R$ not using contradiction.

I've been thinking how to prove the above statement not using contradiction.

My idea for a proof by contradiction is as follows: Suppose for contradiction that $x \neq a$ then the equality fails so $x = a$.

Suppose $a,x$ are integers or rational numbers, then proving the above statement for real numbers prove the statement for these classes of numbers also, since we can view them as real numbers ? Or should one be careful, since the ordering is defined different ?

Answer 1

$x\le a$ means $x<a$ or $x=a$. But $x\ge a$ means $x>a$ or $x=a$, so by trichotomy it is not true that $x<a$. Hence $x=a$.

Answer 2

$\le$ is a total order relation. One of the axioms of these is antisymmetry:

If a ≤ b and b ≤ a then a = b

Apply this to $a\le x$ and $x\le a$.

Answer 3

Let us start with two definitions

Definition: Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following two properties

1. If $x\in S$ and $y\in S$ the one and only one of the statements$$x<y,\quad y<x,\quad x=y$$ is true.
2. If $x,y,z\in S$, if $x<y$ and $y<z$, then $x<z$.

Definition: An ordered set is a set $S$ in which order is defined.

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Note that $\mathbb{R}$ is an ordered set.

We have $a,x\in \mathbb{R}$ satisfying $a\le x \implies a<x$ or $a=x$ and $x\le a \implies x<a$ or $a=x$.

From the definition of order one and only one among the statement can be true and as $a=x$ is common to both implication that is the one statement which is true