How to prove this statement about the ordered field of the real number?

Question

There isn't a number $a\in\mathbb{R}$ such that $x\leq a$; for all $x\in\mathbb{R}$. I think the proof would have to be done by contradiction because if the inequality is correct that implies that the numbers and not equivalent, and if the equality then it is not and inequality.

Answer 1

Like stated in the comments, let's prove by contradiction.

So suppose there is an $a \in \mathbb{R}$ where $\forall{x \in \mathbb{R}}, x \le a $

Now notice that $a + 1 \in \mathbb{R}$ because, by the axioms of a field, $\mathbb{R}$ is closed under addition. And hopefully it is clear to see that $a + 1 \gt a$. This is a contradiction and completes the proof.