Proof that a certain bounded set has no maximum seems wrong

Question

if $A = (0, 1)$ then $1$ is the supremum, but there is no maximum! To show this, assume for a contradiction that $x \in A$ is the maximum of $A$. Then also $$(1+x)/2 \in A \quad \quad (1)$$ and $$(1+x)/2 > x \quad \quad (2).$$

I don't see how this is necessarily correct.

It is clear that (1) is correct. How do you deduce that $(1+x)/2 > x$?

$1 + x > x$, is true. $(1 + x) > 1$, true. $(1 + x)/2 > 1/2$, true. How does it follow that $(1+x)/2 > x$ ?

Sorry if this is basic question.

Answer 1

From $x<1$ it follows that $2x=x+x<1+x$ and so $x<\tfrac{1+x}{2}$.