Rational Cut has Smallest Upper Number

From ProofWiki
Jump to navigation Jump to search


Let $r \in \Q$ be rational.

Let $\alpha$ be the rational cut consisting of all rational numbers $p$ such that $p < r$.

Then $\alpha$ is indeed a cut, and has a smallest upper number that is $r$.


That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a cut follows directly from that definition.

Then it is noted that if $p \in \alpha$ then from Mediant is Between:

$p < \dfrac {p + r} 2 < r$

and so $p$ cannot be the greatest element of $\alpha$.

Hence it is seen that $\alpha$ fulfils all conditions to be a cut.

Because it is not the case that $r < r$, it follows that $r \notin \alpha$.

Thus $r$ is an upper number of $\alpha$.

Aiming for a contradiction, suppose $q \in \Q$ is an upper number of $\alpha$ such that $q < r$.

But then by definition of a cut, $q \in \alpha$.

That is, $q$ is not an upper number of $\alpha$.

Hence by Proof by Contradiction it follows that there is no upper number of $\alpha$ which is smaller than $r$.

That is, $r$ is the smallest upper number of $\alpha$.