Set of Rationals Greater than Root 2 has no Smallest Element

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Theorem

Let $B$ be the set of all positive rational numbers $p$ such that $p^2 > 2$.


Then $B$ has no smallest element.


Proof

Aiming for a contradiction, suppose $p \in B$ is the smallest element of $B$.

Then by definition of $B$:

$p^2 > 2$


Let $q = p - \dfrac {p_2 - 2} {2 p}$.

Then $q > p$, and:

\(\ds q\) \(=\) \(\ds p - \dfrac {p_2 - 2} {2 p}\)
\(\ds \) \(=\) \(\ds \dfrac p 2 + \dfrac 1 p\)

Hence:

$0 < q < p$

and so:

\(\ds q^2\) \(=\) \(\ds p^2 - \dfrac {p^2 - 2} + \paren {\dfrac {p^2 - 2} {2 p} }^2\)
\(\ds \) \(>\) \(\ds p^2 - \paren {p^2 - 2}\)
\(\ds \) \(=\) \(\ds 2\)

That means $q \in B$.

This contradicts our assertion that $p$ is the smallest element of $B$.

It follows that $B$ can have no smallest element.

$\blacksquare$


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