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.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Introduction: $1.1$ Example