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## Top 10 Wanted Proofs

 Frobenius's Theorem --Matt Westwood 18:38, 8 November 2008 (UTC) Wedderburn's Theorem --Matt Westwood 18:38, 8 November 2008 (UTC) Whitney Immersion Theorem --Matt Westwood 08:09, 18 January 2009 (UTC) Whitney Embedding Theorem --Matt Westwood 08:09, 18 January 2009 (UTC) Topological h-Cobordism Theorem --Matt Westwood 08:09, 18 January 2009 (UTC) Thurston's Geometrization Conjecture --Matt Westwood 08:09, 18 January 2009 (UTC) Tartaglia's Formula --Matt Westwood 06:39, 15 March 2009 (UTC) Burnside's Theorem --Joe (talk) 16:50, 16 March 2009 (UTC) Abel-Ruffini Theorem [1] --Matt Westwood 20:45, 16 March 2009 (UTC) Central Limit Theorem --HrMeyer 20:53, 23 April 2009 (UTC)

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## Proof of the Week

Between Every Two Reals Exists a Rational

## Theorem

Let $a, b \in \R$ be real numbers such that $a < b$.

Then:

$\exists r \in \Q: a < r < b$

### Corollary: Reals are Close Packed

$\exists c \in \R: a < c < b$

## Proof

Suppose that $a \ge 0$.

As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.

Thus:

$\dfrac 1 {b - a} \in \R$

By the Archimedean Principle:

$\exists n \in \N: n > \dfrac 1 {b - a}$

Let $M := \left\{{x \in \N: \dfrac x n > a}\right\}$.

By the Well-Ordering Principle, there exists $m \in \N$ such that $m$ is the smallest element of $M$.

That is:

$m > a n$

and, by definition of smallest element:

$m - 1 \le a n$

As $n > \dfrac 1 {b - a}$, from Ordering of Reciprocals, it follows that $\dfrac 1 n < b - a$

Thus:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle m - 1$$ $$\le$$ $$\displaystyle$$ $$\displaystyle a n$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle m$$ $$\le$$ $$\displaystyle$$ $$\displaystyle a n + 1$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \frac m n$$ $$\le$$ $$\displaystyle$$ $$\displaystyle a + \frac 1 n$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$<$$ $$\displaystyle$$ $$\displaystyle a + \left({b - a}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle b$$ $$\displaystyle$$ $$\displaystyle$$

Thus we have shown that $a < \dfrac m n < b$.

That is:

$\exists r \in \Q: a < r < b$

such that $r = \dfrac m n$.

Now suppose $a < 0$.

If $b > 0$ then $0 = r$ is a rational number such that $a < r < b$.

Otherwise we have $a < b \le 0$.

Then $0 \le -b < -a$ and there exists $r \in \Q$ such that:

$-b < r < -a$

where $r$ can be found as above.

That is:

$a < -r < b$

All cases have been covered, and the result follows.

$\blacksquare$