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## Friday, June 23, 2017

### #MathChops Episode 2: Proof That the Irrationals Are a Dense Set Within the Reals

The first conception of this episode was to prove that the rationals are a dense within the reals, which is an algebraic proof showing that between any two real numbers, there is a rational number. This proof does not define the real numbers, and treats them as some empirical fact that you know; yet, once the real numbers are constructed, the proof is really trivial. The proof used in this episode utilizes an analytic definition of dense sets: if a set A’ along with its limit points equals the B’, then A’ is a dense set within B’. You will see that we construct the reals in such a way that the rationals are dense within the reals. But first, a little background.

First, we construct the natural numbers using Peano’s Axioms, and the integers can be constructed many different ways from the natural numbers (think including additive inverses). From the integers, the rational numbers are all ratios of two integers. These ratios can be thought of as finite decimal expansions, and we will construct the real numbers using Dedekind cuts. To define a real number, we chop the number line at the end of an infinite decimal expansion, and call the set of all rational numbers less than that cut the real number. Now of course, this is defining any real number as the limit point of a rational sequence, making the closure of the rationals (the rationals along with their limit points) the reals. The proof that the irrationals are a dense set within the reals is less obvious.

 Construction of the rationals, from AMS.
We need to find an irrational sequence that converges to a rational number (let’s choose 1,  and get any rational number by multiplying our sequence). After some thought, the sequence
a_n = 1 + \frac{\sqrt{2}}{n} is a sequence of irrational terms whose limit point is a rational number. Thus, the irrationals are a dense set within the reals.

Proof