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Density of Irrational Numbers Theorem

Given any two real numbers α, β in R, α<β, there is an irrational number γ such that α<γ<β.

Proof: We will first show that if r and s, s non-zero, are rational, then r+s√2 is irrational. Let r and s, s non-zero, be rational numbers. Suppose r+s√2 were rational. Then it could be expressed in some fashion as x/y, where x and y are integers. In addition, since r is rational, it can be expressed as a/b, where a and b are integers, and s can be expressed as c/d, where c and d are integers. Then:

{(x/y)=(a/b)+(c/d) √2}*(ybd)
bdx=yda+cby√2
bdx-yda=cby√2
(bdx-yda)/cby=√2

However, we know that √2 is irrational, so we have our contradiction. r + s √2 is irrational.

Since α<β, β-α>0. √2>0 as well. We may use the Archimedean property to conclude that there is some integer m so √2<m(β-α), or equivalently,

mα+√2<mβ.

Let n be the largest integer such that n≤mα. Adding √2 to both sides gives

n+√2≤mα+√2<mβ.

But since n is the largest integer less than or equal to mα, we know that mα<n+√2 and therefore that

mα<n+√2<mβ or

α<(n+√2)/m<β.

We know from our above results that (n/m)+(1/m)√2 cannot be rational, so we have shown that there is an irrational number between α and β.

NOTE: Please see Euclid's Proof that 2^.5 is Irrational for an explanation of the irrational nature of √2.

I like this proof because it’s simplistic and low on vocabulary; I suppose it’s more of a hoi polloi-ish proof than the professors would prefer we use. The Archimedean property is the most sophisticated tool you need to understand this, and there’s a good write-up on that. This proof is fantastic for someone being introduced to the study of analysis or a non-major “stuck” taking a single semester of the stuff.

Taken from a homework assignment from a class titled "Fundamental Properties of Spaces and Functions: Part I" at the University of Iowa.

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