**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.