Density of Q in R
For the definitions of an upper bound and the least upper bound of a set, see this.
For the Archimedean Principle, see this
Proof:
Let \(a\) and \(b\) be real numbers with with \(a < b\). We will produce \(q = \frac{m}{n} \in \mathbf{Q}\) where \(m \in \mathbf{Z}\) and \(n \in \mathbf{N}\) such that
We want to pick \(n\) such that it is large enough where a step of size \(1/n\) won’t cross over the entire interval of \([a,b]\). (side note: think of the numbers 0.5 and 1. You don’t want \(n\) to be 1. since any fraction \(m/1\) is never going to be inside the interval). To pick such \(n\), we can use the Archimedean Principle to pick an \(n \in \mathbf{N}\) such that \(n > \frac{1}{b - a}\). Re-arrange the terms to get
So now all we need is to find \(m\) such that \(a < \frac{m}{n} < b\). Multiply the inequality by \(n\) to get
Clearly \(m\) needs to be an integer greater than \(an\) and less than \(bn\). The idea in the book’s proof was to pick \(m\) to be the smallest integer greater than \(na\). In other words,
So now we need to prove that this choice of \(m\) will satisfy the inequality \(an < m < bn\). Clearly, \(na < m\) satisfies the right inequality and we’re done with part. Next, we want to see how \(m - 1 \leq na\) will satisfy \(m < bn\). To see this, we will re-write the inequality \(\frac{1}{n} > b - a\) to isolate \(a\)
Now we can plug this in
From this we see that this choice of \(m\) indeed satisfies \(an < m < bn\) as required.
\(\blacksquare\)
Notes: So in a sense \(Q\) must be dense since we have a rational number between no matter what two real numbers are given.
References: