Deduce $\mathbb Q$ is a dense set $($in real numbers$)$

Question

I am trying to deduce $\mathbb Q$ is a dense set $($in real numbers$)$ i.e. $x, y \in\mathbb Q$, there exists $\alpha$ in $\mathbb Q$ such that $x < \alpha < y$. I have let $x$ and $y$ be real numbers s.t. $x<y$.

Then $x-2^{1/2}$ and $y-2^{1/2}$ are real numbers. There is a real number $r$, $x-2^{1/2}<r<y-2^{1/2}$. If you add $2^{1/2}$ to both sides then $x<r+2^{1/2}<y$. Since $2^{1/2}$ is irrational the $r+2^{1/2}$ is irrational. Thus there is an irrational number $\alpha$ s.t $x<\alpha<y$. Hence $\mathbb Q$ is a dense set in the real numbers.

I feel like I am missing a step or two at the end to get to the conclusion. If someone could help that would be great.

Answer 1

You have to use the definition of reals that a number $x,y$ are a real numbers if there are sequences of rationals $q_i \to x$ and $r_i \to y$.

You need to find some $q_i$ and $r_j$ where the average of $q_i, r_j$ is between $x$ and $y$. In other words, so that $x < \frac {q_i + r_j}2 < y$. Then as $\frac {q_i + r_j}2 \in \mathbb Q$ you will be done.

Now for any $\epsilon$ we can find $|q_i - x| < \epsilon$ and $|r_j - x| < \epsilon$.

This means that $x-\epsilon < q_i < x + \epsilon$ and $y - \epsilon < r_j < y + \epsilon$ and so

$\frac {x+y}2 -\frac \epsilon 2 < \frac {q_i + r_j}2 < \frac {x+y}2 + \frac \epsilon 2$.

How can we use that to guarantee that $x < \frac {q_i + r_j}2 < y$?

By making sure that $x < \frac {x+y}2 -\frac \epsilon 2$ and $y < \frac {x+y}2 +\frac \epsilon 2$.

And we do that by making sure $2x < x + y - \epsilon$ and $2y < x+y + \epsilon$ or in other words $\epsilon < y-x$.

....

So....

If we let $\epsilon < y -x$ then we can find a $q_i$ so that $x-\epsilon < q_i <x + \epsilon$ and $r_j$ so that $y-\epsilon < r_j < y + \epsilon$.

$x+y -\epsilon < q_i + r_j < x+y + \epsilon $

$(x+y) - (y-x) < x+y -\epsilon < q_i + r_j < x+y + \epsilon < (x+y) + (y-x)$

$2x < q_i + r_j < 2y$

$x < \frac {q_i + r_j}2 < y$.

And $\frac {q_i + r_j}2 \in \mathbb Q$.