How does $y=\frac{2x+2}{x+2}$ contribute to demonstrating the Dedekind cut for $\sqrt{2}$? In a previous question a respondent corrected my description a Dedekind cut of $\sqrt{2}$. This Wikipedia article states: "Showing that it is a cut requires showing that for any positive rational $x$ with $x^2 < 2$, there is a rational $y$ with $x < y$ and $y^2 < 2$." I can follow this easily enough. I get lost on the next statement: "The choice $y=\frac{2x+2}{x+2}$ works."  How so? I've tried squaring the expression and substituting $2$ for $x^2$, but I'm having trouble seeing it?  Can anyone help me ask through this?

Let $x$ be a positive rational number with $x^2<2$.  First note that
$$x < \frac{2x+2}{x+2}$$
is equivalent to
$$x^2+2x<2x+2$$
which in turn is equivalent to $x^2<2$, which is true by hypothesis.  So $x<y$.
Now note that
$$\left( \frac{2x+2}{x+2} \right)^2< 2$$
is equivalent to
$$(2x+2)^2 < 2(x+2)^2$$
which simplifies to
$$4x^2+8x+4 < 2x^2 +8x + 8$$
which simple algebra shows is equivalent to $x^2 < 2$, again.  So $y^2<2$, as required.
EDITED TO ADD:
There is nothing particularly magic about this specific choice of $y$.  Here, for example, is an alternative way of constructing a (different) rational number $y$ with $x<y$ and $y^2<2$.
Let $x$ be a positive rational number with $x^2 < 2$.  If we define $z = \frac{2}{x}$, then $z > 2$.  Now construct the average of $x$ and $z$, and call that $w$.
Note first that $w$ lies strictly between $x$ and $z$.  Moreover, we have (by the AM-GM inequality) that $2 = xz < w^2$.
Finally, define $y = \frac{2}{w}$.  It's straightforward to verify that $x<y$ and $y^2 < 2$.
If you follow the trail of breadcrumbs, you can show that this choice of $y$ is given by $y = \frac{4x}{x^2 + 2}$.  You can also prove directly from this formula (using methods similar to the ones at the top of this answer) that this choice of $y$ has the properties that $x<y$ and $y^2 < 2$.
