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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?

MathAdam
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  • Check that $x<y$. This uses that $x^2<2$. Then show that $y^2<2$. This also uses that $x^2<2$. – Andrés E. Caicedo May 01 '17 at 13:55
  • @Adam Hrankowski: Is it important to you that this particular choice of $y$ works or can it by any other suitable (and perhaps more accessible) value? – Moritz May 01 '17 at 14:11

1 Answers1

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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 $y^2 < 2$.

mweiss
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