1. Apr 22, 2014

### mesa

1. The problem statement, all variables and given/known data
I don't take issue with the 'cuts' but am having difficulty understanding an example given in my text for √2. After going over the right and left classes the text jumps into,

"Then this section is such that the R-class has no least member and the L-class has no greatest member; for, if 'x' be any positive rational fraction and,

$$y=\frac{x(x^2+6)}{3x^2+2}$$ then $$y-x=\frac{2x(2-x^2)}{3x^2+2}$$ and $$y^2-2=\frac{(x^2-2)^3}{(3x^2+2)^2}$$, so x^2,y^2 and 2 are in order of magnitude"

2. Relevant equations
Given above

3. The attempt at a solution
A look on wikipedia shows that the last part (where it states, '...are in order of magnitude') seems to represent x^2<y^2<2, other than that I don't understand what the author is getting at or where these equations come from.

***EDITED*** To correct a mistake pointed out by Petek.

Last edited: Apr 22, 2014
2. Apr 22, 2014

### Petek

There appears to be a typo in your above equation for $y^2$. Both $x$ and $y$ are supposed to be rational numbers. However, if $x=1$, then $$y^2= \frac{-1}{25}$$ so $y$ can't even be real.

3. Apr 22, 2014

### mesa

You are correct, I missed a minus 2. Thanks for the correction!

4. Apr 22, 2014

### jbunniii

I think the author is asserting in somewhat obscure (Hardyish?) language that either $x^2 < y^2 < 2$ or $2 < y^2 < x^2$, depending on whether $x \in L$ or $x \in R$.

To see that this is true, consider the two possibilities:

Case 1: $x \in L$

Then $x^2 < 2$, so the formulas for $y-x$ and $y^2 - 2$ imply that $y-x > 0$ and $y^2 - 2 < 0$.

Case 2: $x \in R$

Then $x^2 > 2$, so the formulas imply that $y-x < 0$ and $y^2 - 2 > 0$.