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For ${\displaystyle A_1=\{a\in \mathbb {Q} :a^{2}<2{\text{ or }}a<0\},} A_2=\{b\in {\mathbb {Q}}:b^{2}>2{\text{ and }}b>0\}$, when coming to prove that there is neither in the class $A_1$ a greatest, nor in the class $A_2$ a least number, I was astonished by the following Dedekind's proof at page 7 of his book that he surprisingly put forward the special expression of $y$ without any clues. I think it is not easy for most of us to come up such a complex expression to aid our proof in practice, so is there a common or general way to prove there is neither in the class $A_1$ a greatest, nor in the class$A_2$ a least number only within the rational number system ? enter image description here

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  • $\begingroup$ I wouldn't say it was magical. One can find it by saying $D=\sqrt 2$ and working through the algebra, pretending (for now) that $\sqrt 2$ exists and has the algebraic properties it should. $\endgroup$ Commented Jun 8, 2017 at 15:06
  • $\begingroup$ @RossMillikan you have to prove the conclusion within the rational number system, $\sqrt 2$ is already out of the scope of rational numbers $\endgroup$
    – iMath
    Commented Jun 8, 2017 at 15:30
  • $\begingroup$ That is true, but you can use it informally to find the expressions Dedekind did. Once you find them, you recast the argument completely in terms of rationals. That was the point of my saying that you pretend it exists. $\endgroup$ Commented Jun 8, 2017 at 15:42

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One approach is to follow the Babylonian square root algorithm. Once the approximation is high, it stays high, so given $x$ with $x^2 \gt 2$ we compute $y=\frac 12(x+\frac 2x)$ Now you can show $y\lt x$ and $y^2 \gt 2$ and we have a rational with $2 \lt y^2 \lt x$. Then take $\frac 2y$ as a rational with $(\frac 2x)^2 \lt (\frac 2y)^2 \lt 2$

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  • $\begingroup$ Thanks, but your method also involved one special expression of $y$, although the expression does make the proof a bit more simple than Dedekind 's $\endgroup$
    – iMath
    Commented Jun 9, 2017 at 8:38
  • $\begingroup$ @iMath: but I gave a motivation one to try it. You don't have to make it up out of the blue, you do some playing to identify an expression that will probably work, then do the proof and see if it does. $\endgroup$ Commented Jun 9, 2017 at 13:47
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I think I have found a general solution From here

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