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I am working on the formal construction of the positive real numbers. I have the positive real number 2 as the set of positive rational numbers less than 2. I have $\sqrt{2}$ as the set of positive rational numbers the squares of which are less than 2.

Multiplication of positive real numbers x and y is defined as the set of all positive rational numbers $z=a\times b$ such that $a\in x$ and $b\in y$.

Question: Using these constructions, how can I prove that the $\sqrt{2}\times\sqrt{2} = 2$?

Any help -- hints or online references -- would be appreciated.

Restating question: If $x$ is a positive rational number less than 2, how can I prove there exists positive rational numbers $a$ and $b$ such that $a\times a\lt 2$, $b\times b \lt 2$ and $x=a\times b$ ?

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    You haven't told us how you are defining multiplication. How do you prove that $2\times2=4$?2012-03-29
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    @Gerry Myerson: See my definition inserted above.2012-03-29
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    OK, you've added the definition of multiplication, but you haven't told us how you prove $2\times2=4$. It may be that if you can answer that question, you can answer the one about $\sqrt2$.2012-03-29
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    For the harder part, we can use the fact that for any positive rational $c<2$, there is a rational $r$ such that $c\le r^2<2$.2012-03-29
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    @AndréNicolas: How do you establish this fact? And to apply it, won't I have to change my definition of multiplication with $z\leq a\times b$?2012-03-30
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    For your restated question, simply pick any $b$ with $\sqrt{x} < b < \sqrt{2}$.2012-03-30
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    @Hurkyl, I think you've misread - OP wants $x\lt2$, not $x\lt\sqrt2$. Also, I suspect OP wants a formula for such $a,b$ in terms of $x$.2012-03-30
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    @Gerry: Ah, you caught my typo before I fixed it. :(2012-03-30
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    Dan, you **still** haven't told me how you prove $2\times2=4$. Doesn't that require the existence, given positive rational $x\lt4$, of positive rationals $a,b$ such that $a^2\lt4$, $b^2\lt4$, and $x=ab$? And is that any easier than the $\sqrt2$ question?2012-03-30
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    @Dan Christensen: Sorry for the delay, was away from computer for quite a while. Have written an answer, specifically for $\sqrt{2}$ but easily modified for $\sqrt{x}$.2012-03-30

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You seem to be using a somewhat non-standard definition of cut; but the idea below can be easily adapted to more standard definitions.

We want to prove that $\sqrt{2}\times \sqrt{2}=2$. Of course we mean that the product is equal to the cut "$2$" but it would be a nuisance to give that a new name.

The easy part is to show that every element of $\sqrt{2}\times \sqrt{2}$ is in $2$. For if $a$ and $b$ are positive rationals such that $a^2<2$ and $b^2<2$, then $ab<2$.

Next we need to show that every positive rational $r$ in the cut $2$ can be expressed as $ab$, where $a$ and $b$ are positive rationals and $a^2<2$, $b^2<2$.

The idea is to show that there is a positive rational $b$ such that $r. Then let $a=r/b$. Certainly $ab=r$. Also, $a^2=r^2/b^2=r(r/b^2)<(2)(1)=2$.

We construct $b$. There are many ways to do the job. For instance we can use the estimates associated with the usual Newton Method procedure for approximating $\sqrt{2}$. However, we choose to use an argument that is closer in spirit to the continuity proof for $\sqrt{x}$.

Let $\epsilon=\min(2-r, 1)$, and let $n$ be a positive integer such that $10n/n^2<\epsilon$. There is a perfect square $m^2$ such that $2n^2-10n\le m^2<2n^2$. Let $b=m/n$. Then $r, and we are done.