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From: Glowing Face Man on 8 Jan 2010 09:49
Hi,

I'm trying to formally prove that
S = { x in Q : x < 0 or x^2 < 2 }
is the square root of 2, in the Dedekind Cut sense.
This is claimed without proof all over the place, e.g. in the
Wikipedia article "Construction of the real numbers".

By definition, S^2 = { xy : x,y >= 0 are in S } U { x in Q : x < 0 }.

And 2 = { x in Q : x < 2 } (the 2 on the RHS is the *rational* number
2)

Then 2 contains the number 13/7. But 13/7 cannot possibly be in S^2.
It would have to factor as one of
13/7 * 1/1, 13/1 * 1/7, 1/7 * 13/1, or 1/1 * 13/7.
None of these factorizations put it in S^2. It seems like 13/7 is NOT
contained in S^2, so S^2 not= 2.

Am I doing something wrong? Is Wikipedia doing something wrong?

Thanks
-Sam
From: G. A. Edgar on 8 Jan 2010 10:11
In article
<b45c604f-313c-488d-b2b8-37da5eee630a(a)z7g2000vbl.googlegroups.com>,
Glowing Face Man <glowingfaceman(a)gmail.com> wrote:

> Hi,
>
> I'm trying to formally prove that
> S = { x in Q : x < 0 or x^2 < 2 }
> is the square root of 2, in the Dedekind Cut sense.
> This is claimed without proof all over the place, e.g. in the
> Wikipedia article "Construction of the real numbers".
>
> By definition, S^2 = { xy : x,y >= 0 are in S } U { x in Q : x < 0 }.
>
> And 2 = { x in Q : x < 2 } (the 2 on the RHS is the *rational* number
> 2)
>
> Then 2 contains the number 13/7. But 13/7 cannot possibly be in S^2.
> It would have to factor as one of
> 13/7 * 1/1, 13/1 * 1/7, 1/7 * 13/1, or 1/1 * 13/7.
> None of these factorizations put it in S^2. It seems like 13/7 is NOT
> contained in S^2, so S^2 not= 2.
>
> Am I doing something wrong? Is Wikipedia doing something wrong?
>
> Thanks
> -Sam

13/7 = (4/3) * (39/28) right?

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
From: Glowing Face Man on 8 Jan 2010 10:12
On Jan 8, 10:11 am, "G. A. Edgar" <ed...(a)math.ohio-state.edu.invalid>
wrote:
> In article
> <b45c604f-313c-488d-b2b8-37da5eee6...(a)z7g2000vbl.googlegroups.com>,
> Glowing Face Man <glowingface...(a)gmail.com> wrote:
>
>
>
> > Hi,
>
> > I'm trying to formally prove that
> > S = { x in Q : x < 0 or x^2 < 2 }
> > is the square root of 2, in the Dedekind Cut sense.
> > This is claimed without proof all over the place, e.g. in the
> > Wikipedia article "Construction of the real numbers".
>
> > By definition, S^2 = { xy : x,y >= 0 are in S } U { x in Q : x <  0 }.
>
> > And 2 = { x in Q : x < 2 } (the 2 on the RHS is the *rational* number
> > 2)
>
> > Then 2 contains the number 13/7.  But 13/7 cannot possibly be in S^2.
> > It would have to factor as one of
> > 13/7 * 1/1,  13/1 * 1/7,  1/7 * 13/1,  or 1/1 * 13/7.
> > None of these factorizations put it in S^2.  It seems like 13/7 is NOT
> > contained in S^2, so S^2 not= 2.
>
> > Am I doing something wrong?  Is Wikipedia doing something wrong?
>
> > Thanks
> > -Sam
>
> 13/7 = (4/3) * (39/28) right?
>
> --
> G. A. Edgar                              http://www.math.ohio-state.edu/~edgar/

Ahh... How stupid I feel now! :P
From: Glowing Face Man on 11 Jan 2010 21:28
On Jan 8, 10:11 am, "G. A. Edgar" <ed...(a)math.ohio-state.edu.invalid>
wrote:
> In article
> <b45c604f-313c-488d-b2b8-37da5eee6...(a)z7g2000vbl.googlegroups.com>,
> Glowing Face Man <glowingface...(a)gmail.com> wrote:
>
>
>
> > Hi,
>
> > I'm trying to formally prove that
> > S = { x in Q : x < 0 or x^2 < 2 }
> > is the square root of 2, in the Dedekind Cut sense.

I'm still finding this surprisingly hard to do directly.
It boils down to showing: if 0<x<2 is rational, then x=pq for some
p,q>0 rational with p^2,q^2<2.

Of course, to establish that sqrt(2) exists, it's enough to backtrack
out of this direct approach, prove that Dedekind cuts define an
ordered field with least upper bound property, and then argue that sup
{x in R:x^2<2} is sqrt(2). But I'd really like to get a more direct
proof. I'm having a lot of trouble proving the above claim without
circularly assuming that sqrt(2) exists. Can anyone give me some
pointers or just tell me how the proof would go? Thanks :)

It's not homework.
-Sam
From: David Bernier on 12 Jan 2010 03:25
Glowing Face Man wrote:
> On Jan 8, 10:11 am, "G. A. Edgar" <ed...(a)math.ohio-state.edu.invalid>
> wrote:
>> In article
>> <b45c604f-313c-488d-b2b8-37da5eee6...(a)z7g2000vbl.googlegroups.com>,
>> Glowing Face Man <glowingface...(a)gmail.com> wrote:
>>
>>
>>
>>> Hi,
>>> I'm trying to formally prove that
>>> S = { x in Q : x < 0 or x^2 < 2 }
>>> is the square root of 2, in the Dedekind Cut sense.
>
> I'm still finding this surprisingly hard to do directly.
> It boils down to showing: if 0<x<2 is rational, then x=pq for some
> p,q>0 rational with p^2,q^2<2.
>
> Of course, to establish that sqrt(2) exists, it's enough to backtrack
> out of this direct approach, prove that Dedekind cuts define an
> ordered field with least upper bound property, and then argue that sup

If S_1, S_2, ... S_i .... are left Dedekind cuts with an upper bound L
in the reals, then the union of S_1, S_2 ... S_i ... , T, will be
a left Dedekind cut which is a l.u.b. for S_1, S_2, ... S_i ...

Actually, if we look at:
<
http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_by_Dedekind_cuts>

Condition 4.:
"r contains no greatest element."

Accordingly, if the union of S_1, S_2, ... S_i ... := T has no greatest
element, then T is a Dedekind cut. After some thought, I believe that
if S_1, ... S_i satisfy condition 4., then T will automatically satisfy
condition 4.

What you propose to do next, "prove that Dedekind cuts define an
ordered field with least upper bound property" is the same as
proving the correctness of one construction of the real numbers ...
( it takes quite a bit of time).


> {x in R:x^2<2} is sqrt(2). But I'd really like to get a more direct
> proof. I'm having a lot of trouble proving the above claim without
> circularly assuming that sqrt(2) exists. Can anyone give me some
> pointers or just tell me how the proof would go? Thanks :)

Although S = { x in Q : x < 0 or x^2 < 2 }, members of S such as
-1000 have to be treated with care when taking the square of S.

Actually, since S contains the rational number 1, S >= 0, or
S is "positive" in the notation used in defining multiplication
of Dedekind cuts:

<
http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_by_Dedekind_cuts>

So the definition of AxB for A, B >= 0 can be used.


Suppose S' = {x in Q: x>=0 and x^2 < 2}. If y is in S', is there
a limit on how close y^2 can get to 2? [no]

You could look at integer squares less than 200, 20,000, 2,000,000.

For example, 1414^2 = 1,999,396 so:
(1414/1000)^2 ~= 1.999396 < 2.0 . Also,
141421356237309504880^2 = 19999999999999999999952235666390743814400 or
(141421356237309504880/100000000000000000000) ~=
1.9999999999999999999952 < 2 .

Besides, ((n+1)/n)^2 = (1 + 1/n)^2 = 1+ 2/n + 1/(n^2) .
So lim_{n -> oo} ((n+1)/n)^2 = 1.

This suggests that squares of positive rationals can be
in ]2-1/n, 2[ for any integer n>=1.


There remains to prove that some Dedekind cut D has
DxD = (-oo, 2[ . Starting from S', you can form
D = { y in Q s.t. y<= x, for some x in S'}.

David Bernier



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