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Topic: HELP - SQRT(3)
Replies: 16   Last Post: Dec 1, 1997 7:24 PM

 Messages: [ Previous | Next ]
 Zdislav V. Kovarik Posts: 3,419 Registered: 12/6/04
Re: HELP - SQRT(3)
Posted: Nov 21, 1997 3:30 PM

In article <346CC27B.398@netvision.net.il>,
Emanuel Binnun <e_binnun@netvision.net.il> wrote:
>Hi
>
>

Adding a few extra lines to the standard proof, we can get a
"constructive" proof of the irrationality of sqrt(3): for every (positive)
rational number x, we can find another positive rational number r such
that abs(x - sqrt(3)) > r.

Consider x = p/q where p, q are positive integers, which we can choose
without a common factor (greater than 1). We first show tat

(*) abs(p^2 - 3 * q^2) >= 1.

If to the contrary, abs(p^2 - 3 * q^2) < 1 then p^2 = 3 * q^2.

So, discussing the cases p=3n, p=3n+1, p=3n+2 separately, we find that p
must be divisible by 3: p=3n. Then we cancel 3's and obtain
q^2 = 3 * n^2, and q would be divisible by 3. We would see that p and q
are both divisible by 3, contrary to our assumption (no common factor).

Here the standard proof concludes that p^2/q^2 is never 3. We go further:
Rationalize at the right moment and use sqrt(3) < 2 to get

abs(p/q - sqrt(3)) = abs((p - q * sqrt(3))/q)
= abs((p^2 - 3 * q^2) / (q * (p + q * sqrt(3))))
> 1/(q*(p+2*q))

Now r = 1/(q*(p+2*q)) satisfies the claim.

(We showed that p/q is computably too far from sqrt(3) to have a chance
to be equal to it.)

(For example, test p=5042, q=2911: your calculator with 8-digit display
shows p/q approximately 1.7320508, same as the 8-digit approximation of
sqrt(3). But in fact,

abs(5042/2911 - sqrt(3)) > 1/(2911*10864) = 1/31625104.)

Cheers, ZVK (Slavek).

Date Subject Author
11/14/97 EmanuelBinnun
11/14/97 Norman Walker
11/17/97 Robin Chapman
11/21/97 r3769@aol.com
11/21/97 Robin Chapman
12/1/97 Mike McCarty
11/15/97 Jean-Pierre MERX
11/16/97 William L. Bahn
11/17/97 Sam @ The NIMP Team
11/17/97 Wilbert Dijkhof
11/17/97 Sam @ The NIMP Team
11/17/97 William L. Bahn
11/18/97 Sam @ The NIMP Team
11/17/97 Wilbert Dijkhof
11/17/97 Pete Vermeire
11/18/97 Brian Hutchings
11/21/97 Zdislav V. Kovarik