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

 Messages: [ Previous | Next ]
 Pete Vermeire Posts: 23 Registered: 12/12/04
Re: HELP - SQRT(3)
Posted: Nov 17, 1997 7:34 AM

On Sat, 15 Nov 1997, Jean-Pierre MERX wrote:

> Emanuel Binnun wrote:
> >
> > Hi
> >
> >
> > Thanks
> > Boaz
> > Israel

> Hi Boaz,
>
> suppose that SQRT(3) is a rationnal number... Then your can write 3 =
> p^2/q^2 where you can suppose that p and q have no commun divisor
> (p^q=1).
> So p^2 = 3*q^2 and 3 divides p^2 (3|p). Because 3 is a prime number 3|p.
> So p=3p' and 9p'^2=3q^2 or 3p'^2=q^2. From what you can derive that
> 3|q contradicting the fact that p^q=1. SO SQRT(3) IS NOT A RATIONNAL.
>
> Regards, Jean-Pierre.
>

This is working a bit too hard for a particular case. All you need to
notice is that the square of a non-integer rational number is again a
non-integer rational number. Then we have:

Let x be a positive integer, and assume sqrt(x) is not an integer (e.g.
x is not 1,4,9,16,25,...). If sqrt(x) were rational, we could square
both sides and get that x is a non-integer rational. Hence sqrt(x) is
irrational.

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