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

 Messages: [ Previous | Next ]
 William L. Bahn Posts: 555 Registered: 12/6/04
Re: HELP - SQRT(3)
Posted: Nov 16, 1997 10:36 PM

Jean-Pierre MERX <Jean-Pierre.Merx@wanadoo.fr> wrote in article
> 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.

I'm having a hell of a time following this.

First, what does it mean that since p and q have no common divisor that
p^q=1?

What if p=3 and q=4? They have no common divisor yet 3^4 definitely does
not equal 1.

Then you say, "Because 3 is a prime number 3|p." What does this mean? Are
you trying to say that since 3 is a prime number that 3/p cannot be an
integer if p is an integer (and you never did state that p and q are
integers, but it's probably reasonable for us to assume that's what you
meant).

What does "3|p" mean? Are you saying that 3 is evenly divisible by p? Or
that p is evenly divisible by 3?

And just because 3 divides p^2 doesn't mean that it divides p. After all,
12 divides 6^2 just fine but it doesn't divide 6? I agree that the fact
that 3 is prime precludes these cases. But sqrt(8) is also an irrational
number and we should be able to use the same procedure using 8 instead of 3
- i.e., don't insist that the user recognize that sqrt(8) = 2sqrt(2) ).

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