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

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 Wilbert Dijkhof Posts: 201 Registered: 12/8/04
Re: HELP - SQRT(3)
Posted: Nov 17, 1997 5:29 AM
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> > > Please help me to proove that Square root of 3 is an irrational number
> > 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?

I think he meant gcd(p,q)=1 (greatest common divisor of p and q is 1), although
I'm not familiar with his notation p^q=1 (if no typing error of course).

> 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).

Yes, p and q integers and (q<>0).
a|b (a,b integers and a>0) means b is a multiple of a. Thus b=ar with r
some integer.

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

The latter.

> 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) ).

Indeed the trick is that 3 is prime, thus 3|p => 3|p^2.
Suppose we look at sqrt(8), thus q^2=8*p^2=2(2p)^2.
The right side is even, so q^2 and thus q must be even. Suppose q=2t
(with t integer) then you get (2t)^2=2(2p)^2 and thus t^2=2p^2.
So you reduced it to a prime case (ie the irrationality of 2).

I think you are familiar for the proof that sqrt(2) is not rational, but I
will state it: Suppose gcd(t,p)=1 and t,p integers with p<>0 and sqrt(2)
is rational (ie sqrt(2)=t/p).
Now 2p^2 is even and thus t^2 and t is even. Suppose t=2v, then you get
(2v)^2=2p^2 and thus 2v^2=p^2. Similar, the left side is even, so is p^2 and thus
p. Thus 2|t and 2|p, thus gcd(t,p)=2. Contradiction. Hence sqrt(2) is not rational.
But then, sqrt(8) can't be rational.

Wilbert

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

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