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




Re: HELP  SQRT(3)
Posted:
Nov 17, 1997 5:29 AM


> > > 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 (3p). Because 3 is a prime number 3p. > > So p=3p' and 9p'^2=3q^2 or 3p'^2=q^2. From what you can derive that > > 3q 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 3p." 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). ab (a,b integers and a>0) means b is a multiple of a. Thus b=ar with r some integer. > What does "3p" 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 3p => 3p^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 2t and 2p, thus gcd(t,p)=2. Contradiction. Hence sqrt(2) is not rational. But then, sqrt(8) can't be rational.
Wilbert



