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




Re: HELP  SQRT(3)
Posted:
Nov 16, 1997 10:36 PM


JeanPierre MERX <JeanPierre.Merx@wanadoo.fr> wrote in article <346DAA97.69E3@wanadoo.fr>... > Emanuel Binnun wrote: > > > > Hi > > > > Please help me to proove that Square root of 3 is an irrational number > > > > 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 (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?
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 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).
What does "3p" 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) ).



