
Re: HELP  SQRT(3)
Posted:
Nov 17, 1997 7:34 AM


On Sat, 15 Nov 1997, JeanPierre MERX wrote:
> 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. > > Regards, JeanPierre. >
This is working a bit too hard for a particular case. All you need to notice is that the square of a noninteger rational number is again a noninteger 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 noninteger rational. Hence sqrt(x) is irrational.

