Proof that Sqrt(3) is IrrationalDate: 08/14/97 at 14:03:47 From: Stephen Gardner Subject: Proof: no Q expr. for sqrt(3) I am a third year pure mathematics major, and I have a question pertaining to number theory. I know a proof why sqrt(2) cannot be expressed as a rational number; however, it involves the use of "evenness" of the denominator and numerator; i.e., it suggests that if sqrt(2) = a/b, and we let gcd(a,b) = 1; i.e., the fraction is in its reduced (simplest) form, then 2 = a^2 / b^2, a^2 = 2b^2, therefore 2 | a^2, therefore 2 | a. Then we can express a as 2c => a^2 = 4c^2 = 2b^2 => b^2 is ever => 2 | b => contradiction. I was excited when I learned this proof, and naturally it led to the question, How does one prove that sqrt(3) is irrational? or others? Is there a general algorithm for proving such things? How about just for primes? Date: 08/14/97 at 19:40:01 From: Doctor Anthony Subject: Re: Proof: no Q expr. for sqrt(3) A proof can be conducted along much the same lines as when proving that sqrt(2) is irrational. Let sqrt(3) = a/b where a and b are integers and co-prime. (fraction in its lowest terms) squaring 3 = a^2/b^2 and so a^2 = 3b^2 Now a^2 must have its factors as even powers of primes. So it must have 3^2 as one of its factors, and so a must have 3 as a factor. Now 3b^2, as just seen, has a factor 3^2, this means that b^2 has a factor 3. But if b^2 has a factor 3, then since powers must be even, it has 3^2 as a factor. This means that b must have 3 as a factor. So we have shown that both a and b have 3 as a factor. But we started by saying that a/b was a fraction in its lowest terms, (i.e. no common factors of a and b). Thus we have a contradiction, and it follows that we cannot express sqrt(3) as a rational fraction. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/