Question

Prove that sqrt(3) is an irrational number

Answer 1

Assume 3 = a^2/b^2 (a, b in lowest terms) and try to get a contradiction.

Answer 2

http://in.answers.yahoo.com/question/index?qid=20070526070353AAufN0M

Answer 3

If I'm doing this right, I get to a^2/b^2=3b^2/b^2, conclude that because this can be simplified the fraction a/b is not in its lowest terms. But this would work the same if you used 4 instead of 3.

Answer 4

http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm

Answer 5

ok @dr25161

Answer 6

fine @dr25161

Answer 7

thanks. that works

Answer 8

ok welcome

Answer 9

3 = a^2/b^2 -> 3b^2 = a^2

So both a^2 and b^2 are divisible by 3 and not in lowest terms.