Forbiddensinner SG Anyone know how to prove that sqrt 3 is irrational via contradiction?Proving for the sqrt of an even integer is easy, but I haven't figure out how so for sqrt of odd integers.Thanks in advance.
eagle SG Assume that 3 = p^2/q^2 where p and q are integers and p/q is in lowest terms. So 3 q^2 = p^2 So p^2 is divisible by 3. That means that p must be as well, so p^2 is divisible by nine. So q^2 = p^2/3 and q^2 is divisible by three. But that means that p and q are both divisible by three, so they weren't in lowest terms, which is a contradiction.Taken from website http://mathforum.org/library/drmath/view/52619.html
Forbiddensinner SG Assume that 3 = p^2/q^2 where p and q are integers and p/q is in lowest terms. So 3 q^2 = p^2 So p^2 is divisible by 3. That means that p must be as well, so p^2 is divisible by nine. So q^2 = p^2/3 and q^2 is divisible by three. But that means that p and q are both divisible by three, so they weren't in lowest terms, which is a contradiction.Taken from website http://mathforum.org/library/drmath/view/52619.htmlThanks.
