Maths Question

Forbiddensinner

21 Jan 10, 18:33

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

22 Jan 10, 09:47

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

23 Jan 10, 03:26

Originally posted by eagle:

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

Thanks.