how to prove √X is irrational number

[SOHAWONG]

when X is even number,it's easy to prove
but how about the condition which X is odd number?
I have no idea of this

[Hurkyl]

\sqrt{4} is irrational?

[SOHAWONG]

\sqrt{4} is irrational?
no,i may add despite 1,4,9,16,25...etc

[Char. Limit]

So in other words...

\sqrt{x} is irrational iff x=/=n^2 for n belonging to the integer set.

[SOHAWONG]

So in other words...

\sqrt{x} is irrational iff x=/=n^2 for n belonging to the integer set.
yes, but how to prove?:confused:

[Tinyboss]

Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.

[SOHAWONG]

Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.

what does gcd mean?

[Tinyboss]

Greatest common divisor. If gcd(p,q)=1, it means the fraction p/q is in lowest terms.

Look at the proof for sqrt(2), and adapt it. Remember that "even" just means "is divisible by 2", so that if you're checking a number other than 2, you won't be thinking about "even" anymore.

[skeptic2]

Although right now you're probably more interested in just getting the right answer, you might want to check out the following Wikipedia entries:
http://en.wikipedia.org/wiki/Square_root
http://en.wikipedia.org/wiki/Square_root_of_2

And for an interesting history of the discovery of irrational numbers look at
http://en.wikipedia.org/wiki/Irrational_number