How to prove √X is irrational number

AI Thread Summary
To prove that √X is irrational when X is an odd number, it is established that √X is irrational if X is not a perfect square (n^2 for any integer n). The discussion emphasizes using the Fundamental Theorem of Arithmetic, assuming p^2/q^2 = x with gcd(p, q) = 1, to explore divisibility. Clarification on gcd indicates it refers to the greatest common divisor, ensuring the fraction is in lowest terms. Adapting the proof for √2 can help in understanding the general case. The conversation suggests further reading on square roots and the history of irrational numbers for deeper insights.
SOHAWONG
Messages
16
Reaction score
0
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
 
Mathematics news on Phys.org
\sqrt{4} is irrational?
 
Hurkyl said:
\sqrt{4} is irrational?
no,i may add despite 1,4,9,16,25...etc
 
So in other words...

\sqrt{x} is irrational iff x=/=n^2 for n belonging to the integer set.
 
Char. Limit said:
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:
 
Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.
 
Tinyboss said:
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?
 
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.
 
Back
Top