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Yes it is. The proof is as follows:

We prove the statement by contradiction i.e. Assume that sqrt(6) is a rational number.
Then there exist positive integers p and q with gcd(p,q) = 1 such that p/q = sqrt(6).
Square both sides: p^2 / q^2 = 6,
p^2 = 6q^2.
Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).
This is only possible if 2 is a factor of p.
Let p =2k. Then k is a positive integer as well.
Thus, 4k^2 = 6q^2,
2k^2 = 3q^2.
As 2 is a factor of the LHS, 2 is also a factor of the RHS.
But this is only possible if 2 is a factor of q.
=> gcd(p,q) >= 2. Contradiction!

Thus sqrt(6) is irrational.



yes it is

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Q: Is the square root of 6 an irrational number?
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