Use proof by contradiction. Assume sqrt(2) is rational.
This means that sqrt(2) = p/q for some integers p and q, with q <>0.
We assume p and q are in lowest terms.
Square both side and we get:
2 = p^2/q^2
p^2 = 2q^2
This means p^2 must be an even number which means p is also even since the square of an odd number is odd.
So we have p = 2k for some integer k. From this, it follows that:
2q^2 = p^2 = (2k)^2 = 4k^2
2q^2 = 4k^2
q^2 = 2k^2
q^2 is also even, therefore q must be even.
So both p and q are even.
This contradicts are assumption that p and q were in lowest terms.
So sqrt(2) cannot be rational.
This means that sqrt(2) = p/q for some integers p and q, with q <>0.
We assume p and q are in lowest terms.
Square both side and we get:
2 = p^2/q^2
p^2 = 2q^2
This means p^2 must be an even number which means p is also even since the square of an odd number is odd.
So we have p = 2k for some integer k. From this, it follows that:
2q^2 = p^2 = (2k)^2 = 4k^2
2q^2 = 4k^2
q^2 = 2k^2
q^2 is also even, therefore q must be even.
So both p and q are even.
This contradicts are assumption that p and q were in lowest terms.
So sqrt(2) cannot be rational.
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