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)

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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