How can we prove that the square root of 2 is an irrational number?

The proof that the square root of 2 is an irrational number is typically presented using a method known as proof by contradiction. This method highlights the absurdity of assuming that the square root of 2 can be expressed as a fraction of two integers.

Let’s assume, for the sake of contradiction, that √2 is rational. This means we can express it as a fraction in simplest form:

√2 = p/q

Here, p and q are integers with no common factors (other than 1), and q is not zero. By squaring both sides, we derive:

2 = p2 / q2

Multiplying both sides by q2 gives us:

p2 = 2q2

This equation indicates that p2 is even since it is equal to 2 times another integer (2q2). Hence, if p2 is even, it follows that p must also be even (as the square of an odd number is always odd).

Since p is even, we can say:

p = 2k for some integer k.

Substituting this into our earlier equation provides:

p2 = (2k)2 = 4k2

We can replace this in the equation:

4k2 = 2q2

Dividing both sides by 2 results in:

2k2 = q2

This equation tells us that q2 is also even, which implies that q must be even as well. Now we have established that both p and q are even, meaning they share at least the common factor of 2.

This conclusion contradicts the assumption that p/q was in its simplest form with no common factors, so our original assumption that √2 is rational must be false. Therefore, we conclude:

√2 is an irrational number.

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