Question

Prove that the square root of 2 is irrational.

Answer 1

To prove that the square root of 2 is irrational, we will use a proof by contradiction. Suppose that the square root of 2 is rational, which means that it can be expressed as a fraction a/b, where a and b are integers with no common factors. Then, we can square both sides of this equation to get:

2 = (a/b)^2 = a^2 / b^2

Multiplying both sides by b^2, we get:

2b^2 = a^2

This means that a^2 is an even number, because it is equal to 2 times b^2. Therefore, a must be an even number, because the square of an odd number is always odd. Let a = 2c, where c is an integer. Substituting this into the previous equation, we get:

2b^2 = (2c)^2 = 4c^2

Dividing both sides by 2, we get:

b^2 = 2c^2

This means that b^2 is an even number, and therefore b must also be an even number. However, this contradicts our original assumption that a and b have no common factors. Therefore, our assumption that the square root of 2 is rational must be false, and the square root of 2 is indeed an irrational number.

Answer 2

The proof that the square root of 2 is irrational can be shown using proof by contradiction.

Assume that the square root of 2 is rational, meaning that it can be expressed as the ratio of two integers (p/q, where p and q are integers and q is not equal to zero).

This means that:

√2 = p/q, where p and q are integers and q is not equal to zero.

Squaring both sides of the equation:

2 = p^2/q^2

multiply both sides by q^2

2q^2 = p^2

Now, since p and q are integers, p^2 and q^2 are also integers.

Therefore, 2q^2 is an even integer.

However, if 2q^2 is an even integer, q^2 is also an even integer, which means that q is even.

Therefore, p and q have a common factor of 2. But, this contradicts the assumption that p and q have no common factors other than 1.

Therefore, our assumption that the square root of 2 is rational must be false.

Hence, we can conclude that the square root of 2 is irrational.