Final answer:
To prove that the square root of 3 is not rational, we can use a proof by contradiction. Assuming that the square root of 3 is rational, we can derive a contradiction. By showing that the assumption leads to a situation where the numerator and denominator have a common factor, we can conclude that the square root of 3 is irrational.
Step-by-step explanation:
To prove that the square root of 3 is not rational using a proof by contradiction, we assume that √3 is rational. This means that it can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1. Squaring both sides of the equation (√3)^2 = (a/b)^2, we get 3 = (a^2)/(b^2). From this equation, we can conclude that 3b^2 = a^2. It implies that a^2 is divisible by 3, and therefore, a must also be divisible by 3.
Let's assume a = 3c, where c is an integer. Substituting this into the equation 3b^2 = a^2, we get 3b^2 = (3c)^2 = 9c^2. Dividing both sides of the equation by 3, we have b^2 = 3c^2. This means that b^2 is also divisible by 3, and hence, b must also be divisible by 3.
However, this contradicts our assumption that a and b have no common factors other than 1. Therefore, our assumption that √3 is rational must be incorrect. Hence, the square root of 3 is not rational.