Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) Prove that \(\sqrt{3}\) is irrational.

Short answer

The assumption that \( \sqrt{3} \) is rational leads to a contradiction. Hence, \( \sqrt{3} \) is irrational.

Step-by-step solution

Rational Number Definition

Assume to the contrary that \( \sqrt{3} \) is rational. That means it can be expressed as a ratio of two integers. Let's put \( \sqrt{3} = \frac{a}{b} \) where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \), and then simplify the equation to get \( 3 = \frac{a^2}{b^2} \) or \( a^2 = 3b^2 \).

Formulate the Contradiction

Notice that since \( a^2 = 3b^2 \), the square of \( a \) is a multiple of 3. This implies that \( a \) is also a multiple of 3, because the square of a non-multiple of 3 is also not a multiple of 3. So we let \( a = 3c \), where \( c \) is an integer. Substituting \( a \) with 3c in the equation yields \( 9c^2 = 3b^2 \) or \( 3c^2 = b^2 \). The similar reasoning from before tells us that \( b \) is also a multiple of 3.

Contradiction and Conclusion

However, in Step 1, we stated that \( a \) and \( b \) are co-prime, meaning they have no common factors other than 1, but in Step 2, we derived that both \( a \) and \( b \) are multiples of 3, which means they have at least one common factor, 3, which contradicts our original assumption that \( a \) and \( b \) are co-prime. Therefore, due to contradiction, our initial assumption that \( \sqrt{3} \) is rational is false. The proof concludes that \( \sqrt{3} \) is irrational.