Prove that the square root of 5 is irrational.

Steps to Solve

1. Assume the opposite We will start by assuming that $\sqrt{5}$ is a rational number. According to this assumption, it can be written as a fraction of two integers in the form $a/b$, where $a$ and $b$ are integers with no common factors (i.e., the fraction is in simplest form).

2. Set up the equation From our assumption, we can express the equation as: $$ \sqrt{5} = \frac{a}{b} $$ By squaring both sides to eliminate the square root, we get: $$ 5 = \frac{a^2}{b^2} $$

3. Rearrange the equation Next, multiply both sides by $b^2$ to eliminate the fraction: $$ 5b^2 = a^2 $$

4. Analyze the implications This equation tells us that $a^2$ is a multiple of 5. From number theory, if $a^2$ is divisible by a prime (in this case, 5), then $a$ itself must also be divisible by that prime. Thus, we can express $a$ as: $$ a = 5k $$ for some integer $k$.

5. Substitute back into the equation Now substitute $a = 5k$ into the equation $5b^2 = a^2$: $$ 5b^2 = (5k)^2 $$ This simplifies to: $$ 5b^2 = 25k^2 $$

6. Further simplify the equation Divide both sides by 5: $$ b^2 = 5k^2 $$

7. Conclude that $b$ is also divisible by 5 Following the same reasoning as before, since $b^2$ is divisible by 5, $b$ must also be divisible by 5.

8. Reach a contradiction At this point, we have shown that both $a$ and $b$ are divisible by 5. This contradicts our initial assumption that $a/b$ was in simplest form (with no common factors). Hence, our assumption must be false.

9. Conclude that $\sqrt{5}$ is irrational Therefore, we conclude that $\sqrt{5}$ cannot be expressed as a fraction of two integers, which means it is irrational.