Show that √3 is irrational.

Sure, here is a proof by contradiction that √3 is irrational:

1. Assume, to the contrary, that √3 is rational. This means that it can be expressed as a ratio of two integers. Let's say √3 = a/b, where a and b are coprime integers (they have no common factors except 1) and b ≠ 0.

2. Squaring both sides, we get 3 = a²/b², or a² = 3b².

3. This implies that a² is divisible by 3, and hence a must also be divisible by 3 (since the square of a number that is not divisible by 3 is also not divisible by 3). So we can write a = 3k for some integer k.

4. Substituting a = 3k into the equation a² = 3b², we get (3k)² = 3b², or 9k² = 3b², or b² = 3k².

5. This implies that b² is divisible by 3, and hence b must also be divisible by 3 (for the same reason as above).

6. But this contradicts our initial assumption that a and b are coprime (they have no common factors except 1). Therefore, our initial assumption that √3 is rational must be false.

7. Hence, √3 is irrational.

Sure, here is a proof by contradiction that √3 is irrational:

1. Assume, to the contrary, that √3 is rational. This means that it can be expressed as a ratio of two integers. Let's say √3 = a/b, where a and b are coprime integers (they have no common factors except 1) and b ≠ 0.

2. Squaring both sides, we get 3 = a²/b², or a² = 3b².

3. This implies that a² is divisible by 3, and hence a must also be divisible by 3 (since the square of a number that is not divisible by 3 is also not divisible by 3). So, we can write a = 3k for some integer k.

4. Substituting a = 3k into the equation a² = 3b², we get (3k)² = 3b², or 9k² = 3b², or b² = 3k².

5. This implies that b² is divisible by 3, and hence b must also be divisible by 3 (for the same reason as above).

6. But this contradicts our initial assumption that a and b are coprime (they have no common factors except 1). Therefore, our initial assumption that √3 is rational must be false.

7. Hence, √3 is irrational.