Sure, here is a proof by contradiction that √2 is irrational:

1. Assume to the contrary that √2 is rational. This means that it can be expressed as a ratio of two integers, say a/b, where a and b have no common factors other than 1 (i.e., they are coprime), and b ≠ 0.

2. If √2 = a/b, then squaring both sides gives 2 = a²/b², or a² = 2b².

3. This implies that a² is even (since it's 2 times something), and therefore a must also be even (since only even numbers have even squares).

4. If a is even, then we can write a = 2k for some integer k. Substituting this into the equation a² = 2b² gives (2k)² = 2b², or 4k² = 2b², or b² = 2k².

5. This implies that b² is even, and therefore b is also even (for the same reason as before).

6. But if a and b are both even, then they have 2 as a common factor, which contradicts our assumption that they have no common factors other than 1.

7. Therefore, our original assumption that √2 is rational must be false. Hence, √2 is irrational.

Question

Sure, here is a proof by contradiction that √2 is irrational:

1. Assume to the contrary that √2 is rational. This means that it can be expressed as a ratio of two integers, say a/b, where a and b have no common factors other than 1 (i.e., they are coprime), and b ≠ 0.

2. If √2 = a/b, then squaring both sides gives 2 = a²/b², or a² = 2b².

3. This implies that a² is even (since it's 2 times something), and therefore a must also be even (since only even numbers have even squares).

4. If a is even, then we can write a = 2k for some integer k. Substituting this into the equation a² = 2b² gives (2k)² = 2b², or 4k² = 2b², or b² = 2k².

5. This implies that b² is even, and therefore b is also even (for the same reason as before).

6. But if a and b are both even, then they have 2 as a common factor, which contradicts our assumption that they have no common factors other than 1.

7. Therefore, our original assumption that √2 is rational must be false. Hence, √2 is irrational.

Knowee AI · Accepted Answer