Prove by contradiction that √2 is irrational. Use that 2|k2implies that 2|k for any integer k.

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

Step 1: Assume the opposite, 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.

Step 2: If √2 is rational, then (√2)^2 = (a/b)^2, or 2 = a^2/b^2, or a^2 = 2b^2. This means that a^2 is an even number (since it's 2 times something).

Step 3: From the given, if 2 divides a^2 (i.e., a^2 is even), then 2 also divides a (i.e., a is even). This means that a can be written as 2k for some integer k.

Step 4: Substituting a = 2k into the equation a^2 = 2b^2 gives (2k)^2 = 2b^2, or 4k^2 = 2b^2, or b^2 = 2k^2. This means that b^2 is also even, and so b is also even.

Step 5: But if a and b are both even, then they have 2 as a common factor, which contradicts our initial assumption that a and b have no common factors other than 1.

Step 6: Therefore, our initial assumption that √2 is rational must be false. Hence, √2 is irrational.