Is X² is an even integer?
Since k is an integer, 2k2 is also an integer, so we can write x2 = 2l for the integer l = 2k2. By definition, this means x2 is an even number.
We show that if x is even then x2 is even using a direct proof (the only if part or necessity). If x is even then x = 2k for some integer k. Hence x2 = 4k2 = 2(2k2 ) which is even since it is an integer divisible by 2. This completes the proof of case 1.
