prove that for all intergers a and b if the product of a and .b is even a or b is even
To prove this statement, we will use a proof by contradiction.
Suppose there exist integers a and b such that the product of a and b is even, but both a and b are odd.
Let's assume a = 2n + 1 and b = 2m + 1, where n and m are integers.
Now, let's calculate the product of a and b:
ab = (2n + 1)(2m + 1) = 4nm + 2n + 2m + 1.
Since 4nm + 2n + 2m is an even number (it can be expressed as 2k, where k is an integer), we can rewrite the product as:
ab = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1.
Now, let's examine the expression 2(2nm + n + m). This is an even number multiplied by 2, which results in an even number.
Therefore, ab is the sum of an even number 2(2nm + n + m) and the odd number 1. An odd number added to an even number is always an odd number.
However, we initially assumed that ab is even, which creates a contradiction. Thus, our assumption that both a and b are odd numbers must be false.
Hence, we can conclude that if the product of a and b is even, at least one of the integers a or b must be even.