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The product of two odd numbers is always an odd number.
Let's take a look at this. For any integer n, 2n always be even, then the next consecutive number 2n + 1 must be odd. Let add them first, 2n + 2n + 1 = 4n + 1 = 2(2n) + 1 So their sum is odd, since every even number multiplied by 2 is also even. Let's multiplied them, 2n(2n + 1) = (2n)^2 + 2n Their product is even, since every even number raised in the second power is also even, and the sum of two even numbers is even too. So the answer is that when the sum of two numbers can be odd, their product is an even number. (note that the sum of two odd numbers is even)
One possible conjecture: The product is always an odd number. Another possible conjecture: The product is always greater than either of them. Another possible conjecture: Both odd numbers are always factors of the product. Another possible conjecture: The product is never a multiple of ' 2 '. Another possible conjecture: The product is always a real, rational number. Another possible conjecture: The product is always an integer.
the product of two integers is odd if and only if the two factors are odd
Easily. Indeed, it might be empty. Consider the set of positive odd numbers, and the set of positive even numbers. Both are countably infinite, but their intersection is the empty set. For a non-empty intersection, consider the set of positive odd numbers, and 2, and the set of positive even numbers. Both are still countably infinite, but their intersection is {2}.