How do you prove an expression is even?

If is an integer (a whole number), then the expression represents an even number, because even numbers are the multiples of 2. The expressions 2 n − 1 and 2 n + 1 can represent odd numbers, as an odd number is one less, or one more than an even number.

How do you prove that N 3 is odd?

If n is odd, then n3 is odd. n3 = (2k + 1)3 = 8k3 + 12k2 + 6k + 1 = 2(4k3 + 6k2 + 3k)+1. By the closure of the integers under addition and multiplication, we know that 4k3 + 6k2 + 3k is an integer. Call this integer m, so that we have n3 = 2m + 1.

Is N 3 odd or even?

It follows that n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1. (4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an even integer. Therefore n3 is odd.

Is m and n are odd then Mn is even?

Proof by Contraposition: Assume that it is not true that m is even or n is even. Then both m and n are odd. Proof by Contradiction: Assume that mn is even and that m and n are both odd. Since the product of two odd numbers is an odd number, mn is odd, so we have a contradiction: mn is even and mn is odd.

Is n even if 7N+4 is even?

Original Statement: if n is a positive integer then n is even if and only if is 7n+4 is even. Contrapositive: If n is negative integer then n is odd if and only if 7n+4 is odd.

Is 7N + 4 – (6N+ 4) – (7N+4) odd?

Let n be an integer and assume that 7n + 4 is odd. Now 6n + 4 is obviously even, and the difference between an odd and an even number is odd, therefore (7n + 4) − (6n + 4) is odd. That is, n is odd. This completes the proof.

Is N = 7N+4 = 2k+1?

Contrapositive: If n is negative integer then n is odd if and only if 7n+4 is odd. Therefore by definition of odd: n = 2k+1 Substitute n: =7(2k+1)+4 =14k+7+4 =14k+11 =2(7k)+11 Therefore, n is odd and 7n+4 is odd. Thats as far as i got and i dont even know if what i did above is even right though. Thanks. discrete-mathematicsproof-writing Share Cite