Question

Morgan makes a conjecture that the perimeter of a square whose sides are an odd integer is always even. Which choice is the best proof of her conjecture?

Answer 1

Morgan's conjecture that the perimeter of a square with odd integer sides is always even is correct because multiplying an odd integer by 4 (the number of sides in a square) results in an even number.

Step-by-step explanation:

To prove that the perimeter of a square with sides of an odd integer length is always even, we must remember that the perimeter of a square is calculated by adding the lengths of all four sides together. Since each side is of equal length, the formula for the perimeter (P) is P = 4 × side_length. If the side_length is odd, let's represent it as 2n+1, where n is any integer. Then, the perimeter will be P = 4 × (2n+1) = 4 × 2n + 4 × 1 = 8n + 4. Since 8n is obviously even (as 8 is even), and 4 is also even, their sum is likewise even. Thus, the perimeter of this square is always even, regardless of which odd integer length is chosen for the sides, confirming Morgan's conjecture.