For any integer n, which of the following are alwaysa) two consecutive even numbers?b) two consecutive odd numbers?n and n+2n−1 and n+12n and 2n+22n and 2n+12n−1 and 2n+12n+1 and 2n+2
Q. For any integer n, which of the following are alwaysa) two consecutive even numbers?b) two consecutive odd numbers?n and n+2n−1 and n+12n and 2n+22n and 2n+12n−1 and 2n+12n+1 and 2n+2
First Option Analysis: Let's consider the first option: n and n+2. If n is an even number, then n+2 is also even because adding 2 to an even number results in another even number. Similarly, if n is an odd number, then n+2 is also odd because adding 2 to an odd number results in another odd number. Therefore, n and n+2 are either both even or both odd, depending on the parity of n.
Second Option Evaluation: Now let's consider the second option: n−1 and n+1. If n is an even number, then n−1 is odd and n+1 is odd because subtracting 1 from an even number results in an odd number and adding 1 to an even number also results in an odd number. If n is an odd number, then n−1 is even and n+1 is even because subtracting 1 from an odd number results in an even number and adding 1 to an odd number also results in an even number. Therefore, n−1 and n+1 are always consecutive odd or even numbers, but they are not the same parity as each other.
Third Option Assessment: Next, let's evaluate the third option: 2n and 2n+2. Since 2n is always even for any integer n, and adding 2 to an even number results in another even number, 2n+2 is also even. Therefore, 2n and 2n+2 are two consecutive even numbers.
Fourth Option Examination: Now let's look at the fourth option: 2n and 2n+1. Since 2n is always even for any integer n, and adding 1 to an even number results in an odd number, 2n+1 is odd. Therefore, 2n and 2n+1 are not two consecutive even or odd numbers because they have different parities.
Fifth Option Consideration: Let's consider the fifth option: 2n−1 and 2n+1. Since 2n is always even for any integer n, subtracting 1 from an even number results in an odd number, so 2n−1 is odd. Similarly, adding 1 to an even number results in an odd number, so 2n+1 is also odd. Therefore, 2n−1 and 2n+1 are two consecutive odd numbers.
Last Option Review: Finally, let's evaluate the last option: 2n+1 and 2n+2. Since 2n+1 is always odd for any integer n, and adding 1 to an odd number results in an even number, 2n+2 is even. Therefore, 2n+1 and 2n+2 are not two consecutive even or odd numbers because they have different parities.