The product of the successor and predecessor of an even natural number is always

Question

Accepted Answer

To explain why Option 4 is the correct answer, we need to analyze the product of the successor and predecessor of an even natural number.

Let's denote the even natural number as $ n $. The successor of $ n $ is $ n + 1 $, and the predecessor is $ n - 1 $. The product of the successor and predecessor is:

$$
(n - 1) \times (n + 1) = n^2 - 1
$$

This expression, $ n^2 - 1 $, is known as the difference of squares, which can be factored as:

$$
(n - 1)(n + 1) = n^2 - 1 = (n - 1)(n + 1)
$$

Now, let's evaluate why this product is always an odd number:

1. **Even Natural Number**: Since $ n $ is even, it can be expressed as $ n = 2k $ for some integer $ k $. Therefore, $ n^2 = (2k)^2 = 4k^2 $, which is clearly even.

2. **Subtracting 1**: When you subtract 1 from an even number, the result is always odd. Therefore, $ n^2 - 1 $ is odd.

Thus, the product of the successor and predecessor of an even natural number is always an odd number, confirming that Option 4 is correct.

Now, let's explain why the other options are incorrect:

1. **Option 1: Divisible by 2** - Since $ n^2 - 1 $ is odd, it cannot be divisible by 2. An odd number is not divisible by 2.

2. **Option 2: Divisible by 3** - There is no general rule that states $ n^2 - 1 $ is divisible by 3 for all even $ n $. While it might be true for specific cases, it is not universally true for all even natural numbers.

3. **Option 3: Divisible by 4** - Similarly, $ n^2 - 1 $ is not guaranteed to be divisible by 4. For example, if $ n = 2 $, then $ n^2 - 1 = 3 $, which is not divisible by 4.

In conclusion, the product of the successor and predecessor of an even natural number is always an odd number, making Option 4 the correct answer.