Explain why the product of any two consecutive natural numbers is even. Prove your conjecture.
one of the numbers is even.
an even multiple of any number is even.
To understand why the product of any two consecutive natural numbers is even, we need to consider the properties of even and odd numbers.
Even numbers: These are numbers that are divisible by 2, which means they can be expressed as 2 multiplied by some other integer. For example, 2, 4, 6, 8, and so on.
Odd numbers: These are numbers that are not divisible by 2. They cannot be expressed as 2 multiplied by any other integer. For example, 1, 3, 5, 7, and so on.
Now, let's consider any two consecutive natural numbers. Let's call them n and n+1.
The first case is when n is an even number. In this case, we can express n as 2 multiplied by some integer. So, n = 2k, where k is an integer. The next consecutive number, n+1, will be (2k+1), which is odd.
The product of two consecutive numbers (n)(n+1) = (2k)(2k+1) = 4k^2 + 2k.
We can see that the product has the factor 2 in it. By factoring out 2, we get:
(2)(2k^2 + k).
Since 2 is a factor, the product of any two consecutive numbers will always be even.
The second case is when n is an odd number. In this case, we can express n as 2k+1, where k is an integer. The next consecutive number, n+1, will be (2k+2), which is even.
The product of the two consecutive numbers (n)(n+1) = (2k+1)(2k+2) = 4k^2 + 6k + 2.
Again, we can see that the product has the factor 2 in it. By factoring out 2, we get:
(2)(2k^2 + 3k + 1).
Therefore, even if n is odd, the product of any two consecutive numbers will also be even.
In conclusion, we have shown that regardless of whether the first number is even or odd, the product of any two consecutive natural numbers will always be even.