Let \[$ N \$\] Be A Positive Integer.(i) Explain Why \[$ N(n-1) \$\] Must Be An Even Number.

Let $n$ be a positive integer: Explaining why $n(n-1)$ must be an even number

In mathematics, the properties of integers and their operations are crucial in understanding various mathematical concepts. One such property is the evenness or oddness of a number, which is determined by its divisibility by 2. In this article, we will explore why the product of a positive integer $n$ and its predecessor $n-1$ must be an even number.

Understanding Even and Odd Numbers

Before we dive into the explanation, let's briefly review what even and odd numbers are. An even number is any integer that can be written in the form $2k$, where $k$ is an integer. On the other hand, an odd number is any integer that can be written in the form $2k+1$, where $k$ is an integer.

Why $n(n-1)$ must be an even number

Now, let's consider the product $n(n-1)$. To understand why this product must be an even number, we need to examine the possible cases for the values of $n$.

Case 1: $n$ is even

If $n$ is even, then it can be written in the form $2k$, where $k$ is an integer. In this case, $n-1$ will be odd, as it can be written in the form $2k-1$. The product $n(n-1)$ will then be:

$$n(n-1) = (2k)(2k-1)$$

Since both $2k$ and $2k-1$ are integers, their product will also be an integer. Moreover, since $2k$ is even and $2k-1$ is odd, their product will be even.

Case 2: $n$ is odd

If $n$ is odd, then it can be written in the form $2k+1$, where $k$ is an integer. In this case, $n-1$ will be even, as it can be written in the form $2k$. The product $n(n-1)$ will then be:

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

Since both $2k+1$ and $2k$ are integers, their product will also be an integer. Moreover, since $2k$ is even and $2k+1$ is odd, their product will be even.

In conclusion, regardless of whether $n$ is even or odd, the product $n(n-1)$ will always be an even number. This is because in both cases, the product of an even number and an odd number (or vice versa) will always be even.

The fact that $n(n-1)$ must be an even number has important implications in various areas of mathematics, such as number theory and combinatorics. For example, it can be used to prove the existence of certain types of numbers or to derive formulas for counting certain types of objects.

Here are a few examples of how the fact that $n(n-1)$ must be an even number can be used:

• To prove that the product of two consecutive integers is always even.
• To derive a formula for the number of ways to arrange $n$ objects in a circle.
• To show that the sum of the first $n$ positive integers is always even.

For those interested in learning more about the properties of integers and their operations, we recommend the following resources:

• "Introduction to Number Theory" by G.H. Hardy and E.M. Wright
• "Combinatorics: Topics, Techniques, Algorithms" by Peter J. Cameron
• "The Art of Proof: Basic Training for Deeper Mathematics" by Matthias Beck and Ross Geoghegan

By understanding the properties of integers and their operations, we can gain a deeper appreciation for the beauty and complexity of mathematics.
Let $n$ be a positive integer: Explaining why $n(n-1)$ must be an even number

In this article, we will address some of the most frequently asked questions related to the topic of why $n(n-1)$ must be an even number.

Q: What if $n$ is a negative integer?

A: The statement that $n(n-1)$ must be an even number only applies to positive integers. If $n$ is a negative integer, then $n(n-1)$ may or may not be an even number.

Q: Can you provide an example of a negative integer $n$ for which $n(n-1)$ is even?

A: Yes, consider the case where $n = -2$. In this case, $n(n-1) = (-2)(-3) = 6$, which is an even number.

Q: What if $n$ is a non-integer value?

A: The statement that $n(n-1)$ must be an even number only applies to integer values of $n$. If $n$ is a non-integer value, then $n(n-1)$ may or may not be an even number.

Q: Can you provide an example of a non-integer value $n$ for which $n(n-1)$ is even?

A: Yes, consider the case where $n = 2.5$. In this case, $n(n-1) = (2.5)(1.5) = 3.75$, which is not an even number.

Q: Is there a general formula for determining whether $n(n-1)$ is even or odd?

A: Yes, there is a general formula for determining whether $n(n-1)$ is even or odd. If $n$ is an integer, then $n(n-1)$ is even if and only if $n$ is even. If $n$ is not an integer, then $n(n-1)$ may or may not be an even number.

Q: Can you provide a proof of the statement that $n(n-1)$ must be an even number?

A: Yes, the proof of the statement that $n(n-1)$ must be an even number is as follows:

• If $n$ is even, then $n = 2k$ for some integer $k$. In this case, $n(n-1) = (2k)(2k-1)$, which is an even number.
• If $n$ is odd, then $n = 2k+1$ for some integer $k$. In this case, $n(n-1) = (2k+1)(2k)$, which is an even number.

In conclusion, the statement that $n(n-1)$ must be an even number is a fundamental property of integers that has important implications in various areas of mathematics. We hope that this article has provided a clear and concise explanation of this property, as well as some examples and counterexamples to illustrate its validity.

For those interested in learning more about the properties of integers and their operations, we recommend the following resources:

• "Introduction to Number Theory" by G.H. Hardy and E.M. Wright
• "Combinatorics: Topics, Techniques, Algorithms" by Peter J. Cameron
• "The Art of Proof: Basic Training for Deeper Mathematics" by Matthias Beck and Ross Geoghegan

By understanding the properties of integers and their operations, we can gain a deeper appreciation for the beauty and complexity of mathematics.