Even and odd numbers

Welcome to the fascinating world of mathematics, where even and odd numbers reign supreme! These two categories of numbers are the foundation upon which countless mathematical concepts and theorems are built. Whether you’re a mathematician or simply someone who wants to deepen their understanding of the mathematical universe, understanding even and odd numbers is essential.

In this article, we’ll take a deep dive into the mesmerizing world of even and odd numbers, uncovering their unique properties, exploring their practical applications, and discovering the hidden beauty within their numerical patterns. Get ready to be amazed by the wonders of mathematics

Defining Even and Odd Numbers

First, let’s get clear on the two unique groups of integers we’re dealing with here:

Even numbers

Let’s talk about even numbers! These are the integers that can be divided by 2 without leaving any leftovers. So, if you can express a number “$n$” as $2$ multiplied by another integer “$k$”, then it’s considered even: $n=2k$. Easy peasy, right? Some examples of even numbers are $-2, 0, 4, 8,$ and even the ultimate answer to life, the universe, and everything $- 42$!

Odd Numbers

Did you know that odd numbers are integers that just can’t be divided by $2$ evenly? It’s true! If you have an integer “$n$” and it can be expressed as $n=2k+1$ with “k” being an integer, then it’s considered odd. Some examples of odd numbers include $-3, 1, 7, 15 $, and $33$. So next time you come across a number that can’t be divided by $2$ evenly, you’ll know it’s an odd one!

Properties of Even and Odd Numbers

Even numbers possess distinctive characteristics when subjected to arithmetic operations. The addition or subtraction of two even numbers invariably results in an even number. Conversely, the addition or subtraction of two odd numbers yields an even number. However, the multiplication of even numbers always yields an even product.

The arithmetic operations of addition, subtraction, and multiplication applied to a pair of even numbers result in an even number.
Let $a$ and $b$ be even integers. It follows that there exist two integers $p$ and $q$ such that $a=2p$ and $b=2q$. Consequently, we have that $$ a+b=2(p+q),\quad a-b=2(p-q),\quad ab=2(2pq).$$ Hence, $a+b$, $a-b$, and $ab$ are all even integers.
An odd number is obtained when an even number is added to or subtracted from another odd number.
The proof is concluded by demonstrating that the following equations hold true for any integers $k$ and $k’$: $$ 2k+(2k’+1)=2(k+k’)+1,\quad 2k-(2k’+1)=2(k-k’)+1.$$
When an even number is multiplied by an odd number, the resulting product is an even number.
In fact, this follows for the equation $$ (2k)(2k’+1)=2(2kk’+k’).$$

Worksheets on even and odd numbers

The objective of this inquiry is to ascertain whether the value of $c$, which is given by the expression $(2n+1)^2+2n-1$, is an even or odd number.
We can write \begin{align*}c&=(2n+1)^2+2n-1\cr &= 4n^2+4n+1+2n-1\cr &= 4n^2+6n\cr &= 2(2n^2+3n).\end{align*} This shows that $c$ is an even number.
From the following list of numbers, determine which are even: $$ 5^2,\quad \frac{378}{3},\quad 15^2-8.$$
We have
  1. $5^2=5\times 5=25$, it is an odd number.
  2. $\frac{378}{3}=126=2\times 63,$ it is an even number.
  3. $15^2-8=225-8=217=2\times 108+1$, it is an odd number.
Demonstrate that the square of an even integer is also an even integer.
Assuming $N$ to be an even integer, it can be expressed as $N=2n$, where $n$ is a natural number. By squaring both sides of the aforementioned equation, we obtain $N^2=(2n)^2=4n^2=2(2n^2)$. Consequently, $N^2$ is an even integer
Demonstrate that the product of two consecutive integers is an even number.

Let $k$ be an integer and consider the selection of $N=k(k+1)$, which is the product of two consecutive integers. There are two possible cases for $k$: when $k$ is even or odd.

First, let us assume that $k$ is even. In this case, we can express $k$ as $k=2n$, where $n$ is an integer. Substituting this value of $k$ into the expression for $N$, we obtain $N=2n(2n+1)=2p$, where $p=2n^2+n$ is an integer. Therefore, $N$ is an even number.

Second, let us assume that $k$ is odd. In this case, we can express $k$ as $k=2n+1$, where $n$ is an integer. Substituting this value of $k$ into the expression for $N$, we obtain $N=(2n+1)(2n+1+1)=(2n+1)(2n+2)$. This can be rewritten as $N=2q$, where $q=(2n+1)(n+1)$ is an integer. Therefore, $N$ is also an even number.

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