# Use a direct proof to show that the product of two odd numbers is odd.

This article aims to prove that product of two odd numbers is an odd number. This article uses the concept of odd numbers. Odd numbers are any number that cannot be divided by two. In other words, numbers of the form $2 k + 1$, where $k$ is an integer, are called odd numbers. It should be noted that the numbers or sets of integers on the number line can be either odd or even.

If $n$ and $m$ are odd number , then $n * m$ is odd.

$n$ and $m$ are real numbers.

$n = 2 a + 1$

$n$ is an odd number.

$m = 2 b + 1$

Calculate $n . m$

$n . m = ( 2 a + 1) . ( 2 b + 1)$

$n . m = 4 a b + 2 a + 2 b + 1$

$n . m = 2 ( 2 a b + a + b ) + 1$

$Odd \: integer = 2 k + 1$

$n . m = 2 k + 1$

Where

$k = 2 a b + a + b = integer$

Hence, $n$ and $m$ are odd.

We can also check if the product of two odd numbers is odd by taking any two odd numbers and multiplying them to see if their product is odd or even. Odd numbers cannot be exactly divided into pairs; that is, they leave a remainder when divided by two. Odd numbers have digits $1$, $3$, $5$, $7$, and $9$ in the units place. Even numbers are those numbers that are exactly divisible by $2$. Even numbers can have the digits $0$, $2$, $4$, $6$, $8$ and $10$ in the units place.

## Numerical Result

If two numbers $n$ and $m$ are odd, then their product $n . m$ is also odd.

## Example

Prove that product of two even numbers is even.

Solution

Let $x$ and $y$ be two even integers.

By the definition of even numbers, we have:

$x = 2 m$

$y = 2 n$

$x . y = ( 2 m ). (2 n) = 4 n m$

Where $n m = k = integer$

Therefore, the product of two even numbers is even.

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