This **article aims**Â to prove that **product of two odd numbers**Â is an**Â odd number. **This article uses the**Â concept of odd numbers**. **Odd numbers**Â 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**Â can be either odd or even.

**Expert Answer**

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

$ n $ and $ m $ are **real numbers.**

\[ n = 2 a + 1 \]

$ n $ is an **odd number.**

\[ m Â = 2 b + 1 \]

**Calculate**Â $ n . m $

\[ n . m = ( 2 a + 1) . ( 2 b + 1) \]

\[ Â n . m = Â 4 a b Â + 2 a + 2 b + 1 \]

\[ n . m = 2 ( Â 2 a b Â + a + b ) + 1 \]

\[ Odd \: integer = 2 k + 1 \]

\[n . m = 2 k + 1 \]

**Where **

\[ k Â = 2 a b + a + b = integer \]

Hence, $ n $ and $ m $ are**Â odd.**

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

**Numerical Result**

If **two numbers**Â $ n $ and $ m $ are**Â 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:

\[ x = 2 m \]

\[ y = 2 n \]

\[x . y = ( 2 m ). (2 n) Â = 4 n m \]

Where $ n m = k Â = integer $

Therefore, the**Â product of two even numbers is even**.