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.

**Expert Answer**

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.**

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\[ m = 2 b + 1 \]

**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**.