To show that the sum is an even number, we need to demonstrate that it can be written in the form 2 × integer. We can achieve this by factoring out 2 from the simplified sum.

Prove that the sum of two odd numbers is always even

Answer: The sum of two odd numbers is always an even number.

Step-by-step solution

Step 1: Define Odd Numbers

To prove this statement, we first need to define what an odd number is. An odd number is any integer that cannot be divided exactly by 2. Mathematically, we can represent any odd number as $2k+1$ , where $k$ is an integer. We will use this definition for our two odd numbers, $m$ and $n$ .

Step 2: Represent the Odd Numbers Algebraically

Since $m$ and $n$ are two different odd numbers, we can represent them using the definition of an odd number. We use different integer variables, $k_1$ and $k_2$ , to ensure that $m$ and $n$ can be any two odd numbers, not necessarily the same one.

Step 3: Find the Sum of the Two Odd Numbers

Now that we have algebraic expressions for our two odd numbers, $m$ and $n$ , we can find their sum by adding these expressions together. This is the first step towards simplifying the sum.

Step 4: Simplify the Sum

We can simplify the expression for the sum by combining the constant terms. This brings us closer to identifying the form of an even number.

Step 5: Factor out 2

To show that the sum is an even number, we need to demonstrate that it can be written in the form $2 \times \text{integer}$ . We can achieve this by factoring out 2 from the simplified sum.

Step 6: Conclude the Proof

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer. Adding 1 to this sum, $K = k_1 + k_2 + 1$ , also results in an integer. Thus, the sum $m+n$ can be expressed as $2K$ , which is the definition of an even number. This completes the proof.

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