To prove that the sum of two odd numbers is even, we will begin by expressing odd numbers algebraically and then add them together.

Step 1: Definition of Odd Numbers

An odd number can be written in the form: $n = 2k + 1$ where $k$ is an integer. This expression ensures that $n$ is an odd number because when divided by 2, it leaves a remainder of 1.

Step 2: Adding Two Odd Numbers

Let the two odd numbers be represented by: $n_1 = 2k_1 + 1 \quad \text{and} \quad n_2 = 2k_2 + 1$ where $k_1$ and $k_2$ are integers.

Now, let's add these two odd numbers together: $n_1 + n_2 = (2k_1 + 1) + (2k_2 + 1)$ Simplify the right-hand side: $n_1 + n_2 = 2k_1 + 2k_2 + 2$ Factor out the 2: $n_1 + n_2 = 2(k_1 + k_2 + 1)$ Let $m = k_1 + k_2 + 1$, where $m$ is an integer (since the sum of integers is also an integer).

Thus, we have: $n_1 + n_2 = 2m$ which is in the form of an even number (since any number that is a multiple of 2 is even).

Conclusion:

We have shown that the sum of two odd numbers is always even.

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Would you like more details on any part of the proof?

Here are some related questions:

1. Why is a number of the form $2k + 1$ always odd?
2. Can you prove that the sum of two even numbers is also even?
3. What happens if you add an odd number to an even number?
4. How can you prove that the product of two odd numbers is odd?
5. Can you derive a proof for the difference of two odd numbers?

Tip: Always express odd and even numbers in terms of $2k$ (for even) and $2k + 1$ (for odd) to simplify proofs.