Question

Show that if m and n are odd numbers then m+n is also odd

Answer 1

To show that the sum of two odd numbers \( m \) and \( n \) is also odd, we start by recalling the definition of an odd number.

An integer \( m \) is considered odd if it can be expressed in the form:
\[
m = 2k + 1
\]
for some integer \( k \). Similarly, another integer \( n \) is odd and can be expressed as:
\[
n = 2j + 1
\]
for some integer \( j \).

Now, we consider the sum \( m + n \):
\[
m + n = (2k + 1) + (2j + 1)
\]

Combining the terms, we can rewrite this as:
\[
m + n = 2k + 2j + 1 + 1
\]
\[
m + n = 2k + 2j + 2
\]

Factoring out the common factor of 2 gives:
\[
m + n = 2(k + j + 1)
\]

Let \( p = k + j + 1 \). Since \( k \) and \( j \) are integers, \( p \) is also an integer. Thus, we can express \( m + n \) as:
\[
m + n = 2p
\]

This indicates that \( m + n \) is an even number because it is expresible as two times an integer. Since we know that both \( m \) and \( n \) are odd numbers, and we've shown that their sum is even, we need to confirm that the final result represents an odd sum.

It seems there might have been a misunderstanding in the task, as the conclusion here is that the sum \( m + n \) is actually even (not odd) when both \( m \) and \( n \) are odd. The correction follows thus:

If \( m \) and \( n \) are odd integers, their sum \( m+n \) is indeed even.

To clarify:
- The statement should correctly conclude that if \( m \) and \( n \) are both odd, then \( m+n \) is even, not odd.

In conclusion, \( m+n \) is even as shown.