Letm,n∈Z. Provethatm+nisoddiffexactlyoneofmandnisodd

To prove the statement: 

**Statement:** Let \( m, n \in \mathbb{Z} \) (i.e., \( m \) and \( n \) are integers). Then \( m + n \) is odd if and only if exactly one of \( m \) or \( n \) is odd.

### Proof:

#### **1. Proof of the "if" direction**: 
(If exactly one of \( m \) or \( n \) is odd, then \( m + n \) is odd)

Assume \( m \) is odd and \( n \) is even (the argument is symmetric, so the result holds if \( m \) is even and \( n \) is odd).

- **An odd integer** can be expressed as \( m = 2k + 1 \) for some integer \( k \).
- **An even integer** can be expressed as \( n = 2l \) for some integer \( l \).

Now, consider the sum \( m + n \):

\[
m + n = (2k + 1) + 2l = 2(k + l) + 1
\]

This expression is of the form \( 2 \times \text{(integer)} + 1 \), which is odd by definition.

Hence, if one of \( m \) or \( n \) is odd and the other is even, then \( m + n \) is odd.

#### **2. Proof of the "only if" direction**:
(If \( m + n \) is odd, then exactly one of \( m \) or \( n \) is odd)

Assume \( m + n \) is odd. We will prove that this implies exactly one of \( m \) or \( n \) is odd.

- Suppose \( m \) is even. Then \( m = 2k \) for some integer \( k \).
- If \( n \) were also even, then \( n = 2l \) for some integer \( l \).

Then:

\[
m + n = 2k + 2l = 2(k + l)
\]

This expression is of the form \( 2 \times \text{(integer)} \), which is even, contradicting the assumption that \( m + n \) is odd.

- Suppose instead that \( m \) is odd, i.e., \( m = 2k + 1 \) for some integer \( k \).
- If \( n \) were also odd, then \( n = 2l + 1 \) for some integer \( l \).

Then:

\[
m + n = (2k + 1) + (2l + 1) = 2k + 2l + 2 = 2(k + l + 1)
\]

This expression is also of the form \( 2 \times \text{(integer)} \), which is even, again contradicting the assumption that \( m + n \) is odd.

Therefore, if \( m + n \) is odd, then one of \( m \) or \( n \) must be even and the other must be odd.

### Conclusion:

We have shown both directions:

- If exactly one of \( m \) or \( n \) is odd, then \( m + n \) is odd.
- If \( m + n \) is odd, then exactly one of \( m \) or \( n \) is odd.

Thus, we have proven that \( m + n \) is odd if and only if exactly one of \( m \) or \( n \) is odd.

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Would you like further clarification on any part of this proof? Here are some related questions you might find interesting:

1. How can you prove similar results for other modular arithmetic properties?
2. What is the significance of odd and even numbers in number theory?
3. Can you prove the above statement using truth tables and logical operations?
4. How does this proof relate to parity functions in computer science?
5. Can this method of proof be applied to more complex number properties?

**Tip:** When working with modular arithmetic or parity (odd/even properties), expressing numbers in terms of \( 2k \) or \( 2k + 1 \) helps in simplifying and proving many statements.

Learn how to prove the statement that m + n is odd if and only if exactly one of m or n is odd with a detailed mathematical proof. This problem involves concepts from number theory, modular arithmetic, and the parity of odd and even numbers. Suitable for high school students and anyone interested in number properties.

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