1. **State the problem:** We need to prove two statements: - If the sum of two integers is even, then at least one of the summands is even. - The difference of any two even integers is even. 2. **Recall definitions and properties:** - An integer is even if it can be written as $2k$ for some integer $k$. - An integer is odd if it can be written as $2k+1$ for some integer $k$. - Sum or difference of integers follows usual arithmetic rules. 3. **Proof of first statement:** - Let the two integers be $a$ and $b$. - Suppose their sum $a+b$ is even. - We want to prove that at least one of $a$ or $b$ is even. 4. **Use proof by contrapositive:** - The contrapositive statement is: If both $a$ and $b$ are odd, then $a+b$ is odd. - Assume $a=2m+1$ and $b=2n+1$ for some integers $m,n$ (both odd). - Then, $$a+b = (2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)$$ - Since $m+n+1$ is an integer, $a+b$ is even. - This contradicts the contrapositive assumption that $a+b$ is odd. 5. **Re-examine step 4:** - Actually, the sum of two odd numbers is even, so the contrapositive is false. - The original statement "If sum is even, then one summand is even" is false. 6. **Counterexample:** - Take $a=1$ (odd), $b=3$ (odd), sum $=4$ (even), but neither summand is even. 7. **Conclusion for first statement:** - The statement is false; the sum of two integers being even does not guarantee one summand is even. 8. **Proof of second statement:** - Let $a=2m$ and $b=2n$ be two even integers. - Their difference is $$a - b = 2m - 2n = 2(m-n)$$ - Since $m-n$ is an integer, $a-b$ is even. 9. **Final conclusion:** - The difference of any two even integers is even. **Answer:** - The first statement is false. - The second statement is true and proven above.