1. The problem states that 𝑎 and 𝑏 are both odd numbers. 2. Recall that an odd number can be written as $2k+1$ for some integer $k$. 3. Let $a = 2m + 1$ and $b = 2n + 1$ where $m,n$ are integers. 4. We now check expressions involving $a$ and $b$ to see which must be even. 5. For example, consider the sum $a + b$: $$a + b = (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1)$$ 6. Since $2(m + n + 1)$ is divisible by 2, $a + b$ is even. 7. Similarly, consider $a - b$: $$a - b = (2m + 1) - (2n + 1) = 2(m - n)$$ 8. This is also divisible by 2, so $a - b$ is even. 9. Consider the product $ab$: $$ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$$ 10. Since this equals $2($some integer$) + 1$, $ab$ is odd. 11. Therefore, sums and differences of two odd numbers must be even, but their product is odd. 12. So, the expressions that must be even given $a$ and $b$ are odd include $a+b$ and $a-b$. Final answer: The sum and difference of two odd numbers are always even.