Please help me spot the error in my "proof" that the sum of two irrational numbers must be irrational
Solution 1:
To say that it is not true that all swans are white does not mean that all swans are non-white; it only means that at least one swan is non-white.
Similarly, to say that it is not true that every sum of two irrational numbers is irrational does not mean that every sum of two irrational numbers is rational; it only means that at least one sum of two irrational numbers is rational.
You start by assuming, not that the sum of (every) two irrational numbers is rational, but rather that the sum of two irrational numbers $a$ and $b$ is rational, i.e. that there is one instance of two irrational numbers whose sum is rational.
That assumption is true. For example: If $a=\pi$ and $b=4-\pi,$ then the sum of the two irrational numbers $a$ and $b$ is the rational number $4.$ And the sum of the two irrational numbers $a$ and $-b$ is the irrational number $2\pi-4.$ The fact that the sum of two irrational numbers $a$ and $b$ is rational does not mean that the sum of the two irrational numbers $a$ and $-b$ is rational, nor that any other sum of two irrational numbers is rational.
Solution 2:
You have assumed $a+b$ and $b-a$ are rational and arrived at a contradiction. Therefore, the strongest conclusion your proof can make is at least one of $a+b$ and $b-a$ must be irrational.
Solution 3:
The initial assumption is, "Assume that the sum of two irrational numbers $a$ and $b$ is rational"; later on you say, "from our assumption that the sum of two irrational numbers is rational...", but these are not the same statement. You've cunningly morphed from an existential statement to a universal statement. That is: from an assumption that there are at least two numbers for which it is true (which is true), to an assumption that it's true for any such numbers (which is false).
A specific counterexample would be $a = -\sqrt 2$, $b = \sqrt 2$. In this case $a + b = 0$ is indeed rational, but $(b + (-1)(a)) = 2 \sqrt 2$ is not.