There must be an error in my "proof" since it is evident that the sum of two irrational numbers may be rational, but I am struggling to spot it. A hint would be appreciated.
The "proof" is by contradiction:
Assume that the sum of two irrational numbers a and b is rational.
Then we can write
$$ a + b = \frac{x}{y} $$
$$ \implies a + b + a – a = \frac{x}{y} $$
$$ \implies 2a + (b – a) = \frac{x}{y} $$
$$ \implies 2a = \frac{x}{y} + (-1)(b + (-1)(a)) $$
-> from our assumption that the sum of two irrational numbers is rational, it follows that $(b + (-1)(a))$ is rational
-> therefore, the right side is rational, being the sum of two rational numbers
-> but the left side, $2a$, is irrational, because the product of a rational and irrational number is irrational
-> this is a contradiction; since assuming that the sum of two irrational numbers is rational leads to a contradiction, the sum of two irrational numbers must be irrational.
Best Answer
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.