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.
from Hot Weekly Questions - Mathematics Stack Exchange
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