Prove that for all real numbers $x$ and $y$, if $x+y \geq 100$, then $x \geq 50$ or $y \geq 50$. I'm confused about the following question in my math textbook.
Prove that for all real numbers $x$ and $y$, if $x+y \geq 100$, then $x \geq 50$ or $y \geq 50.$
The or is what gets me. For or to be true don't we need only one of the statements in the operation to be true? Couldn't we have $x = 12, y = 55, x+y = 67$ is there something I'm missing here? Shouldn't it be an and instead of an or?

The problem is with the connective 'or'. It is not (always) exclusive. Suppose you took $x=12$, then to satisfy the inequality $x+y\geq 100$ you need $y \geq 88$, which does satisfy $y \geq 50$.
Hint: Prove the contrapositive.
Edit: Answering the comment below, no, because $x+y=72 < 100$. Your hypothesis is that $x+y \geq 100$. In other words, you have to think in this order:


*

*Pick numbers such that $x+y \geq 100$.

*Check if $x \geq 50$.

*Check if $y \geq 50$.


If 2 or 3 is verified then you strengthen your belief that it is true. Otherwise you have found a counterexample, but it is imperative that $x+y \geq 100$ holds.
