I am helping out a friend who can't seem to get these proofs; unfortunately, I can't find them either. Can someone tell me how to solve this or point me in the right direction with resources?
Question 1:
Prove that for all real numbers x, y, and z, if x + y + z greater than or equal
to 3, then either x greater than or equal to 1 or y greater than or equal to 1
or z greater than or equal to 1.
$$\forall x,y,z\in \mathbb R, \quad x+y+z \geq 3 \implies x \geq 1 \lor y \geq 1 \lor z\geq1 $$
Question 2:
Prove that for all real numbers x and y, if xy less than or equal to 2, then
either x less than or equal to square root of 2 or y less than or equal to
square root of 2.
$$\forall x,y \in \mathbb R,\quad xy \leq 2 \implies x \leq \sqrt 2 \lor y \leq \sqrt 2$$
Best Answer
Here are three ways you can choose to prove conditionals (if-then statements). There are other ways but these are most common.
Direct Proof
If $A$, then $B$.
$(A \implies B)$
Assume $A$ is true, then show that when $A$ is true that $B$ must also be true. This seems like it should be the simplest way, but that is not always the case.
Contrapositive
If not $B$, then not $A$.
$(\neg B \implies \neg A)$
This is when you assume that $B$ is false, then show that when $B$ is false then $A$ must also be false. This works because $(\neg B \implies \neg A)$ is logically equivalent to $(A \implies B)$.
Contradiction
$A$ and not $B$.
$(A \land \neg B)$
This is when you assume that $A$ is true and that $B$ is false. Then go on with attempting to prove that assumption just to arrive at a contradiction, or an absurdity in the proof. By showing $(A \land \neg B)$ is a contradiction you have done enough to prove that $(A \implies B)$ so you can end your proof there.
I know this is all formal symbolic logic but it's quite useful in math proofs. Hope that helps.