Understanding connection of $=$ and $>$ relations in proofs.

Question

I would like to ask about a certain pattern I see in some proofs. This is an example taken from the book: Lang, Serge & Murrow, Gene. "Geometry - Second Edition" (p. 69)

10. An obtuse angle is an angle which has more than 90°. Prove (in a sentence or two) that a triangle cannot have more than one obtuse angle.

Proof: Suppose triangle ABC has two obtuse angles, $\angle A$ and $\angle B$. $m(\angle A) + m(\angle B) > 180°$ contradicting the theorem that the sum of angles in a triangle equals 180°. Therefore, no triangle can contain two obtuse angles.

Is $=$ relation sometimes taken as the negation of $>$ relation ? Am I missing something ?

Answer 1

$=$ and $>$ are incompatible. But that does not mean they are negations of each other.

If $a =b$ then $a < b$ and $a > b$ are impossible.

If $a>b$ then $a= b$ and $a< b$ are impossible.

If $a < b$ then $a=b$ and $a > b$ are impossible.

If $m \angle A + m \angle B > 180^\circ$ then $m\angle A + m\angle B =180^\circ$ is impossible.

You don't need a negation to do a proof by contradiction.

Suppose you want to prove Lucy is not a lion, and you manage to prove that Lucy is lizard. Being a lizard means Lucy can't be a lion. So you are done. QED. But if Lucy is not a lion that doesn't mean she is a lizard. There are more than two options. But those two options, being a lizard and being a lion, are incompatible.

Answer 2

$P\not>Q$ or $P\leq Q$ is the negation of $P>Q$.

$P=Q$ simply implies that $P>Q$ is false, but does not constitute a negation.

Answer 3

As we have $\neg (\forall x \in A)P(x) = (\exists x \in A) \neg P(x)$, then $\neg (\forall x \in A)(x >a) = (\exists x \in A) \neg (x >a) = (\exists x \in A)(x \leqslant a)$. So finding $x$ such as $x=a$ can be take as counterexample for $x>a$, but is not its logical negation.

In stated case is used so called proof by contradiction $$\left[(\neg P \Rightarrow Q) \land (\neg P \Rightarrow \neg Q ) \right] \Rightarrow P $$ $P$ is "amount of obtuse angles x = 1". It's logical negation $\neg P$ is "$x < 1 \lor x>1$", but in proof is used only part of it $x>1$.

$Q$ sentence is "sum of angles = $180^\circ$", which we consider as true, so implication with $Q$ at end is always true i.e. $(x>2) \Rightarrow Q$ is true. So proof reduced to $(x>2) \Rightarrow \neg Q$ which you have above.