Can anyone help me with this homework question of mine? I'm actually new to proofs.
Here's the question, "Prove, by contradiction, that no chord of a circle is longer than a diameter."
My only knowledge on proof by contradiction is on conditional statements, for instance, given $P\rightarrow Q$, then the proof by contradiction will first assume that the hypothesis $(P)$ is true with the denial / negation of the conclusion $(Q)$, then work your way to find a contradiction to the assumption.
Let $A,B$ be two points on the circle radius $r$, and center $O$, and $\angle AOB = \alpha$. Then $AB = \sqrt{r^2+r^2-2r^2\cos \alpha} = 2r\sin (\frac{\alpha}{2})$. If $AB > d = 2r \Rightarrow \sin (\frac{\alpha}{2}) > 1$, contradiction.
Let A and B be two points on a circle and let L be the length of the chord AB joining them. Suppose that L is larger than the diameter D of the circle. Consider the diameter through A, and call X the second point where it cuts the circle, so that AX is a diameter, whose length is D. By assumption, L is larger than D, so in particular B and X are different. Now consider the triangle ABX. Because AX is a diameter, this triangle has a right angle at B. Since the length of AB is larger than the one of AX (by assumption), we can find a point in the segment AB, call it B', such that the length of AB' is exactly the length of AX, which is D. This means that the triangle AB'X is isosceles. In particular, its angles at X and B' are equal, thus both are acute. But on the other hand the angle at B' can be seen to be obtuse, because its supplement is one of the acute angles in the right triangle BB'X. This contradiction gives the proof.
(We could also have argued, once we have the right triangle ABX, that the length of AX is larger than the one of AB because it is the hypotenuse, and this contradicts the assumption).
"My only knowledge on proof by contradiction is on conditional statements, for instance, given (P)→(Q) , then the proof by contradiction will first assume that the hypothesis (P) is true with the denial / negation of the conclusion (Q) , then work your way to find a contradiction to the assumption."
That's a good start. Let's see what this looks like with the simple illustration exercise provided.
To avoid a technicality that is not good here, let's rely on a drawing :
With obvious notations,
(P):"$a$ a chord of a circle of diameter $c=DD'$"
(Q):"$a\leq c$
As simple as it is, hoping to demonstrate $P\implies Q$ still requires some knowledges. Let's choose the following knowledges: (1)$CC'B$ is a right triangle; (2)therefore, in $CC'B, BC\leq CC' \color{red}{(*)}$.
As you wrote,let us "assume that (P) is true with the denial non(Q)$$\text{(nonQ):}c<a$$ of the conclusion (Q) , then work your way to find a contradiction."
Here, it's immediate to contradict $\color{red}{(*)}$ and to conclude by having used the tool that is prove by contradiction.