Proving propositions involving "if and only if"

[Mr Davis 97]

I am usually pretty good about interpreting what a question is asking when it is in the form, "prove that if p, then q," where p and q are statements. However, I cannot seem to understand how to interpret when it is in the form "prove that p if and only if q." The statement I am working with currently is this: "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle."

How do I begin proving this? What I mean is how can I decompose this proposition to clearer propositions that can be proved individually? I looked online and I couldn't find much about iff statements, so help would be appreciated.

 

[S.G. Janssens]

"P if and only if Q" means that you need to show that P and Q are either both true or both false. Hence, you need to exclude the possibility that one is true while the other is false. That is, you need to show that "P implies Q" as well as "Q implies P".

So, for your example,

P = a radius of a circle is perpendicular to a chord
Q = the chord is bisected by the radius

Now prove the implication "if P, then Q". Next, prove the implication "if Q, then P". For both implications, use any method of your choice (= direct proof, contraposition, or contradiction).

 

[member 587159]

Generally, if you have to prove: p <=> q, you have to proof p=> q and q=>p.

In logic, p<=> q is equivalent to (p=> q) ∧ (q =>p) (∧ is the logic symbol for 'and').

In your case, when you have to proof that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius, you need to show that:

If a radius of a circle is perpendicular to a chord, then the chord is bisected by the radius. (=>)

AND

If the chord is bisected by the radius, then a radius of a circle is perpendicular to a chord. (<=)

 

[fresh_42]

If you're good at $p → q$ then also at $q → p$. Both together is $p ↔ q$.
What makes it a bit of messy is the additional condition $(c)$ "assume the chord isn't a diameter". Does it belong to $p$ or to $q$?
Such additional conditions are often not very well stated and one has to decide where to place them:
$c → (p ↔ q)$ or
$(c ∧ p) ↔ q$ or
$(c ∧ p) → q → p$ or
$p ↔ (c ∧ q)$ or
$(c ∧ q) → p → q$ or
$(p ∧ c) ↔ (q ∧ c)$?

(Remark @ logicians: I haven't checked whether some of these are already equivalent.)
Often it is clear by the context, or $c$ is a special, often trivial case of either $p$ or $q$.

Here it is (I think) $(c ∧ p) → q → p$, where $c= \text{ (chord is no diameter) }$ , $p= \text{ (radius bisects chord) }$ and $q= \text{ (radius perpendicular to chord) }$.

 

[mfb]

Here it is (I think) $(c ∧ p) → q → p$, where $c= \text{ (chord is no diameter) }$ , $p= \text{ (radius bisects chord) }$ and $q= \text{ (radius perpendicular to chord) }$.

That is the best you can prove, but from the way the problem statement is written I would expect $(p ∧ c) ↔ (q ∧ c)$ which is equivalent to $c → (p ↔ q)$

 

[Mr Davis 97]

That all helps a ton. But given the statement "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle," I am still not completely sure which part to assign p and which part to assign q. In simpler cases where, for example, we have "If I am a mammal, then I am human," it is obvious that p = I am a mammal, and q = I am a human. But when we have something like "I am mammal if and only if I am a human," I can't really decipher which part of the statement is assigned to p and which is assigned to q.

 

[fresh_42]

But when we have something like "I am mammal if and only if I am a human," I can't really decipher which part of the statement is assigned to p and which is assigned to q.

That doesn't matter in "if and only if", aka "iff" statements. It is symmetric:
$(p ↔ q) ↔ ((p→q) ∧ (q→p))$

 

[Mr Davis 97]

So for "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle," I could just assign P to "a radius of a circle is perpendicular to a chord" and Q to "the chord is bisected by the radius," and then go on to show that "if P then Q" and "if Q then P" in any order I choose?

 

[mfb]

Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius

The assignment becomes obvious if you highlight they keywords.

 

[fresh_42]

Yes. Plus you have "a chord is not a diameter" as given at hand whenever needed or only in one direction as mentioned by mfb in #5.

 

[Mr Davis 97]

Alright. Also, for the proposition "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius," is there an "if" part and an "only if" part? In proving this statement, my book starts with "For the 'if' part, we prove that if a radius of a circle is perpendicular to a chord, then the chord is bisected by the radius. For the 'only if' part, we prove that if a chord is bisected by a radius then the radius of a circle is perpendicular to the chord." What does he mean by "if" and "only if" parts?

 

[mfb]

If-part:

Prove that a radius of a circle is perpendicular to a chord if the chord is bisected by the radiusOnly if part:

Prove that a radius of a circle is perpendicular to a chord only if the chord is bisected by the radius
Which is easier to understand if you swap the order:
Prove that [a] chord is bisected by the radius if a radius of a circle is perpendicular to [the] chord

 

[fresh_42]

You may restructure those sentences as

"Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius,"
⇔
"Prove that a radius of a circle is perpendicular to a chord ⇔ the chord is bisected by the radius,"
⇔
If "a radius of a circle is perpendicular to a chord" then "the chord is bisected by the radius," and only if "the chord is bisected by the radius," then "a radius of a circle is perpendicular to a chord"
⇔
a) If "a radius of a circle is perpendicular to a chord" then "the chord is bisected by the radius,"
and
b) if "the chord is bisected by the radius," then "a radius of a circle is perpendicular to a chord"

To call it if-part (⇒) and only if-part (⇐) is IMO sometimes confusing. Usually it gets clear by the proof. You can also say that the if-part (⇒) shows, that the statement on right hand side is a necessary condition (for the condition on the left to hold), and the only if-part (⇐) that this statement on the right hand side is also a sufficient condition (for the condition on the left to hold).

 

[Mr Davis 97]

So let's see. Let me see how I can apply to this next proof: "Prove that the product of two integers is odd if and only if both of the integers are odd."

The "if" part:
the product of two integers is odd if both of the integers are odd

The "only if" part:
the product of two integers is odd only if both of the integers are odd
or
both of the integers are odd if the product of the two integers is odd

So if I were to prove the "if" part, how would it look in terms of P and Q (such that if P then Q)? Would P = both of the integers are odd and Q = the product of two integers is odd?

 

[mfb]

Right.

 

[Mr Davis 97]

Why would it not be P = the product of two integers is odd and Q = both of the integers are odd?

 

[S.G. Janssens]

Why would it not be P = the product of two integers is odd and Q = both of the integers are odd?

Because, as fresh_42 wrote in post #7, there is a symmetry in P and Q. (The irony, however, is that in that post, equivalence itself is used in the explanation.) In words: Do you agree that "if and only if" means that you need to prove

"P implies Q" and "Q implies P"

? If you do agree, then notice what happens when P and Q swap places: We prove: "Q implies P" and "P implies Q", which is exactly the same as in the line above. So, it does not matter which statement is called "P" and which is called "Q".

 

[Mr Davis 97]

Alright, I think it all makes sense. Just to be clear, the irrelevance of which part is P or Q is only for biconditional propositions, right?

For example, if we just had to prove the conditional statement that "the product of two integers is odd if both of the integers are odd," P would have to be "both of the integers are odd" and Q would have to be "the product of two integers is odd," such that if P then Q?

 

[fresh_42]

So let's see. Let me see how I can apply to this next proof: "Prove that the product of two integers is odd if and only if both of the integers are odd."

The "if" part:
the product of two integers is odd if both of the integers are odd

⇒ That is if both of the integers are odd then the product of two integers is odd

So far so good.

The "only if" part:
the product of two integers is odd only if both of the integers are odd

⇐ That is if the product of two integers is odd then both of the integers are odd

Yes. But as a stand-alone version it is confusing.

or
both of the integers are odd if the product of the two integers is odd

⇐ That is if the product of two integers is odd then both of the integers are odd

Yes, and better to read / see.

So if I were to prove the "if" part, how would it look in terms of P and Q (such that if P then Q)? Would P = both of the integers are odd and Q = the product of two integers is odd?

Again. It is symmetric. No matter what you choose P and Q to be.
Usually one would take it as it is:
"Prove that the product of two integers is odd (P) if and only if both of the integers are odd (Q)."

In this case your splitting above reads:
a) Q ⇒ P (if part = P is necessary for Q)
b) P ⇒ Q (only if part = P is sufficient for Q)

 

[mfb]

Just to be clear, the irrelevance of which part is P or Q is only for biconditional propositions, right?

Right.

For example, if we just had to prove the conditional statement that "the product of two integers is odd if both of the integers are odd," P would have to be "both of the integers are odd" and Q would have to be "the product of two integers is odd," such that if P then Q?

Right.

You can assign P and Q in the opposite way but then you have to prove Q=>P instead of P=>Q. They are just variable names, they don't have a deeper meaning.