This is the third post in a series on logic, with a focus on how it is expressed in English. We’ve looked at basic ideas of translating between English and logical symbols, and in particular at negation (stating the opposite). Now we are ready to consider how to change a given statement into one of three related statements.
A conditional statement and its converse
We’ll start with a question from 1999 that introduces the concepts:
Math Logic - Determining Truth A number divisible by 2 is divisible by 4. I'm suppose to figure out the hypothesis, the conclusion, and a converse statement, say whether the converse statement is true or false, and if it is false give a counterexample. I don't understand.
Ricky has been asked to break down the statement, “A number divisible by 2 is divisible by 4,” into its component parts, and then rearrange them to find the converse of the statement. I took the question:
You're asking about the terminology of logic, which is important in math to help us talk about proofs and how we know something is true. Words such as "converse" allow us to talk about our reasoning and see whether we are really making sense. A statement such as "any number divisible by 2 is divisible by 4" (I've changed "a" to "any" to clarify the statement a little) can be rewritten as IF a number N is divisible by 2, THEN the number N is divisible by 4 The hypothesis, or premise, is what is given or supposed, the "if": a number N is divisible by 2 The conclusion is what is concluded from that, the "then": the number N is divisible by 4
We commonly write such a statement symbolically as “\(p\rightarrow q\)“, where the hypothesis is p and the conclusion is q. I rewrote each part slightly to allow it to exist outside of the sentence, naming the number N to avoid needing pronouns. What was important was to rewrite the statement in if/then form.
The converse of this statement swaps the hypothesis and conclusion, making “\(q\rightarrow p\)“:
The converse of the statement "IF a THEN b" is "IF b THEN a", turning the statement around so that the conclusion becomes the hypothesis and the hypothesis becomes the conclusion. In this case, the converse is IF a number N is divisible by 4, THEN the number N is divisible by 2
Ricky was asked to decide whether the converse is true or not, and then prove it, whichever way it goes. This part goes beyond mere logic and enters the realm of “number theory”; but commonly this sort of question is first asked in cases where the proof is not too hard, which is the case here.
Now we have to consider whether either statement is true. A statement and its converse may be either both true, or both false, or one true and the other false; knowing whether one is true says nothing about whether the other is true. In this case, the original statement is false. (This makes me wonder if you copied the problem wrong; it doesn't sound like this possibility was considered in the question.) How do I know it's false? Because I can give a counterexample: a number N for which the hypothesis is true but the conclusion is false. Can you see what I can use for N, which is even but not divisible by 4?
To show that a statement is not always true, we only need to find an example for which it is false. In this case, an easy example is 2, or we could use 6, or 102, or whatever we like.
But the question was about the converse:
However, the converse is true. See if you can see why. You might just try listing lots of numbers that are divisible by 4, and see whether they are all even. If all your examples are even, you haven't proven anything; but the list may suggest to you a reason why you will never be able to find a counterexample. That reason would be the basis of a proof.
I didn’t give a proof, in part because Ricky needed to think about that for himself, but also because I didn’t know what level of proof Ricky is expected to handle. One approach is to see that any multiple of 4 can be written as 4k for some integer k; but that can be written as 2(2k), which is clearly a multiple of 2.
Converse, inverse, and contrapositive
Now we can review the meanings of all three terms, in this 1999 question, which again uses an example from basic number theory:
Contrapositive, Converse, Inverse Let m and n be whole numbers, and consider the statement p implies q given by "if m + n is even, then m and n are even." A) Express the contrapositive, the converse and the inverse of the given conditional. B) For the statements that are true, give a proof. C) For the statements that are false, give a counterexample. I have part A (I think) but I'm having trouble deciding which statements are true and which are false, and I'm completely lost on the proofs.
Doctor Kate could have asked Hollye for her answers to part A, to make sure she understands that part; but she chose to provide them:
I'll give you what I got for the first part, to see if it's the same as what you got. First, though, here's what my "p" and "q" are:
p is "m + n is even"
q is "m and n are even"
~p is "m + n is odd" ("~p" means "NOT p")
~q is "either m or n is odd"
It’s important to identify the parts of a conditional statement (if p then q); and since two of the new statements require negations, that also might as well be done early. Notice that the negation of “is even” could have been written as “is not even”, but since every number (integer) is either odd or even, writing “is odd” is cleaner. Also, the negation of “both are even” is “at least one is not even”; this is an application of De Morgan’s law, or can be seen by considering that if it is not true that both are even, then there must be one that is not even. These ideas were discussed last time.
Now here are the new statements:
A. Contrapositive (if ~q then ~p): "If either m or n is odd, then m + n is odd." B. Converse (if q then p): "If m and n are even, then m + n is even." C. Inverse (if ~p then ~q): "If m + n is odd, then either m or n is odd."
We saw the converse above; there we just swap p and q. The inverse keeps each part in place, but negates it. The contrapositive both swaps and negates the parts.
To check out which of these are true, it's best to experiment a little. Try some numbers. Let's look at and pick some numbers where m or n is odd: 2 and 3 3 and 7 1 and 8 Notice that I tried to pick a variety of numbers - sometimes both odd, sometimes only one. That is because the opposite of "m and n are even" is "at least one of m or n is odd, and maybe both are." You can figure that out by imagining all sorts of things that don't satisfy "m and n are even." It could be really false (both m and n are not even) or just a bit false (only n is not even or only m is not even). Anyway, let's take a look at these numbers. Is 2 + 3 odd? Yes. Is 3 + 7 odd? That's 10... no, it's not. Wait, statement A says 3 + 7 WOULD be odd. This is a counterexample.
So now we know that the contrapositive, “If either m or n is odd, then m + n is odd,” is false, because there is at least one case, 3 and 7, where the hypothesis is true but the conclusion is false.
Remember that a statement like "<BLAH> is always true" can be proven false by just one example of when <BLAH> could be false. If I claim all dogs are black, all you have to do is bring me a Dalmatian, and I am wrong, even if a lot of dogs are black. Statement A is claiming that ALL the time, if one or both of n or m is odd, n+m is ALWAYS odd. But look, we found an example where it isn't. So statement A is false.
That’s the essence of a counterexample.
Doctor Kate continued, showing a way to prove that B and C (the converse and inverse) are both false. You can read that on your own, since my goal here is just to look at the logic. (We’ll have a series on proofs some time in the future.)
Rewriting the statement
Continuing, here is a similar question, where statements must first be written in conditional form:
Converse, Inverse, Contrapositive For the directions it says "Write the converse, inverse, and contrapositive of each conditional. Determine if the converse, inverse, and contrapositive are true or false. If false, give a counterexample." I can't seem to do these: All squares are quadrilaterals. If a ray bisects an angle, then the two angles formed are congruent. Vertical angles are congruent. Thank you!
The second statement is straightforward, but the others need thought. Doctor Achilles first defined the three forms, as we’ve already seen, and then dealt with the first case:
The problem with your questions are that they don't neatly fit into the "if p, then q" format, so you need to first find EQUIVALENT sentences that are "if p, then q." Your first example says "all squares are quadrilaterals." That is the same as saying "if x is a square, then x is a quadrilateral."
Thus, “all” (the universal quantifier) translates directly to a conditional. The answer, left for Hana to do, will be:
- Converse: “If x is a quadrilateral, then x is a square”; i.e. “Any quadrilateral is a square.”
- Inverse: “If x is not a square, then x is not a quadrilateral”; i.e. “Anything that is not a square is not a quadrilateral.”
- Contrapositive: “If x is not a quadrilateral, then x is not a square”; i.e. “Anything that is not a quadrilateral is not a square.”
The original statement, and the contrapositive, are true, because a square is a kind of quadrilateral; the converse and inverse are false, and a counterexample would be an oblong rectangle, which is not a square but is a quadrilateral.
The questions so far, where they dealt with truth at all, only asked about specific examples. Our last two questions will look more broadly at when these statements are equivalent.
Which can I use in a proof?
Consider this question, from 2002:
Contrapositive I have a logic proof that I'm trying to solve. I'm up to the point after I've written down all my givens. One of the givens is p-->q. I want to say ~p-->~q, with my reason being inverse. Am I allowed to do this?
If we know a statement is true, can we conclude that the inverse is true? Doctor TWE answered with a counterexample:
No. Although the statement ~p --> ~q is called the inverse of p --> q, it does not necessarily follow.
Let's look at an example. Suppose that:
p = "X is 2"
and
q = "X is an even number"
Clearly, p --> q is true ("If X is 2 then X is an even number."). But is the inverse, ~p --> ~q, also true? This statement reads, "If X is NOT 2 then X is NOT an even number." Suppose X = 4. Then the "if" part, X is NOT 2, is true, but the "then" part, X is NOT an even number, is false. So the statement as a whole is false.
Here we are using logic to talk about logic: The statement “For all p and q, \((p\rightarrow q)\rightarrow(\lnot p\rightarrow\lnot q)\)” is false! Sometimes both original and inverse are true, but we can’t conclude the latter from the former.
What you *are* allowed to use in a logic proof is the contrapositive. The contrapositive of p --> q is ~q --> ~p. It turns out that any conditional proposition ("if-then" statement) and its contrapositive are logically equivalent. In our example, the contrapositive of "If X is 2 then X is an even number" would read, "If X is NOT an even number then X is NOT 2." We can see that this is also true.
Giving one example where the contrapositive is true does not prove that it is always equivalent; we’ll prove it below.
A third possible "switching" of the statement p --> q is q --> p. This is called the converse, but like the inverse, it does not follow logically from the original statement. The converse of our original statement would read, "If X is an even number then X is 2." Clearly, not all even numbers are 2. So the converse statement is false. (It turns out that the inverse and converse statements are logically equivalent to each other - but not logically equivalent to the original statement.) To summarize, given the statement p --> q: The inverse is q --> p, NOT equivalent to p --> q The converse is ~p --> ~q, NOT equivalent to p --> q The contrapositive is ~q --> ~p, IS equivalent to p --> q
In fact, the converse and inverse turn out to be equivalent to one another, though not to the original.
Why is the contrapositive equivalent?
Let’s look at one more, from 2003:
Truth of the Contrapositive The inverse of a statement's converse is the statement's contrapositive. True, but why? I don't know how to explain it. I tried an example: p: I like cats. q: I have cats. Converse If I have cats, then I like cats. Inverse If I don't like cats, then I don't have cats. Contrapositive If I don't have cats, then I don't like cats. I still can't explain the answer "true" that I came up with. Maybe it is wrong.
The opening statement describes the contrapositive as the inverse of the converse. What that means is this: Suppose we start with “\(p\rightarrow q\)“. Its converse is “\(q\rightarrow p\)” (swapping the order), and the inverse of that is “\(\lnot q\rightarrow\lnot p\)” (negating each part). This is the contrapositive. In the example, the converse of “If I like cats, then I have cats” is “If I have cats, then I like cats”, and the inverse of that is “If I don’t have cats, then I don’t like cats”, which is the contrapositive.
Doctor Achilles, perhaps misreading the question, answered the bigger question: Which of these are true?
The contrapositive is true if and only if the original statement is true. It is false if and only if the original statement is false. So it is logically equivalent to the original statement. Let's say you have a conditional statement: "if I like cats, then I have cats." What does this mean? When is it true? When is it false? Well, for starters, if you like cats and you have cats, then the conditional will come out true. That is, (P -> Q) is true when P and Q are both true. Also, if you don't like cats and you don't have cats, then the conditional will come out true. That is, (P -> Q) is true when P and Q are both false. Also, if you don't like cats and you have cats, then the conditional still comes out true. Remember, it says that if you like cats, then you will have them; it makes NO claim at all about what will happen if you don't like cats. So, (P -> Q) is true when P is false and Q is true. However, if you like cats and you don't have cats, then the conditional will come out false. It says that if you like cats, then you will have them. So it is proven wrong if you like cats, but you still don't have any. So, (P -> Q) is false when P is true and Q is false.
In effect, he has made a truth table:
P Q P->Q --- --- ------ T T T F F T F T T T F F
If you are unconvinced by any of the reasoning, see Why, in Logic, Does False Imply Anything?.
So to review, (P -> Q) is true under any of these conditions: P is true and Q is true P is false and Q is true P is false and Q is false And it is only false under this one condition: P is true and Q is false You can say that another way, using ~P and ~Q (not-P and not-Q). (P -> Q) is true under any of these conditions: ~P is false and ~Q is false ~P is true and ~Q is false ~P is true and ~Q is true And it is only false when: ~P is false and ~Q is true Is there another sentence that uses P and Q that is only false when ~P is false and ~Q is true? Yes, the sentence is: (~Q -> ~P) You can go through the same analysis of this sentence as I did for (P -> Q) and you'll find that it has the same truth conditions.
So the truth table for the contrapositive is that same as for the original; this is what we mean when we say that two statements are logically equivalent.
We can instead just think through the example:
You can also understand this more intuitively: The sentence: "If I like cats, then I have cats." says that as long as the first part, "I like cats," is true, the second part, "I have cats," will definitely be true. In this case, what does "I don't have cats" mean? The only way "I don't have cats" can happen is if "I like cats" is false. That is, the only way "I don't have cats" can be true is if "I don't like cats" is true. Therefore, "If I don't have cats, then I don't like cats."
Which is more convincing? That depends upon you.