0

Suppose we want to prove the implication P implies Q. Now we can use proof by contradiction and show that there is no case were P is true and not Q is true. But this does not imply that there is a case where P is true and Q is true. Maybe P is always false. In general in mathematics we have implications that the hypothesis can be true. So does it make sense to prove implications only in systems where there is at least one case where the hypothesis can be true?

  • 2
    If P is always false the implication P → Q is true for any Q, but that may not be at all obvious. People in number theory prove many theorems assuming existence of odd perfect numbers. If they do not, in fact, exist then those theorems will help proving it eventually, by allowing to derive a contradiction. So no, it does not make sense to restrict implications to those whose premise can be true, in many cases we simply can not tell if it can or not. – Conifold Jul 9 '20 at 6:06
  • If some of the answers below satisfies you, please accept it. – Mauro ALLEGRANZA Sep 7 '20 at 14:13
2

I'm not sure the fully understand your question, but I'm trying to answer, reading the title-question:

Do mathematicians care about implications where the hypothesis is always false?

as regarding the well-known issues about the truth-functional definition of the "if..., then..." connective, and specifically regarding the fact that "if P, then Q" is considered true when P is false.

The first point must be a sort of disclaim: the propositional connectives are a (very) simple mathematical model of natural language, suited for modelling formal arguments.

Some of them are "proxing" natural language mechanism in a good way (negation and conjunction), while others do it with some arbitrariness (disjunction, inclusive : vel instead of aut). Finally, some do it in a quite "unnatural" way: the conditional.

Having said that, we must take into account the key role played by the truth-functional connective "if ... then" in the inference rule of Modus ponens that allows us to infer from the premises A and "if A, then B", the conclusion B.

We must read it as Gottlob Frege did in his Begriffsschrift (1879):

assuming as true both the premises, the assumption that "if A, then B" is True, rule-out the row T-F in the truth-table for implies, while the assumption that also A is True rule out two other rows (F-F and F-T, respectively). Then, the conclusion that B is True is licensed.

So, assuming the truth-functional definition of "if A, then B", we have that the truth of A is a sufficient condition for that of B.

This "mechanism" is what is used again and again in mathematical proofs: either having assumed some axiom A as true or having available some already proven theorem A, we may prove a new theorem B through a deductive argument showing "if A, then B".

In this sense we may answer:

No, mathematicians do not care about implications where the hypothesis is false.

In other words, a true theorem like: "if 0 =1, then the Continuum hypothesis is true" is not of interest.

0

Contrary to @Mauro ALLEGRANZA's reply, there are cases where mathematicians rely on implication where the premise is false. This occurs in certain theorems where either the statement, or the proof, can be rendered 'cleaner' (in the sense of having fewer assumptions in the statement, or better proof structure) by exploiting this feature of implication.

A simple example lies in the definition of course-of-values induction:

IF: We can establish P(n) whenever P(m) holds for all m<n, THEN: We can conclude P(n) holds for all n.

Notice how this translates when n=0: The clause 'for all m<n' becomes vacuous because there ARE NO m<0! That means the premise 'P(m) holds for all m<0' must be taken to be false. [The practical effect is to require the prover to establish P(0) without recourse to induction.] Without recourse to that feature of implication, course-of-values induction would have to be stated more complexly, thus:

IF: We can establish p(0), and when 0<n we can establish P(n) whenever P(m) holds for all m<n, THEN: We can conclude P(n) holds for all n.

  • can you provide an actual example of this being done in mathematical research? i hope that makes sense [apologies if i misunderstood] – user46524 Jul 8 '20 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.