Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
When $p$ and $q$ are propositions such that "if $p$ then $q$" (which is an implication) is false only in the condition when $p$ (which is hypothesis) is true but $q$ (which is conclusion) is false.
But I couldn't understand why will the implication be true when the hypothesis itself is false. Shouldn't the implication be false when the initial assumption or hypothesis is false ?
This has been answered many many times on this site, but here is a simple reason. If one would use any other truth table for $P\to Q$ it would no longer mean what it is supposed to. If the hypothesis $P$ is true, then everyone agrees that $P\to Q$ should hold if and only if $Q$ does. Now suppose the hypothesis $P$ is false. If one would define (as you suggest) $P\to Q$ to be false in this case, regardless of $Q$, then $P\to Q$ would be true only if both $P$ and $Q$ are true, making it a symmetric operation for which we already have a name (namely "and", written $P\land Q$). If instead when $P$ is false we would define $P\to Q$ to be true only if $Q$ is, then the meaning of $P\to Q$ would ignore the status of the hypothesis $P$ completely, and be equivalent to $Q$. Finally if when $P$ is false we would define $P\to Q$ to be true only if $Q$ is false, then $P\to Q$ would be true if any only if $P$ and $Q$ have the same truth value, again a symmetric relation that does not reflect what we mean by implication. So the only reasonable option is the define, when the hypothesis is false, the relation $P\to Q$ to be true regardless of $Q$.
And this actually captures what one wants to say pretty well: if the hypothesis is true then we claim the conclusion must be true as well, but if the hypothesis is false, we did not mean to claim anything, so we shouldn't be called out for making a false claim in such cases.
The implication is the statement "if p then q". It's true if "every time" p is true, q is also true. Since p is "never" true, it satisfies the statement, so the implication is true. I wrote "every time" and "never" because the value of p and q never changes per se.
It can be also demonstrated with the empty set. We can say that all numbers in the empty set are even. Otherwise it wouldn't be a subset of the set of even numbers. But it is. So this can be reformulated like "if the empty set contains a number A, then A is even." p = "the empty set contains a number A" q = "A is even" So the implication must be true, or the empty set won't be a subset of even numbers.
