Doubt regarding conditional statement in mathematical logic

Question

Conditional statement is represented as $p\to q$. Its truth table is given as:

$$ \begin {array}{|c|c|c|} \hline p & q & p\to q\\ \hline T&T&T\\\hline T&F&F\\\hline F&T&T\\\hline F&F&T\\\hline \hline \end{array}$$

As an explanation, my textbook gave the following example:

If $x$ is an odd integer, then $4$ divides $x^2-1$.
If $x$ is an odd integer, then $4$ does not divide $x^2-1$.
If $x$ is not an odd integer, then $4$ divides $x^2-1$.
If $x$ is not an odd integer, then $4$ does not divide $x^2-1$.

It says that (2) is false, while others are true, because in all the three statements, the conclusion is a fact.

I don't see how (3) is true. $4$ does not divide $x^2-1$ when $x$ is even. What am I missing?

Which textbook is that? Either it is very confusingly written for trying to illustrate the meaning of $\to$ with a statement that has free variables in it, or you must be missing or misquoting something. — hmakholm left over Monica, Jan 08 '15 at 05:00
I don't know if the number of times this question has been asked is countable or uncountable... — user21820, Jan 08 '15 at 05:34

score 2 · Answer 1 · answered Jan 08 '15 at 05:03

To understand $p\to q$, I always think of it as stating the $p\to q$ statement and then cycling through the 4 possibilities of $p$ and $q$.

So let's say $p$ is the statement, "It is raining today."

And $q$ is the statement, "The store is selling umbrellas."

I tell you $p\to q$. This means if $p$ is true, then $q$ must be true. Now, let's go through the possibilities and see if any of them contradict this statement.

It is raining today and the store is selling umbrellas.
It is raining today and the store is not selling umbrellas.
It is not raining today and the store is selling umbrellas.
It is not raining today and the store is not selling umbrellas.

Can you see that #2 is the only one that contradicts $p\to q$?

If $p$ is true, then $q$ must be true. Does this mean that if $p$ is false, then $q$ doesn't necessarily have to be false? — Tejas, Jan 08 '15 at 05:08
Yes, if $p$ is false, then all bets are off. $q$ can be anything. — turkeyhundt, Jan 08 '15 at 05:10

score 0 · Answer 2 · answered Jan 08 '15 at 05:03

Hint: Since $p \rightarrow q$ is false for the if-then statement (2), then (3) is the negation of a false statement (2). Once this point is taken, then one applies the table to the negation of a false if-then statement and the solution will be taken from there.

hmakholm left over Monica · Answer 3 · 2015-01-08T05:10:58.817

The four claims don't really have a truth value before someone decides on a value for $x$.

If, for example, we set $x=7$, then (1), (3) and (4) are indeed true and (2) false -- but not for the reason you quote. The reason why (4) is true here is that the antecedent is false -- namely, "7 is not an odd integer" is false.

On the other hand, if we set $x=6$, then (1) and (2) are automatically true, (4) is true too, but (3) is now false.

Note that the meaning of $\to$ is "designed" to be useful under a $\forall$ quantifier. So we can say that $\forall x\in\mathbb Z:(1)$ and $\forall x\in\mathbb Z:(4)$ are both true, because (1) and (4) are true both when $x$ is odd and when $x$ is even.

But $\forall x\in\mathbb Z:(2)$ and $\forall x\in\mathbb Z:(3)$ are false -- it is only some but not all possible $x$ that make e.g. (2) true.

Yeah. I think the book must have defined two statements before as 'p : $x$ is an odd integer' and 'q : $4$ divides $x^2-1$'. Then, in (3), as $p$ is false, it won't matter whether $q$ is false or true. Then the truth value of (3) would be true. Correct? — Tejas, Jan 08 '15 at 05:26

Doubt regarding conditional statement in mathematical logic

3 Answers3