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In the Princeton Companion to Mathematics' explanation of the logical connective "implies" (ei. "if P then Q"), they give an example where:

"If n is a positive integer, then the statement "n is a perfect square with final digit 7" implies the statement "n is a prime number", not because there is any connection between the two but because no perfect square ends in 7"

Why do we have to conclude that "n is a prime number" -- I can't think of a prime number that could be a perfect square.

Moreover, the way I understand "if P then Q" logical connectives, is that Q is always true when P is always true. but "n is a perfect square with final digit 7" is never true. So how can the statement "n is a prime number" be logically implied by a statement that is always false?

user151959
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    Because in logic the connective "if ..., then ___" is "modelled" truth-functionally with $\rightarrow$; as you said $P \rightarrow Q$ is true when $P$ and $Q$ are both true, but also when $P$ is false. The only case in which $P \rightarrow Q$ is false is when $P$ true and $Q$ false. – Mauro ALLEGRANZA May 19 '14 at 07:02
  • See also this post – Mauro ALLEGRANZA May 19 '14 at 07:08

2 Answers2

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The statement $p\to q$ is false in the case $p$ true, $q$ false, and is true otherwise. In the case you have given, $p$ is the statement "$n$ is a perfect square with final digit $7$". This is always false. And $q$ is the statement "$n$ is prime". Depending on $n$ this could be true or false, but in either case the combined statement is true.

Why is my first sentence the case? There are a few things worth considering here.

  1. The statement is the definition of implication. It's just a statement of how we intend to use the symbol. Therefore there is no question of proving it; it cannot be right or wrong; it can only be sensible or stupid. In my following points I will try to convince you that it is sensible.
  2. Consider the following scenario: you go to the doctor, who tells you, "if you take this medicine then you will get better". Under what circumstances could you reasonably say that this diagnosis was false? Only if you took the medicine and didn't get better. So the only way we could say $p\to q$ is false is when $p$ is true and $q$ is false. You might say that if you don't take the medicine then you have not tested the doctor's statement; so if $p$ is false then $p\to q$ should be "meaningless" rather than true; but this is a luxury we don't have. We need every combination $p\to q$, $p\vee q$ and so on to be definitely either true or false in every case, otherwise we never know that simplifying some complicated statement is not going to end up "meaningless".
  3. How about the following? - "if $0=1$ then $1=2$". Although $0=1$ is obviously false, and so is $1=2$, you could argue that the "if. . . then. . ." refers to the way in which we obtain the second from the first. Since we have added $1$ to both sides, and this is a correct arithmetic operation, it seems reasonable to say that the statement as a whole is true.
  4. The usefulness of a definition is judged by its consequences. Defining $p\to q$ the way we have done leads to conclusions such as: if $p\to q$ and $p$ are both true, then $q$ must be true. This is the way that we intuitively think "if. . . then. . ." should behave, and if it didn't work out then there would be something seriously wrong with our definition.

Hope this is of interest.

David
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Now, any natural number in $N$ can be expressed in one of the following ways $ 10k + 1, 10k +2,....., 10k + 9$ for some $k$ in $N$.

Find out all their squares and this will show you that no number with final digit 7 is TRUE/FALSE. This is your P.

IS the statement Q true/false?

If you consider the negation of Q, what do you get? If it is not a prime then it is composite. If it is composite then it could be a square! IS it a square?

wannadeleteacct
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