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  1. There is one and only one natural number $n$ for which the proposition formula $P(n)$ holds.

  2. There are no integer solutions to the equations $x^2-y^2=11$

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1 Answer 1

up vote 2 down vote accepted

(I think that you meant predicate logic, not propositional logic.)

For (1) the trick is to do it in two parts: first specify that there’s at least one $n$ that works, and then specify that it’s the only $n$ that works. Thus, you want to start with the proposition $$\exists n\big(n\in\mathbb{N}\land P(n)\big)\;.$$ To say that there’s only one such $n$, you need to add a clause saying that every $m$ satisfying $P$ is equal to the specific one whose existence you just asserted: $$\exists n\bigg(n\in\mathbb{N}\land P(n)\land\forall m\Big(\big(m\in\mathbb{N}\land P(m)\big)\to m=n\Big)\bigg)\;.$$ In words this boils down to:

There is a natural number $n$ satisfying $P$ such that if $m$ is any natural number satisfying $P$, then $m=n$.

For (2), consider how you’d express the statement that there is at least one integer solution to the equation $x^2-y^2=11$; that would be

$$\exists n \exists m(n\in\mathbb{N}\land m\in\mathbb{N}\land\dots)\;,\tag{1}$$

where I leave you to fill in the blank. The statement that you actually want says that $(1)$ is false; how do you modify $(1)$ to say that?

(By the way, is (2) true, or false?)

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I'm pretty sure you cannot use quantifications in propositional logic, only in predicate logic. EDIT: did not read your very first sentence. –  user12205 Dec 7 '11 at 3:34
    
@Jeroen: I’m assuming that propositional is an error for predicate, as the question is impossible otherwise. EDIT: It might not have been there when you read the answer: I added it a little after I originally posted. –  Brian M. Scott Dec 7 '11 at 3:36
    
It would be possible if you were to say e.g. P = "There is one and only one natural number n for which the proposition formula $P(n)$ holds." Then you could express 1) as P. It would be totally stupid of course :) –  user12205 Dec 7 '11 at 3:38

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