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I was told to write the following in propositional logic: "The product of two negative numbers if positive" and "every positive real number has exactly two square roots".

I think I know how to do this using predicate logic. The first one should be $\forall x \forall y (x<0 \land y<0)\rightarrow (x*y>0)$ and the second one should be $\forall x \exists y (x>0)\rightarrow (\sqrt{x}=y \land \sqrt{x}=-y)$. Is this correct?

The real problem is I'm told to write this in propositional logic, which I'm not sure how to do since I'm not allowed to use quantifiers? Is this a mistake or am I missing something?

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Your first sentence is correct, your second is not. You've written $\sqrt{x} = y \wedge \sqrt{x} = -y$; by the transitivity of equality, this would require $y = -y$, so $y = 0$ and therefore $x = 0$, which is not what you wanted to say. You should translate "$y$ is a square root of $x$" as "$y \cdot y = x$", not "$y = \sqrt{x}$". Also, even with this modification, your sentence only expresses "every positive real number has two square roots", not "every positive real number has exactly two square roots". You should try naming the two roots $y$ and $z$ and trying to express (1) $y$ and $z$ are both square roots of $x$; (2) $y$ and $z$ are different numbers; and (3) no number other than $y$ and $z$ is a square root of $x$.

As for the predicate/propositional logic issue: This sounds like either a mistake or a trick question. If the author of the problem really meant this to be propositional logic, it would be a very easy problem - since neither sentence involves any Boolean connectives, they'd just translate as $P$ and $Q$ respectively.

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  • $\begingroup$ Ah I see, So is the correction $\forall x \exists y \exists z ((x>0)\Rightarrow (y*y=x)\land (z*z=x)\land (y\neq z))$ satisfactory? Or is it necessary to include a third variable $\neg\exists w (w*w=x \land (w\neq y \land w\neq z))$? $\endgroup$ Commented Jul 5, 2017 at 0:02
  • $\begingroup$ You need the third variable, in order to get "exactly" two. Otherwise, yes, your new answer is correct. $\endgroup$ Commented Jul 5, 2017 at 1:30

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