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By the logic, I can understand that if $P$ is true, $P \lor Q$ is also true whatever $Q$ is. Since $x\ge 0 \iff (x>0 \lor x=0)$, if I know $x>0$, can I conclude $x\ge 0$? Sorry for some sort of stupid question.

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  • $\begingroup$ $$P\Longrightarrow P\vee Q$$ $\endgroup$
    – user0102
    Commented Oct 24, 2019 at 21:41
  • $\begingroup$ Yes you can. Try encasing math symbols in dollar signs: \$x>0\$ to make it look better ($x>0$) $\endgroup$
    – David Diaz
    Commented Oct 24, 2019 at 21:47
  • $\begingroup$ You could also say $x\gt-1$ however, and so your statement is weaker than the original. $\endgroup$
    – JMP
    Commented Oct 24, 2019 at 21:49

4 Answers 4

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Yes of course

$$x>0 \implies x\ge 0$$

since

$$x\ge 0 \equiv \left(x=0\quad \lor \quad x>0\right)$$

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Yes, everything you have written is correct, except for the suggestion that there's something stupid about it. The point that you raise was considered important enough for a footnote beginning with "There is one slightly perplexing feature of the symbols $ \geq$ and $ \leq$."

That's in a book by Michael Spivak that is for first-year University calculus students who would be prepared to continue in math up to and including the Ph.D. level, without having to go back and upgrade their calculus.

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Yes. As a proof:

\begin{array}{lll} 1. & x \geq 0 \Leftrightarrow (x > 0 \lor x = 0) & \geq definition\\ 2. & x > 0 & Given\\ 3. & x > 0 \lor x = 0 & \lor \ Intro \ 2\\ 4. & x \geq 0 & \Leftrightarrow \ Elim\\ \end{array}

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Yes. $P \implies P$ or $Q$.

so if $P$ is $x > 0$ and $Q$ is $x=0$ then if $x> 0$ is true, then $x > 0$ or $x =0$ is true. Which is the same thing as $x \ge 0$ is true.

I can see why this seems weird: We know something specific ($x > 0$) and we are suggesting something less specific and even implying something we know to be false ($x \ne 0$) to possibly be true ("$x \ge 0$??? so that means... $x$ could be $0$???").

But consider this. Suppose you knew $x = 3$. Can we conclude that $x \ge 0$? Well, of course we can! $3 \ge 0$ and as $x = 3$ we know $x \ge 0$. But $x \ge 0$ implies that $x$ could be $15$. Are we saying that $x$ could be $15$? Are we saying that $3$ could be $15$?!?!?!

Well, no. The statement that $x \ge 0$ isn't necessarily a statement of all potential possible things[1] and the only thing we know for certain is that $x < 0$ is false. The statement that $x \ge 0$ is literally just a statement of fact; it is the case that $x$ is greater than or equal to $0$. We may or may not know any more than that but that much is true.[2]

Another analogy is "Penguins are animals". Well, "animals" can be many things including camels or elephants or eagles. But "Penguins are animals" doesn't mean that all those possibilities ("Penguins might be camels"? No Way!) are open.

[1] although it could be.

[2] actually $x$ must be something. Its not possible that is both $x = 0$ and $x > 0$. So it is never a case that all possibilities are open. $x$ must be something and if its that, it can't be anything else. We might not know what it is, but it can't be more than one unknown option. But that's okay. Saying $x$ is one of several things isn't implying that all are equally valid.

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