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  1. $∀x ∃y P(x, y)$

    For every $x$ there exists a $y…$

    This is true, because for every number $x$ there exists at least one number $y$ for the statement to be true. For example, we can choose $x=100$ and $y =10.$

  2. $∀y ∃x P(x, y)$

    ?

  3. $∃x ∀y P(x, y)$

    There exists a $x$ (such that) for every $y…,$

    This is true, we can increment $x$ and $y$ to infinite, and every time the statement will stay true. If we set an end to the infinite and give the largest number possible to $x,$ that number will hold true for any possible $y.$

  4. $∃y ∀ x P(x, y)$

    There exists a $y$ (such that) for every $x…,$ means that there is a $y$ that is less than $x^2$ for every possible $x.$ This is true because we can give $y$ a negative value, whatever value we will give $x$ will still be more than any negative value because it is quadratic.

Is this how you correctly understand quantifiers? Or am I understanding wrongly?

ryang
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ili
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2 Answers2

3

P.S. Forgot to add that x and y are REAL numbers

It isn't necessary to specify this, since complex numbers cannot be ordered anyway.

  1. $∀x ∃y P(x, y)$

    For every $x$ there exists a $y…$

    This is true, because...

Yes.

  1. $∀y ∃x P(x, y)$

This is true, because no matter what $y$ is, there is always some $x$ whose square is bigger, since $\mathbb R$ has no upper bound.

  1. $∃x ∀y P(x, y)$

There exists a $x$ (such that) for every $y…,$

This is true, we can increment $x$ and $y$ to infinite, and every time the statement will stay true. If we set an end to the infinite and give the largest number possible to $x,$ that number will hold true for any possible $y.$

Proposition $3$ is false, because no such $x$ exists, precisely because we cannot "set an end to the infinite".

  1. $∃y ∀ x P(x, y)$

There exists a $y$ (such that) for every $x…,$ means that the is a $y$ that is less than $x^2$ for every possible $x.$ This is true because...

Yes. But do be careful though: "there is a $y$ that is less than $x^2$ for every possible $x$" is ambiguous (it contains a hanging quantifier), and can be interpreted either as $∃y ∀x P(x, y)$ or $∀x ∃y P(x, y).$

ryang
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  • Thank you for this Answer, would it help if I draw the function? Does the visual help? – ili Oct 16 '21 at 19:59
  • @ili $P(x,y)$ is satisfied everywhere underneath the parabola; I feel that sketching this region isn't terribly useful, since your trouble is not with understanding that simple inequality, but with getting used to parsing first-order-logic sentences (in particular, discerning the difference in meaning when quantifiers switch order). – ryang Oct 16 '21 at 20:25
  • @ili As for md2perpe's comment, I wouldn't worry too much about his (not incorrect) point, since the important thing is that you do actually understand proposition $(1)$ and were merely citing an example, albeit in a way that seemed like you were explaining a different proposition. – ryang Oct 16 '21 at 20:29
  • Thank you for the help – ili Oct 16 '21 at 20:32
2

Your understanding and examples for your first and last sentence sounds right.

As for the sentence $\forall y \exists x P(x, y)$ in your case, it reads "for any $y \in \mathbb{R}$ there's some $x \in \mathbb{R}$ such that $y<x^2$" which is certainly true as there's no upper bound of the term $x^2$.

As for the sentence $\exists x \forall y P(x, y)$ in your case, it reads "there's some $x \in \mathbb{R}$ for any $y \in \mathbb{R}$ it's the case that $y<x^2$" which is clearly false since if you happens to find such $x$, you can always find a positive $y>x^2$.

In conclusion order matters for mixed quantifiers with different types...

cinch
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  • Thank you for this Answer, would it help if I draw the function? Does the visual help? – ili Oct 16 '21 at 19:59
  • @ili Yes it always does no harm... You can directly visualize above conclusions easily for your example with the right shaded region in your graph though I guess translated English sentence should be good enough here. – cinch Oct 16 '21 at 20:09
  • The statement I give for ∀x ∃y P(x, y), doesn't that translate to ∃x ∃y P(x, y), since I am not stating an arbitrary x? – ili Oct 16 '21 at 20:18
  • @ili ∀x ∃y P(x, y) has certainly different meanings than ∃x ∃y P(x, y) from their obvious syntactic difference, your above sentence 1. is translated correctly by your own effort. If we want to translate ∃x ∃y P(x, y) for your case, then it should read as "There's some x and y such that x is less than the square of y" which is much weaker than ∀x ∃y P(x, y) . So in general you can prove rigorously ∀x ∃y P(x, y) $\to$ ∃x ∃y P(x, y) if you've learned some mathematical logic. – cinch Oct 16 '21 at 20:27
  • Thank you for the help – ili Oct 16 '21 at 20:32