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How many of the following quantified statements are true, where the domain of x and y are all real numbers?

∃y∀x(x^2 > y)

∃x∀y(x^2 > y)

∀x∃y(x^2 > y)

∀y∃x(x^2 > y)

Hello, I am currently stuck on this problem and am thinking the answer is three or four. Although I am leaning towards the fact that only three of them are true with the second being false. Please let me know if my logic is correct.

∃y∀x(x^2 > y): This statement says that there exists a real number y such that for all real numbers x, x^2 is greater than y. This is true because, if y is negative than the statement is always true.

∃x∀y(x^2 > y): This statement says that there exists a real number x such that for all real numbers y, x^2 is greater than y. This is not true because, for any x, there is always a y that is greater than x^2 (e.g. y=x^2+1).

∀x∃y(x^2 > y): This statement says that for all real numbers x, there exists a real number y such that x^2 is greater than y. This is true because for any x, we can always find a y that is less than x^2 (e.g. y=x^2-1).

∀y∃x(x^2 > y): This statement says that for all real numbers y, there exists a real number x such that x^2 is greater than y. This is true because, for any y, we can always find an x such that x^2 is greater than y (e.g. x=sqrt(y+1)).

Myle
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  • Your answer to the first one is flat wrong. Your rationale says "because … we can always find an $x$". But that is is "$\exists x$, not "$\forall x$". The formulas has "$\forall x$" which means not just that we can find some $x$ that does what is needed, but that every $x$ does what is needed, and it doesn't matter which one we pick. – MJD Jan 29 '23 at 23:35
  • I suggest you read this, which addresses a similar question: https://math.stackexchange.com/questions/690167/interpretation-of-quantifiers This one is also somewhat similar: https://math.stackexchange.com/questions/1418968/help-with-order-of-quantifiers – MJD Jan 29 '23 at 23:37
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    Hello, thanks for the help. I have read the post and I updated the logic of my first statement. This statement says that there exists a real number y such that for all real numbers x, x^2 is greater than y. This is true because, if y is negative than the statement is always true. I think my other logic for the other three statements are correct though. Can you please check if I am processing the information correctly? – Myle Jan 29 '23 at 23:46
  • I agree with you that three of them are true with the second being false. One tiny problem with the reasoning for the fourth statement, is that $x=\sqrt{y+1}$ can be undefined for some $y$. – peterwhy Jan 30 '23 at 00:14

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