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"The following statements are about positive real numbers. Which one is true? Explain your answer."

$\forall x, \exists y$ such that $xy < y^2$
$\exists x$ such that $\forall y, xy < y^2$

I try understanding this but the English is difficult for this problems. I think first one I say

$$xy<y^2 \iff x<y$$

so if counterexample $y=1$ is true, then $x \geq y$ is false. This statement is false.

For second statement

$$xy < y^2 \iff x<y$$

counterexample $x=1$ is true, then $y \leq x$ is false. Statement is false.

Is correct understanding? I feel doubt about my work.

Pefta Seng
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    I think you aren't reading the quantifiers correctly. The first one, for example, says "for any $x$ we can find some $y$ such that $xy<y^2$." Indeed , since $x>0$ then $y=\frac x2$ works. – lulu Sep 07 '20 at 20:54
  • The second statement reads "for some, particular, positive, real $x$ we have $xy<y^2$ no matter what value we take for $y$." Can you produce a counterexample to that? To stress: it's not enough to pick some particular value for $x$, like $x=1$, and show that the claim is false for that particular choice. You need to argue that there is no possible choice of $x$ that works. – lulu Sep 07 '20 at 20:56
  • I thought $\forall x$ mean "for all x"? – Pefta Seng Sep 07 '20 at 21:02
  • I do not understand first comment. If $y=\frac{x}{2}$, then $xy < y^2 \iff x \frac{x}{2} < (\frac{x}{2})^2 \iff \frac{x^2}{2} < \frac{x^2}{4}$. Does this mean first statement is false? – Pefta Seng Sep 07 '20 at 21:09
  • How can I understand second statement different from first? I think about it but I have trouble understanding different. – Pefta Seng Sep 07 '20 at 21:12
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    They are different, @Pefta. Take more than a few minutes to understand what Lulu is saying. Note the difference between, e.g., "Someone loves everyone" and "Everyone loves someone." – amWhy Sep 07 '20 at 21:15
  • @lulu Thank you for your all help! – Pefta Seng Sep 07 '20 at 21:29
  • @lulu, I think you meant to say "$y=2x$ works," not $y=x/2$. (When you cancel the $y$ in $xy\lt y^2$, you're left with $x\lt y$.) – Barry Cipra Sep 07 '20 at 21:30
  • @BarryCipra Yes, absolutely. Thanks for the correction. – lulu Sep 07 '20 at 21:43
  • @PeftaSeng Note the comment from Barry Cipra. I should have written $y=2x$, not $y=\frac x2$. Indeed, with $y=2x$ we have $xy=2x^2<4x^2=y^2$. – lulu Sep 07 '20 at 21:45
  • Ah thank you! :) – Pefta Seng Sep 07 '20 at 21:47

1 Answers1

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As others have said in the comments, I think there's a misunderstanding of what the quantifiers mean, so the meaning of the statement is lost in translation.

$\forall x$ means "for all x," which also means "given any x, the following condition is true." It's true for all x, so if you pick any one specific x, it will be true.

$\exists x$ means "there exists an x," so it might not be true for all x, but it's true for at least one.

So let's look at the first statement: "For all $x$, there exists $y$ such that $xy<y^2$." Is it true? Can you find such a $y$, given any value of $x$?

Now, the second statement: "There exists $x$ such that, for all $y$, $xy<y^2$." Can you find an $x$ such that, no matter what $y$ you multiply it by, the result is always less than $y^2$?

You've made things harder for yourself by introducing a biconditional in both cases, and there was no reason to do so. My advice would be to either think carefully about the symbols you introduce into your translation, or do away with the symbols entirely and translate it into full sentences.

Stephen Goree
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  • Thank you! This help me understand better. I am sorry if my question was simple. – Pefta Seng Sep 07 '20 at 21:28
  • I still struggle with second statement. If there is an $x = \frac{y}{2}$ that exist, then the proposition is $\frac{y^2}{2} < \frac{y^2}{4}$ is false. Does this mean second statement is false, or that I fail to find "some" x that make statement true? – Pefta Seng Sep 07 '20 at 21:46
  • @PeftaSeng You're on the right track, but your reasoning is off. Choose any arbitrary $x$. Can you find $y$ such that it contradicts the statement? – Stephen Goree Sep 07 '20 at 22:07
  • So if $x=1$, then the proposition become $y < y^2$. I find $y=\frac{1}{2}$ to contradict it. Would that be correct way to reason and disprove second statement? – Pefta Seng Sep 07 '20 at 22:13
  • Not quite. That would be one step in recognizing the pattern, but you need to show that such an $x$ cannot exist. Let $x$ be an arbitrary real number and define $y$ in terms of $x$. There's a really easy solution that I can see right away. – Stephen Goree Sep 07 '20 at 22:34
  • I don't seem to understand :( – Pefta Seng Sep 07 '20 at 22:46
  • Choose $y=x$. Do you see it now? – Stephen Goree Sep 07 '20 at 23:01
  • Oh my. Yes I see now! How I did not see it earlier? Haha thank you for your patience! – Pefta Seng Sep 07 '20 at 23:04