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I need help determining the truth value of each of these statements if the domain of each variable consists of all integers. Justify your answer.

  1. $\forall x\exists y (x = 3y + 1)$
  2. $\exists x\forall y (y^2 > x)$

If I take $x$ as $-5$ and take $y$ as $-2$ then it would be

$ -5 = 3(-2) + 1 $

$ -5 = -5$

True

So what I did is true only if i take those values. If I take $x$ as $-5$ and $y$ as $-5$ then it wouldn't equal so its False

So I am just confused on how to do these type of questions so can someone help me figure it out.

Thank you.

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  1. False. Take $x=2$. Then there is no integer $y$ such that $2 = 3y+1$, for then $3y = 1$. Remember, it has to work for all $x$, not just some.

  2. True. Take any negative number for $x$. Then $y^2$ will always be greater than $x$. So yes, there is some $x$ such that $y^2 >x$ for all $y$.

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For #1 your correct reading is "for all $x$ there exists a $y$ ..."

You correctly say that when $x= -5$ you can take $y= -2$.

Then you make a mistake. The way the sentence is written tells you the value of $y$ is allowed to depend on the value of $x$. All you have to do is find one once you know $x$. That's what you did starting with $x = -5$ to get $y = -2$. You clearly know enough algebra to find a $y$ that goes with any given $x$ you start with. (In this particular case there will be just one value of $y$.) Will that $y$ always be an integer?

For #2 you have to find some particular $x$ for which the statement is true for every possible $y$. Is there such a thing? (Hint: negative numbers are allowed.)

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