If $Q(x,y)$ be the statement $"x+2=3y"$, what are the truth values of

$\forall x\exists yQ(x,y)$,

$ \exists x \forall y(x,y)$,

and $ \forall x \forall yQ(x,y) $?

I know how to do it if $Q(x)$ be the statement $"x+2=3x"$. For example:

For $\forall xQ(x)$ be false and $\exists xQ(x)$ be true as $x+2=3x$ is true if and only if $x=1$, we see that $Q(x)$ is true if and only if $x=1$.

But when $Q(x,y)$, I not quite sure how to get the truth values. Can anyone help me please?

  • 1
    $\begingroup$ Is there supposed to be a $y$ in the equation? Also, what are the universal sets for $x$ and $y$? $\endgroup$ – matt biesecker May 10 '15 at 23:43
  • $\begingroup$ Yes, there is. Sorry, I miss typed it. I have made the correction now. $\endgroup$ – PeaceKing May 10 '15 at 23:50
  • $\begingroup$ One thing you can try is substituting various values for $x,y$ to get a feel for the truth value: For example $Q(4,2): \ 4+2=3(2),$ which is true. On the other hand, $Q(4,3): \ \ 4 + 2 = 3(3)$ is false since $6 \neq 9.$ Thus there are some values of $(x,y)$ for the statement is false. $\endgroup$ – matt biesecker May 10 '15 at 23:53
  • $\begingroup$ I think for the second statement you mean $\exists x \forall y Q(x,y)$. Also, are we saying $\forall y \in \Bbb R$ and $\forall x \in \Bbb R$? $\endgroup$ – layman May 10 '15 at 23:53
  • $\begingroup$ To study the first statement $(\forall x)(\exists y) Q(x,y)$, can you answer the following question: Given any $x\in \mathbb{R},$ can a I find specific $y \in \mathbb{R}$ such that $x+2=3y.$ It appears you can: $y=\frac{1}{3}(x+2)$ $\endgroup$ – matt biesecker May 10 '15 at 23:56

What is the universal set? If the universal set is $R$, then the first one is TRUE. The second and the third are false. For the first one, given any x in R, take y as $\frac{1}{3}(x+2)$. For the second one, given x=0, then y=1 gives a contradiction. The same counter example works for the third one.

  • $\begingroup$ To answer the second question, suppose there is such an x. Then x=3y-2 for all y. in particular, for y=0 and y=1. Thus x=-2 and x=1. i.e -2=1---a contradiction. For the third, x=0, y=1 shows that the proposition is not true. $\endgroup$ – OKPALA MMADUABUCHI May 11 '15 at 0:02
  • $\begingroup$ Prove the first by contradiction. Assume that the first statement is false. Then $\exists x\forall y$, $x+2\neq 3y$. We may assume without loss of generality that $x+2>3y$ for all reals. In particular, for $y=\frac{1}{3}(x+2)$ we obtain that $x+2>(3y=x+2)$ which is a contradiction. $\endgroup$ – OKPALA MMADUABUCHI May 11 '15 at 0:09
  • $\begingroup$ Appreciate your time and help. It was really helpful. $\endgroup$ – PeaceKing May 11 '15 at 0:17

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