Determine if the following quantified formula is true for all natural numbers N = {1,2,3...}. $ ∀_x ∃_y ∀_z $ P(x,y,z): $ xy^2 $ != $ z^2 $

Question

$ ∀_x ∃_y ∀_z $ P(x,y,z) where P(x,y,z): $ xy^2 $ != $ z^2 $

My understanding is that there must be a game approach to solve this formula. What I don't understand is how to determine who has the "winning strategy".

I know my job is determine if there exists a natural number y whose square multipled by another natural number x does not equal the square of a third natural number z. The other game player's job is to determine that there doesn't exist any x and z to satisfy this formula. I just don't really know how to go about it. Any help would be appreciated.

Answer 1

Following along the lines of a game, we have:

• Player 1 picks a natural number $x$.

• Player 2 picks a natural number $y$.

• Player 1 picks a natural number $z$.

Then player 2 wins if $xy^2 \ne z^2$, and player 1 wins if $xy^2 = z^2$. If player 2 wins, then the quantified sentence is true. If player 1 wins, the quantified sentence is false.

Player $1$ can win here. Pick $x$ to be a perfect square. Then player 2 picks some $y$. Can you see what player $1$ picks for $z$, to complete the winning strategy?

Answer 2

It is clearly false.

Let's take $x=1$, then you have to find $y$ such that for all $z$

$$y^2!=z^2.$$

Which is false because $y$ can't depend on $z$.