Mixed Quantifiers: Solving the Puzzle

[ver_mathstats]

Homework Statement

Give your own example of a statement with two different quantifiers which changes its meaning and truth value when the order of the quantifiers is exchanged.

Homework Equations

The Attempt at a Solution

(∀x∈ℝ)(∃x∈ℝ)(xy=0) is true but (∃x∈ℝ)(∀x∈ℝ)(xy=0) is false.

Is this correct?

 

[fresh_42]

Homework Statement

Give your own example of a statement with two different quantifiers which changes its meaning and truth value when the order of the quantifiers is exchanged.

Homework Equations

The Attempt at a Solution

(∀x∈ℝ)(∃x∈ℝ)(xy=0) is true but (∃x∈ℝ)(∀x∈ℝ)(xy=0) is false.

Is this correct?

No. First of all, you have used the same variable $x$ in both quantifiers, and haven't quantified $y$ at all. Then you used a symmetric statement, $xy=0$. How should it depend on the ordering? $x=0$, resp. $y=0$ makes both statements true if chosen in the existence clause.

 

[ver_mathstats]

No. First of all, you have used the same variable $x$ in both quantifiers, and haven't quantified $y$ at all. Then you used a symmetric statement, $xy=0$. How should it depend on the ordering? $x=0$, resp. $y=0$ makes both statements true if chosen in the existence clause.

My apologies for using the same variable in both quantifiers.

Would this one make sense (∀x∈ℝ)(∃y∈ℝ)(x²+y=0) is true but (∃y∈ℝ)(∀x∈ℝ)(x²+y=0)?

 

[fresh_42]

My apologies for using the same variable in both quantifiers.

Would this one make sense (∀x∈ℝ)(∃y∈ℝ)(x²+y=0) is true but (∃y∈ℝ)(∀x∈ℝ)(x²+y=0)?

This works.
I.) For all $x$ is a $y$, namely $y=-x^2$ such that ... is true.
II.) There is a real number $y$ such that for all $x$ ... is false: e.g. $y + 0^2 \neq y + 1^2$ no matter how $y$ is chosen.

Do you see the difference between the two statements? There is a neglect in the notation of the first statement. Do you see it? This isn't directed towards you, since the majority of people neglect this, too. But what would be a better statement I.)?