Existential and Universal Quantifiers

431 Views Asked by At

Quantifiers
(a) Please see below. I cannot work out why one is correct. If $x < 0$, then there's no value $y \in \mathbb{R}$ so that $y^2 = x$.

(b) If I have $\exists$ followed by $\forall$, then does imply that there exists exactly one value applicable for all values in the second $\forall$ quantifier?

Say I have the natural numbers as my domain and:

$(\exists n)(\forall m) 2m=n$

Does it say there's exactly one number $n$, such that for all numbers $m$, $2m=n$? I'm a bit confused here.

Here is the relevant part to (a):

The order of the quantifiers counts.

\begin{align}&(\exists y \in \mathbb{R})(\forall x \in \mathbb{R})x=y^2\\&(\forall x \in \mathbb{R})(\exists y \in \mathbb{R})x=y^2\end{align} The first is false and the second is true.

Edited 20160608: I would like to add another problem I have problems with: enter image description here. The textbook gives TRUE for problem 2.
But, how can I find one value n, that holds true (for all) m?
Doesn't is say that there exists only one value that satisfies all m?
Thanks for your help!

1

There are 1 best solutions below

2
On BEST ANSWER

It always helps to read it in English first and understand it.

(a) The first says that $(\exists y \in \mathbb{R})(\forall x \in \mathbb{R})x=y^2$. In English this says that there exists an element $y$ in $\mathbb{R}$ such that for all $x \in \mathbb{R}$, $x = y^2$. This is clearly false because there is no element $y \in \mathbb{R}$ such that every real number is equal to the square of it!

The second says that $(\forall x \in \mathbb{R})(\exists y \in \mathbb{R})x=y^2$. In English this says that for all $x \in \mathbb{R}$, there exists an element $y \in \mathbb{R}$ such that $x=y^2$. You are absolutely correct that it is not true for $x<0$, so this is probably a mistake in the text. If we look at $x \geq 0$ however, it is true, and this is probably what the text meant.

(b) Yes - almost. You say that "If I have $\exists$ followed by $\forall$, then does this imply that there exists $\color{red}{\text{exactly one}}$ value applicable for all values in the second $\forall$ quantifier?"

There does not need to be exactly one, we can have as many as we like. Related is $\exists!$, which means that there exists a $\color{red}{\text{unique}}$ element, i.e. exactly one.

To answer your question "if $\exists \forall$, does it imply that I need to find exactly one value from many, that fits the $\forall$? That's causing my confusion. Or I should say, the $\exists$ value is fixed, once I have chosen one?"

You want to find a value from all of $\mathbb{R}$ that when fixed, you can apply the $\forall$ part of the statement. So in your case you want to find some real number $y$ such that every real number is equal to the square of it.