Logical Quantifiers order

Question

So for the definition of an inverse I have: "There exists $e$ belonging to $G$ such that for all $x$ belonging to $G$, $x*e=e*x=x$"

I'm aware the order of quantifiers matters but I can't see how the above is different from:

"For all $x$ belonging to $G$, there exists $e$ belonging to $G$ such that $x*e=e*x=x$"

Answer 1

Consider any $G$ with at least two elements and define $a*b=a$. Then for all $x\in G$, there exists $e\in G$ (namely, $e=x$) such that $x*e=e*x=x$. However, there is no $e\in G$ such that for all $x\in G$, we have $e*x=x$ (because $e*x=e$)

Answer 2

You've actually defined an identity element, not an inverse.

Anyway, in your second case, you're saying that the choice of $e$ is dependent upon the chosen $x$, so the identity is not unique. The first says you can find an $e$ for which any $x$ will yield $xe = ex = x$. This $e$ works for any choice of $x \in G.$

Answer 3

First one states that for all $x$ there is an elemnt $e$ satisfing that condition but the Second one is saying any $x$ has its own $e$ with that property. These two ones can be found out through definitions of Continuity and Uniformy Continiuty as well: $$\forall x \in I \, \exists \delta > 0\, \forall y \in I \, ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon,$$ $$\, \forall \varepsilon > 0\, \exists \delta > 0\, \forall x \in I\, \forall y \in I\, ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon$$

Answer 4

For example, consider the set $G=\{a,b,c\}$ with $x*y=y$ for all $x,y\in G$.

Let's first test the group axiom:

There exists an $e\in G$ such that for all $x\in G$, $x∗e=e∗x=x$

So let's check:

• Could it be that $e=a$? No, because $a*b=b\ne a$.
• Could it be that $e=b$? No, because $a*c=c\ne a$.
• Could it be that $e=c$? No, because $c*a=a\ne c$.

So no matter which element we choose for $e$, the condition is not fulfilled for all $x$, so the axiom is not fulfilled for my example.

Now let's check your alternative:

For all $x\in G$, there exists an $e\in G$ such that $x∗e=e∗x=x$

So we have to check all $x\in G$.

• For $x=a$, there exists such an $e$, namely $e=a$, as $a*a=a*a=a$.
• For $x=b$, there exists such an $e$, namely $e=b$, as $b*b=b*b=b$.
• For $x=c$, there exists such an $e$, namely $e=c$, as $c*c=c*c=c$.

So your alternative axiom is fulfilled for my example.

I hope that example clears up the difference between both statements.