How do you prove statements of the following forms?
1) $\forall x\in X: \exists ! Q(x)$
2) $\exists!x \in X: \forall P(x)$
For instance, if I wanted to show that $\exists! x\in \mathbb{N}$ such that $\forall n \in \mathbb{N}, n\geq x$. How would I prove the existence and uniqueness?
and how would I prove $\forall n \in \mathbb{N}$ $\exists x \in \mathbb{N}$ such that $n>x$?
In the second case I could just say let n$\in$ $\mathbb{N}$ and pick $x = n-1$ then clearly $n>n-1$, right? but how would I prove the first one? I know its not in terms of uniqueness, but i'm worried about structure.
First find something with that property (existence). Then show that if two things have the property, they are the same (uniqueness).
For $\exists! x\in \mathbb{N}: \forall n \in \mathbb{N}: n \ge x$, think about what this means. "There is a unique number such that every natural number is at least as big as it." Once you work out what the number is, it's easy to show that it meets that property.
To show that it's unique is a little bit harder: given $x$ and $y$ which meet the property, we need to show $x=y$. The first step towards this is the inequality $x \ge y$, as we put $n := y$ into the given property. Can you see what other inequality we can use?
For the second one, indeed this shows that the order of quantifiers matters. $x := n-1$ works now that we can choose $x$ after choosing $n$.
For the final question, showing that $e$ is the only group element to satisfy $g^2 = g$, we don't need contradiction because we can show it directly. As every group element has an inverse, we can prove directly that $g^2=g \implies g=e$. Can you work out how?