The sets $\{-x,x\}$ form a partition of $\mathbb Z$

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I've been really trying to understand how some of these proofs work; I've spent a majority of my time studying the material for this class, but I'm still performing poorly in it. It doesn't help that the book is very vague; what's worse is that it contains little to no solutions and does not have a solutions manual, so I don't even know if I'm right or wrong half the time.

Anyways, in the problem, we are asked to prove that a set is a partition. A problem from the book:

Prove that $P=\left\{X: X = \{-x,x\} \space \text{and} \space x\in\mathbb{N} \cup\{0\}\right\}$ is a partition on $\mathbb{Z}$.

Recalling that the three criteria of a partition are that:

  1. If $X$ is an element of the partition, $X$ cannot be empty.

  2. If $X$ and $Y$ are elements of the partition, they are equal or pairwise disjoint.

  3. The union of all the elements in the partition are equal to the set we are taking the partition of.

I'd greatly appreciate help. I understand what the criteria demand intuitively, but I just can't seem to connect the logic when I do the proofs.

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The criteria demands that when you and your friends order a pizza and you each take a part of the pizza:

  1. Every one got some part of the pizza.
  2. No two people ate from the same slice.
  3. You ate the entire pizza.

As for the proof, my best advice is once you get the intuitive idea, or even well before that, just work with the definitions carefully and slowly.

You have to prove that if $X\in P$, then $X$ is not empty. How do we do that? Let $X\in P$ be an arbitrary element, then by definition there is some $x\in\Bbb N\cup\{0\}$ such that $X=\{-x,x\}$. In particular $x\in X$, so $X$ is not empty.

You have to prove that if $X\neq Y$ are elements from $P$, then $X\cap Y=\varnothing$. So again, you pick two elements from $P$ which are different, and you show that if $x\in X$, then $x\notin Y$ and vice versa.

Finally you have to prove that if $k\in\Bbb Z$, then for some $X\in P$ we have $k\in X$. And I'll let you try to figure out what to do on this one all alone.