Questions 20 & 22 Post Write

20. Statement: The polynomial ring $\mathbb{Z}[x]$ over the integers is countably infinite.
Definitions:
A countable union of countably infinite sets is countably infinite.
The finite product of infinitely countable sets is infinitely countable.
$\mathbb{Z}[x] := { \sum^n _{i=0} \alpha _i x^i| \alpha_i \in \mathbb{Z} \forall i \wedge n \in \mathbb{N} }$
Proof: We can prove the polynomial ring is countably infinite by showing the union of all elements of $\mathbb{Z}[x]$ ($S_oUS_1US_2U...US_nU... \forall{n}\in\mathbb{N}$) is countable and that each set of $\mathbb{Z}[x]$ is countably infinite.
We use induction to show that $S_n$ is countably infinite $\forall{n}\in\mathbb{N}$. The base case $S_o$ is equal to the integers which are countably infinite. We then assume $S_n$ is countably infinte $\forall n < N$. Next, we want to show that $S_{n+1}$ is countably infinite. $S_{n+1}\simeq \mathbb{Z} \times S_n$. As $\mathbb{Z}$ and $S_n$ are countably infinite, and thus so is $S_{n+1}$.
The union of all elements of $\mathbb{Z}[x]$ can be shown by proving a bijection $f: S_n \rightarrow \mathbb{N}$. $n \in \mathbb{N}$ so all elements in $n$ correspond to one element in $\mathbb{N}$

(Prof. Simmons, 4-2-18: The LaTeX command for union is \cup $\cup$. The argument is basically right, but you're done after showing that $S_{n+1}$ is countable if $S_n$ is because there are obviously countably many $S_i$. At the beginning you can simply say that you will show $\mathbb{Z}[x]$ to be a countable union of countable sets.)

22. Let $(R,0, 1, +, \cdot,\leq)$ be an ordered ring with identity.
a. Statement: If $x>0$, then $-x<0$.
$x + (-x) = 0$ is the definition of an inverse
$x>0$
$x + (-x) > 0 + (-x)$ add -x to both sides (And addition respects inequality in ordered rings)
$0 > -x$ by the definition of an inverse and the additive property of 0.
b. Statement: $1>0$.
$1·n=n$ by the identity property
$1>0$ (This is what you're trying to prove….)
$1·n>0·n$ multiply by n on both sides
$n>0$ by the identity property and the multiplicative identity of zero

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