Question

Show that the set of all real numbers of the form

$a_0+a_1\mathrm{\pi }+{\mathrm{a}}_2{\mathrm{\pi }}^2+......+{\mathrm{a}}_{\mathrm{n}}{\mathrm{\pi }}^{\mathrm{n}}\mathrm{with}\mathrm{n}\ge 0\mathrm{and}{\mathrm{a}}_{\mathrm{i}}\in \mathrm{\mathbb{Z}}$

is a subring of $\mathrm{ℝ}$ that contains both$\mathrm{\mathbb{Z}}$ and $\mathrm{\pi }$.

Short answer

Hence proved, the given set is a subring of$\mathrm{ℝ}$ that contains both $\mathrm{\mathbb{Z}}and\mathrm{\pi }$.

Step-by-step solution

Polynomial Arithmetic: 

If any given function$R\left[x\right]$ is a ring, then the commutative, associative, and distributive laws holds true such that the function $f\left(x\right)+g\left(x\right)$ exists.

Polynomial Operations:

The given set is:

$a_0+a_1\mathrm{\pi }+{\mathrm{a}}_2{\mathrm{\pi }}^2+.....+{\mathrm{a}}_{\mathrm{n}}{\mathrm{\pi }}^{\mathrm{n}},\mathrm{with}\mathrm{n}\ge 0\mathrm{and}{\mathrm{a}}_{\mathrm{i}}\in \mathrm{\mathbb{Z}}$

Rewrite as:

$\mathrm{\mathbb{Z}}\left[\mathrm{\pi }\right]=\left\{\underset{i=0}{\overset{m}{\sum a_i{\mathrm{\pi }}^{\mathrm{i}}\overline{)\mathrm{m}\in \mathrm{\mathbb{N}};{\mathrm{a}}_{\mathrm{i}}\in \mathrm{\mathbb{Z}}}}}\right\}\subset \mathrm{ℝ}$

Now, for subtraction and multiplication for all elements $a,b\in \mathrm{\mathbb{Z}}$, the given subset can be expressed as:

$\begin{array}{rcl}\left\{\underset{i=0}{\overset{m}{\sum a_i{\mathrm{\pi }}^{\mathrm{i}}}}\right\}-\left\{\sum _{j=0}^n{b}_j{\mathrm{\pi }}^{\mathrm{j}}\right\}& =& \sum _{i=0}^{max\left(m,n\right)}\left(a_i-b_i\right){\mathrm{\pi }}^{\mathrm{i}}\in \mathrm{\mathbb{Z}}\left[\mathrm{\pi }\right]\\ \left\{\sum _{\mathrm{i}=0}^{\mathrm{m}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{\pi }}^{\mathrm{i}}\right\}\left\{\sum _{\mathrm{j}=0}^{\mathrm{n}}{\mathrm{b}}_{\mathrm{j}}{\mathrm{\pi }}^{\mathrm{j}}\right\}& =& \sum _{\mathrm{k}=0}^{\mathrm{m}+\mathrm{n}}\left(\underset{\mathrm{i}+\mathrm{j}=\mathrm{k}}{\sum {\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{i}}}\right){\mathrm{\pi }}^{\mathrm{k}}\in \mathrm{\mathbb{Z}}\left[\mathrm{\pi }\right]\\ & & \end{array}$

When $m>n$, we have: $b_{n+1}=b_{n+2}=.....=b_m=0$

And, when$n>m$ , we have: $b_{m+1}=b_{m+2}=......=b_n=0$

Hence proved, the given set is a subring of$\mathrm{ℝ}$ that contains both$\mathrm{\mathbb{Z}}and\mathrm{\pi }$ .