Problem: Show that the set of all real numbers of the form $a_0 +a_1\pi +a_2\pi^2 + \cdot \cdot \cdot + a_n\pi^n $ with $ n \ge 0, a_i \in \mathbb{Z} $ is a subring of $ \mathbb{R} $ that contains both $ \mathbb{Z} $ and $ \pi $.

I think I've successfully shown all criteria of it being a subring. However, it's the statement about showing the it contains both $\mathbb{Z}$ and $\pi$ I'm not sure about.

To show it's a subring must show it's closed under addition, multiplication, $0_R $ in the set, and the solution of $a+x=1_R $ is in the set.

Let $ a_i,b_i\in \mathbb{Z} $ with $ n\ge 0$ then $$(a_0+a_1\pi +a_2\pi^2+\cdot \cdot \cdot + a_n\pi^n)+(b_0+b_1\pi+b_2\pi^2 + \cdot \cdot \cdot + b_n\pi^m $$

$$ = (a_0+b_0)+(a_1+b_1)\pi+(a_2+b_2)\pi^2 + \cdot \cdot \cdot + (a_n+b_n)\pi^{n+m}$$

Similarly, $$(a_0+a_1\pi +a_2\pi^2+\cdot \cdot \cdot + a_n\pi^n)(b_0+b_1\pi+b_2\pi^2 + \cdot \cdot \cdot + b_n\pi^m) $$ $$ = (a_0b_0)+(a_0b_1+a_1b_0)\pi+(a_0b_2+a_1b_1+a_2b_0)\pi^2 + \cdot \cdot \cdot + (a_nb_n)\pi^{n+m}$$

Both of which have the required form by polynomial addition and multiplication.

$0_R = a_0 +a_1\pi+a_2\pi^2 + \cdot \cdot \cdot +a_n\pi^n$ where $(\forall i\in (0,n))(a_i = 0) $

$ -(a_0 +a_1\pi + \cdot \cdot \cdot + a_n\pi^n)$ is in the set since $ -a_i\in \mathbb{Z}$ and hence the solution to $a+x=1_R$ is in the set.

Now, for the last part (assuming all of the above is right showing that its a subring). Is it enough to just state that since each $a_i \in \mathbb{Z}$ it must necessarily contain $\mathbb{Z}$ since each $a_i$ is arbitrary.

Each $a_i$ with a coefficient of $\pi$ isn't in $\mathbb{Z}$ since an integer times an irrational is still irrational, but since $a_0$ is just a constant term with an arbitrary $a_i\in \mathbb{Z}$ it can span the integers and hence the subring contains $\mathbb{Z}$. For the last part it obviously contains multiples of $\pi$

  • $\begingroup$ It is enough to state that for each $m\in\mathbb{Z}$, setting $a_1=a_2=\ldots=0$ and $a_0=m$ retrieves the element $m$. The notation "$R$ contains $\mathbb{Z}$ and $\pi$" is a bit dangerous. I assume they mean $\{\pi\}$. $\endgroup$ – parsiad Feb 20 '16 at 20:57

Proof of a subring is perfect.

Now, $\mathbf{Z}$ is contained in out set because the first coefficient $ a_0$ is an integer, and we can set all other coefficients to zero. Certainly $\pi$ is in our set because we can set all coefficients other than $a_1$ to zero, and let $a_1 = 1$. So you're done, your proof seems flawless to me.

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