# subring proof of real numbers

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 $$-a_i\in \mathbb{Z}$$, hence the solution to $$a+x=0_R$$ 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 $$\pi$$

• 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 $\{\pi\}$. Feb 20 '16 at 20:57

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.