Guided Example

The Set $LaTeX: S=\left\{a+b\sqrt[3]{7}+c\sqrt[3]{3}?x,y,z\in\mathbb{Q}\right\}$ is a

A. ring with respect to addition on $ℝ$
B. ring with respect to multiplication on $ℝ$
C. ring with respect to addition and multiplication on $ℝ$
D. none of the above $ℝ$

The correct answer is C.

Step 1

We start by showing that S is closed with respect to addition.
For $LaTeX: m+n\sqrt[3]{7}+p\sqrt[3]{3},x+y\sqrt[3]{7}+z\sqrt[3]{3\in S}$ ,
$LaTeX: m+n\sqrt[3]{7}+p\sqrt[3]{3}+x+y\sqrt[3]{7}+z\sqrt[3]{3}=\left(m+x\right)+\left(n+y\right)\sqrt[3]{7}+\left(p+z\right)\sqrt[3]{3}\in S$ .

Step 2

Then $LaTeX: -a\ge0$ and $LaTeX: -a+a\ge0+a$ (Addition property of order)

Step 3

Therefore, the set S is closed under addition.
Now we proceed to show that S is closed under multiplication. $LaTeX: \left(m+n\sqrt[3]{7}+p\sqrt[3]{3}\right)\left(x+y\sqrt[3]{7}+z\sqrt[3]{3}\right)=\left(mx+3nz+3py\right)+\left(my+nm+3pz\right)\sqrt[3]{7}+\left(mz+ny+px\right)\sqrt[3]{3\in S}$
So S is also closed under multiplication.

Step 4

Next note that since S is a subset of the ring R the following properties hold.

Step 5

Finally, $LaTeX: 0+0\sqrt[3]{7}+0\sqrt[3]{3}=0\in S$ , and for each $LaTeX: a+b\sqrt[3]{7}+c\sqrt[3]{3}$ there exists
$LaTeX: -m-n\sqrt[3]{7}-p\sqrt[3]{3}\in S$ . Hence S has all the required properties of the ring.