Decide whether the $\{a+b\sqrt{2} \mid a,b \in \Bbb Z\}$ is a ring with usual addition and multiplication. If a ring is formed, state whether the ring is commutative, whether it has unity, and whether it is a field.
I am confused about how this is a ring closed under the binary operation of "+". If a = b = 1, then don't we have some number that is not in the integers? Thanks for your help!
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Consider arbitrary elements $a_1+b_1\sqrt{2}$ and $a_2+b_2\sqrt{2}$ with $a_1,a_2,b_1,b_2 \in \mathbb{Z}$.
\begin{equation} a_1+b_1\sqrt{2} + a_2+b_2\sqrt{2} = (a_1+a_2)+(b_1+b_2)\sqrt{2}. \end{equation}
Since $a_1,a_2,b_1,b_2 \in \mathbb{Z}$, $a_1+a_2 \in \mathbb{Z}$ and $b_1+b_2\in \mathbb{Z}$ (closure of integers under $+$).
Therefore $(a_1+a_2)+(b_1+b_2)\sqrt{2} \in \{a+b\sqrt{2} | a,b \in \mathbb{Z}\}$, and the set is closed under $+$.
The condition is that the coefficients are integers, not that the elements of the ring are integers. The element $1+\sqrt 2$, in your example, is perfectly valid because the coefficients, $a=1$ and $b=1$, are integers. On the other hand, something like $$ \frac{1}{2}-\frac{1}{2}\sqrt2 $$ is not in the ring because it has coefficients that are not integers.