Simple Proof for Commutative Property of Multiplication I'm supposed to show that $a\cdot b=b\cdot a$ for a set $K:=\{s+t\sqrt2:s,t\in\mathbb{Q}\}$ to show that this set is a field.
I was going to set it up like:  
Let $a, b\in K$ such that $a=s_1+t_1\sqrt2$ and $b=s_2+t_2\sqrt2$.
Then $a\cdot b =(s_1+t_1\sqrt2)\cdot(s_2+t_2\sqrt2)$.
Then I would use the FOIL method and multiply them out:
$a\cdot b = s_1s_2+s_1t_2\sqrt2+t_1\sqrt2s_2+2t_1t_2$
But then do I have to write my "method" for factoring them? Could I just write $(s_2+t_2\sqrt2)\cdot(s_1+t_1\sqrt2)$ next?
I guess I'm confused because I could just write something like $a\cdot b =(s_1+t_1\sqrt2)\cdot(s_2+t_2\sqrt2)=(s_2+t_2\sqrt2)\cdot(s_1+t_1\sqrt2) = b \cdot a$ but this seems trivially easy, like just stating that $a\cdot b = b\cdot a$.
 A: What constitutes a valid "proof" depends on exactly how $K$ and the arithmetic operations on $K$ are defined and what facts you are assumed to already know.  I will assume that $K$ is defined as a subset of the real numbers (that is, $s+t\sqrt{2}$ refers to the real number $s+t\sqrt{2}$, rather than just being some formal expression), that $K$ inherits its arithmetic operations from the real numbers, and that basic properties of the real numbers are assumed to be known.  With these assumptions, your proof is fine, since all you're using is basic algebraic manipulation rules for real numbers.  In fact, even more simply, you can just invoke the fact that multiplication of real numbers is commutative, so multiplication of elements of $K$ is automatically commutative, being just a special case of multiplication of real numbers.  You can similarly use known facts about arithmetic of real numbers to prove several of the other field axioms for $K$.
A: It is trivially easy. Your method is fine.  You could also say that the multiplication is commutative because $K\subseteq\mathbb R$. There is more work though to show that $K$ is a field. You need to show that it is closed under both addition and multiplication, and that the additive inverse of any member of $K$ is also in $K$.  You'll need to show the same thing for multiplicative inverse of non-zero elements.
