Showing that $a+b\sqrt{2}$ is associative over multiplication I need to show that $a+b\sqrt{2}$ is associative over multiplication. This is what I have so far. I may be taking a wrong route so just please let me know.
$$(a+b\sqrt{2})*((c+d \sqrt{2})*(e+f \sqrt{2} )) = \\
 (a+b\sqrt{2})*(ce+2df+(cf+de)\sqrt{2}) $$
and then I just opened up the brackets and got an ugly looking thing that I am not sure what to do next. 
$$(ace+2adf+(acf+ade)\sqrt{2}+bce\sqrt{2}+2bdf\sqrt{2}+2(bcf+bde)$$
How would I go from here?
 A: Another viewpoint, if the "abstract algebra" tag is taken seriously, that these "numbers" are identifiable as elements of the quotient ring $\mathbb Q[x]/\langle x^2-2\rangle$, which inherits its associativity from the polynomial ring $\mathbb Q[x]$.
A: You don't need any more  simplifications. Now try the other order of associativity 
$$ \Big( (a+b\sqrt{2})*(c+d \sqrt{2}) \Big) *(e+f \sqrt{2}) = \\ 
\Big( ac + 2bd + (ad+bc) \sqrt{2}   \Big) *(e+f \sqrt{2})  $$
which when you expand it, will give exactly what you have
$$(ace+2adf+(acf+ade)\sqrt{2}+bce\sqrt{2}+2bdf\sqrt{2}+2(bcf+bde)$$
So we can conclude that 
$$ \Big( (a+b\sqrt{2})*(c+d \sqrt{2}) \Big) *(e+f \sqrt{2}) =  (a+b\sqrt{2}) * \Big( (c+d \sqrt{2}) *(e+f \sqrt{2})\Big)
 $$
for all $a, b, c, d, e, f.$
A: You should simplify and rewrite what you have in the form $x+y\sqrt{2}$, so it makes it easier to read.  Next, simply compute $((a+b\sqrt{2})*(c+d\sqrt{2}))*(e+f\sqrt{2})$, and compare it to what you have.  Then if they match, you will have shown what you need.
A: Are you trying to show that ordinary multiplication is associative when applied to numbers of the form $a+b\sqrt2$ (with $a$ and $b$ rational and/or integral)?
That is trivially true because $(pq)r=p(qr)$ holds for all real numbers $p$, $q$, and $r$ -- the case where they all lie in $\mathbb Z+\mathbb Z\sqrt2$ or $\mathbb Q+\mathbb Q\sqrt2$ is just a special case of this general truth.
