# 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?

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Open parentheses in both expressions and show you get the very same in both cases...! Say, $\,(\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma)$ – DonAntonio Aug 19 '12 at 15:04
Thanks. I gotcha :) – math101 Aug 19 '12 at 15:07
What does "associative over multiplication" mean for a binary operation? I know "associative" itself, and "distributes over multiplication", but not your combination here. – Henning Makholm Aug 19 '12 at 15:17
I meant associative with the binary operation, multiplication – math101 Aug 19 '12 at 15:47
This question has the same ambiguity as your prior question. Namely, are you allowed to view this as a subring of a ring already known to be associative (e.g. $\Bbb R$, as in Henning's answer), or, instead, are you adjoining a square-root of $2$ to $\Bbb Q$, (as, in paul's answer)? Also, what is the coefficient (base) ring, i.e. the ring that $\rm\,a,b\,$ range over? – Bill Dubuque Aug 19 '12 at 23:13

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.

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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 $\mathbb Q[x]$.

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LOL Don't take the abstract algebra tag too seriously. This was just a math course that briefed many advanced topics in math such as Number Theory, Group Theory, Graph Theory and many other topics. I was unsure what to tag the question so I just tagged it Abstract Algebra. Thanks – math101 Aug 19 '12 at 17:18
Heh. I did wonder! But it won't hurt anyone to be exposed to this little idea... I think Kronecker was the first proponent of this way of legitimizing algebraic numbers. – paul garrett Aug 19 '12 at 17:25
+1 Good point. Though perhaps doing something like this explicitly might be pedagogically useful as a warm-up exercise before one proves that multiplication of formal polynomials is associative. – Henning Makholm Aug 20 '12 at 1:32

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 $a, b, c, d, e, f.$

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Thanks. Didnt realize it was that simple :) – math101 Aug 19 '12 at 15:07

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.

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Thanks. Stupid me. I should have realized that – math101 Aug 19 '12 at 15:06