Show that multiplication is commutative

**coopsterdude** · August 5th 2010, 10:08 AM

I don't quite understand how to get the last line in this problem.

**Ackbeet** · August 5th 2010, 10:13 AM

I'm assuming the bd = db kind of multiplication is already assumed to be commutative, or else you'd have yourself a circular argument. Multiply out the last line (foil it), and you'll get the second-to-last line. So, from the second-to-last line to the last line is a matter of factoring. Since foiling it out gets you the desired result, the method of factoring, whatever it was, was valid.

**Archie Meade** · August 5th 2010, 10:27 AM

Originally Posted by coopsterdude

I don't quite understand how to get the last line in this problem.

2nd line to the last line....

Split 2 into $2=\sqrt{2}\sqrt{2}$ first

$ac+\left(\sqrt{2}\right)b\left(\sqrt{2}\right)d+\l eft(\sqrt{2}\right)ad+\left(\sqrt{2}\right)bc$

$=a(c+\sqrt{d})+\left(\sqrt{2}\right)b(c+\sqrt{d})$

However, there is no proof of commutativity in any of that example.

**Soroban** · August 5th 2010, 11:24 AM

Hello, coopsterdude!

I don't understand the proof . . . It's very sloppy!

$\begin{array}{cccc}(c+d\sqrt{2})(a+b\sqrt{2}) &= & ca + cb\sqrt{2} + da\sqrt{2} + 2bd & [?]\\ \\ &=& (ac+2bd) + (ad+bc)\sqrt{2} & [?]\\ && \!\!\!\!\uparrow \\ \\[-3mm] &=& (a+b\sqrt{2})(c+d\sqrt{2}) \end{array}$

We are supposed to prove that multiplication is commutive.
But in the proof, they already assumed that mult'n is commutative
. . . as Ackbeet already pointed out.

The first step should look like this:

. . $(c+d\sqrt{2})(a+b\sqrt{2}) \;=\; (c)(a) + (c)(b\sqrt{2}) + (d\sqrt{2})(a) + (d\sqrt{2})(b\sqrt{2})$

They simplified the third term like this:

. . $\begin{array}{ccccccc} (d\!\cdot\!\sqrt{2})\!\cdot\!(a) &=& d\!\cdot\!(\sqrt{2}\!\cdot\!a) & \text{assoc.} \\ & =& d\!\cdot\!(a\!\cdot\!\sqrt{2}) & \text{comm.} \\ & = & (d\!\cdot\!a)\!\cdot\!\sqrt{2} & \text{assoc.} \\ & = & da\sqrt{2}\end{array}$

They simplified the fourth term like this:

. . $\begin{array}{ccccc} (d\!\cdot\!\sqrt{2})\!\cdot\!(b\!\cdot\!\sqrt{2}) &=& d\!\cdot\!(\sqrt{2}\!\cdot\!b)\!\cdot\!\sqrt{2} & \text{ assoc.} \\ & = & d\!\cdot\!(b\!\cdot\!\sqrt{2})\!\cdot\!\sqrt{2} & \text{comm.} \\ & = & (d\!\cdot\!b)\!\cdot\!(\sqrt{2}\!\cdot\sqrt{2}) & \text{assoc.} \\ & = & (b\!\cdot\!d)\!\cdot\!(2) & \text{comm.} \\ & = & 2bd & \text{comm.}\end{array}$

In the next step, they assume that: . $ca \,=\,ac,\;\;da \,=\,ad.$

The second line should be:

. . $ac + bc\sqrt{2} + ad\sqrt{2} + 2bd$

. . . . $=\;(ac + bc\sqrt{2}) + (ad\sqrt{2} + 2bd)$

Factor: . $c(a+b\sqrt{2}) + d\sqrt{2}(a + b\sqrt{2})$

Factor: . $(a + b\sqrt{2})(c + d\sqrt{2})$

Like I said, very sloppy!

**undefined** · August 5th 2010, 11:40 AM

Originally Posted by coopsterdude

I don't quite understand how to get the last line in this problem.

Maybe we're trying to prove multiplication is commutative in the field $\mathbb{Q}(\sqrt{2})$ ? Probably easier than factoring would be to expand separately $(a+b\sqrt{2})(c+d\sqrt{2})$ and $(c+d\sqrt{2})(a+b\sqrt{2})$ (like Ackbeet suggested) and see that they are equal term-for-term.

**coopsterdude** · August 5th 2010, 12:23 PM

Thanks, I didn't think about splitting the 2. Got it.

**coopsterdude** · August 5th 2010, 12:30 PM

I see what you're saying, the last line can only be commutative if we started with your first line. Thanks for the explnation...very happy.

Thanks

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