0
$\begingroup$

$A$ is a set with elements $a$ and $b$. Here, $a = (a_1a_2...a_n)$ and $b = (b_1b_2...b_m)$.

An operation ($*$) known as $\textit{concatenation}$ is defined on the set. That is, $a*b = ab$.

For example,

if $a = \underbrace{aaaaa}_\text{5 times}$ and $b = \underbrace{bbb}_\text{3 times}$ then $a*b = \underbrace{aaaaa}_\text{5 times}\underbrace{bbb}_\text{3 times}$.

Show that the operation is associative.

My working,

For associativity to hold,

$$(x*y)*z = x*(y*z)$$ but since the set only contains $a$ and $b$, then there would be $(2\times2\times2) = 8$ possible operations i.e.

$$(a*b)*a = a*(b*a)$$ $$(a*b)*b = a*(b*b)$$ $$(b*a)*b = b*(a*b)$$ $$(b*a)*a = b*(a*a)$$ $$(a*a)*a = a*(a*a)$$ $$(b*b)*b = b*(b*b)$$ $$(a*a)*b = a*(a*b)$$ $$(b*b)*a = b*(b*a)$$

So if associativity were to hold, would I need to prove all eight cases?

I'm really not sure if I'm on the right track or completely off.

For case 1, $$(a*b)*a = a*(b*a)$$

$$LHS: (a*b)*a = [(a*b)_1(a*b)_2...(a*b)_n](a_1a_2...a_m) $$

$$=[\left\{(a_1a_2...a_n)(b_1b_2...b_m)\right\}_1\left\{(a_1a_2...a_n)(b_1b_2...b_m)\right\}_2...\left\{(a_1a_2...a_n)(b_1b_2...b_m)\right\}_n](a_1a_2...a_m) $$

$$=[(a_1a_2...a_n)_n(b_1b_2...b_m)_n](a_1a_2...a_m) $$

$$RHS: a*(b*a) = (a_1a_2...a_n)[(b*a)_1(b*a)_2...(b*a)_m] $$

and by the same process we arrive at,

$$=(a_1a_2...a_n)[(b_1b_2...b_n)_m(a_1a_2...a_m)_m] $$

  • 0
    What is the connection with complex numbers? – 2017-02-22
  • 0
    I feel like I should note that I did post a similar question yesterday, but deleted it. – 2017-02-22
  • 0
    @mrf I'm not sure what you're talking about it. – 2017-02-22
  • 0
    See, if you feel that the answers to all the cases are similar, then do one case and understand how it changes for the other cases. There is no rigorous need to do all eight of them and fill pages of work in the process. Do one case and show us. – 2017-02-22
  • 0
    You tagged the question complex-numbers, and nothing else. I'm pretty sure this is incorrect. Please fix your tagging. – 2017-02-22
  • 0
    I've provided some of my working for case 1. – 2017-02-22
  • 0
    Yes, you're on the right track. It's a bit longer (eight times longer) track than it need to be, but it certainly leads to the target. – 2017-02-22
  • 0
    @Ron As already noted yesterday, the definition you showed us for the concatenation is different from the one you uses to prove your first case! I suggestes chages yesterday. For the eight cases: Take $a,b,c\in A$ and show it in general. Noone says two of these three can't be equal. – 2017-02-22
  • 0
    I now see what you mean @ M. Winter. Thanks for your help in both instances. – 2017-02-22

1 Answers 1

1

I'm not sure how exactly your definition works, there are some subtleties there (like what exactly $ab$ is?). So let's make it more rigorous:

Let $a=(a_1,\ldots,a_n)$ and $b=(b_1,\ldots,b_m)$ be sequences. For the moment let's forget over what set, it doesn't really matter. I assume that your operation is simply

$$a*b=(a_1,\ldots,a_n,b_1,\ldots, b_m)$$

Now to show that it is associative pick 3 arbitrary sequences $a=(a_1,\ldots,a_n)$, $b=(b_1,\ldots,b_m)$ and $c=(c_1,\ldots,c_k)$. Calculate

$$(a*b)*c = (a_1,\dots, a_n,b_1,\ldots,b_m)*c=(a_1,\dots, a_n,b_1,\ldots,b_m,c_1,\ldots, c_k)$$

On the other hand

$$a*(b*c)=a*(b_1,\ldots,b_m,c_1,\ldots, c_k)=(a_1,\dots, a_n,b_1,\ldots,b_m,c_1,\ldots, c_k)$$

Clearly both sequences are equal which completes the proof. $\Box$

So as you can see you don't have to prove case by case since $a, b, c$ were picked arbitrarly.

Now this general operation is associative. If you restrict it to some smaller set (which it seems you do?) then it still preserves this property.

  • 0
    Thank you @freakish! A lot simpler than the mess I was making. – 2017-02-22