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I'm relatively new to proofing and am wondering if this is an acceptable proof. The book for anyone who would like to reference it is "A Book of Abstract Algebra" by Charles Pinter. It is problem $3$ of problem set D in chapter $2$.

In short, the question defines the operation of concatenation on $A^*$, the set of all sequences of symbols in the alphabet $A$ as "say $a=a_1a_2\cdots a_n$ and $b=b_1b_2\cdots b_m$, then $a\cdot b=a_1a_2\cdots a_nb_1b_2\cdots b_m$." The question then mentions that the the empty sequence is denoted by say $\emptyset$. The problem then asks us to prove that there is an identity element for the operation of $A^*$. My proof is as follows

Theorem: The operation of concatenation denoted by $\cdot$ and defined by $a\cdot b=a_1a_2...a_n b_1b_2...b_n$ on $A^*$, the set of all sequences of symbols in the alphabet $A$, has an identity element.

Proof: Suppose $x\in A^*$.

Suppose $\emptyset$ denotes the empty sequence.

$\emptyset\cdot x=x_1x_2...x_n$ and $x\cdot\emptyset=x_1x_2...x_n$,

Thus, by the definition of an identity element, $\emptyset$ is the identity element of $\cdot$ on $A^*$.

Therefore there is an identity element for the operation of concatenation on the set of all sequences in the alphabet $A$.

My largest concern is that $\emptyset\cdot x=x_1x_2...x_n$ and $x\cdot\emptyset=x_1x_2...x_n$, does not look like $\forall x\in A^*~~~ \emptyset \cdot x=x$ and $x\cdot \emptyset=x$ if that makes any sense.

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    $\begingroup$ Typesetting note: \ is the escape character to use, not /. Further, you would need to enclose things you want to render in MathJax in $'s. E.g. $\emptyset\cdot x = x_1x_2\cdots x_n~\&~x\cdot \emptyset=$ renders $\emptyset\cdot x = x_1x_2\cdots x_n~\&~x\cdot \emptyset=$. You will need to put spaces after each escape sequence if it is followed by a character. $\inA$ creates an error, use $\in A$. Visit this page for a reference on how to type in MathJax and $\LaTeX$ on the site. $\endgroup$
    – JMoravitz
    Commented Aug 10, 2015 at 4:00
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    $\begingroup$ One more typesetting note: you can always learn by looking at what other people have done by rightclicking and then selecting "show math as" to see how it was formatted. This is especially useful if someone is using an uncommon array or symbol that you want to use later, such as $\left(\!\!\binom{5}{2}\!\!\right)$ $\endgroup$
    – JMoravitz
    Commented Aug 10, 2015 at 4:20
  • $\begingroup$ Thank you! I'm assuming it's looked down upon if you never learn and let all the nice editors fix it for you...? $\endgroup$ Commented Aug 10, 2015 at 6:05
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    $\begingroup$ The fact that you had made an attempt at all to typeset it speaks volumes about you in a positive light. Getting good at typesetting things in $\LaTeX$ (or MathJax technically) takes practice, which is one of the less obvious benefits of this site. Spend enough time answering and asking questions here and you'll hone a skill which will help you throughout the rest of your professional career. If you have a specific typesetting question for use on math.se, feel free to ask in the chat, or if you have a larger question, there is TeX.se $\endgroup$
    – JMoravitz
    Commented Aug 10, 2015 at 14:41
  • $\begingroup$ @JMoravitz Thanks :) $\endgroup$ Commented Aug 11, 2015 at 4:45

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The proof is fine as it is written. You could use an extra line immediately following $x\in A^*$ where you state:

Let $x\in A^*$ where $x=x_1x_2\dots x_n$ with each $x_i\in A$.

This will clear any confusion as to what you are referring to when you refer to $x\cdot \emptyset = x_1x_2\dots x_n$ later on since previously the $x_1x_2\dots x_n$ were not defined or mentioned.

The fact that you did not specify anything about $x$ itself apart from that it was an element of $A^*$ implies that $x$ could have been anything. Since the result you showed follows regardless what $x$ is, it must have been true for every $x$.

You did somewhat tacitly assume that $x$ is of length $n\geq 3$ in the way that you formatted your proof (since you wrote it as $x_1x_2\cdots x_n$), but it is easily forgivable and ignorable in this case. It might not hurt to mention briefly as a side note why the proof obviously works if $x$ were a length 0,1, or 2 string.

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  • $\begingroup$ Very helpful, thank you! Especially the first comment about the extra line. I understand what you're saying and the obvious truth about it working in a smaller string, but what exactly would a "side note" like that look like and how would you show that without showing the obvious truth of that in each case of that makes sense $\endgroup$ Commented Aug 10, 2015 at 6:11
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    $\begingroup$ @Liam You could approach it by rephrasing the same line as "Without loss of generality, let $x\in A^*$ where $x=x_1x_2\cdots x_n$ with...." The phrase "without loss of generality" is commonly used in order to avoid needing to repeat essentially identical sections of proof for technically different cases. In using it here, we essentially tell the reader that "hey, I know I used $1,2$ and $n$ as indices, but don't pay too close of attention to them because they don't really matter." $\endgroup$
    – JMoravitz
    Commented Aug 10, 2015 at 14:32
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    $\begingroup$ @Liam alternatively, after stating $\emptyset\cdot x_1x_2\dots x_n=x_1x_2\dots x_n$ and the same for $x\cdot \emptyset$, you can write a line just before the "thus an identity element..." which reads "analogous results hold for $x$ of shorter lengths as well." This again has the purpose of pointing out that we know that our notation was such that it implied $x$ had positive length 3 or greater, but that it doesn't have any effect on the actual proof and $x$ could have been shorter. $\endgroup$
    – JMoravitz
    Commented Aug 10, 2015 at 14:36
  • $\begingroup$ Great thanks! I'm leaning a lot here! $\endgroup$ Commented Aug 10, 2015 at 18:25

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