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I recently bought Pinter's A book of Abstract Algebra and I came to Examples (Chapter 2). Now, I do not understand why are "Checks" in (iii), (iv) needed because we deductively concluded that neutral element $e$ must be $- 1$ and simillary we showed that each inverse $x'$ of $x$ is $x'= - x - 2$ (of course, we concluded all this after assuming that symbol $+$ means regular adittion with all its properties).

Therefore, Checks are just surplus because we’ve already set up to get what we’re “checking” now. (Probably I am just wrong so please correct me.)

Here is the 1st example in part B: enter image description here

1b3b
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    I have to plead with people to not downvote this question. I am sure a lot of secondary school students not only lack the understanding on why checks are needed, but also lack the routine of doing the checks in the first place. Thus, this is the right question at the right place, and might get at least a few people to get checking (and understanding why!) where they previously assumed their algebraic manipulations will just yield the solution(s). –  Dec 26 '20 at 20:37
  • See this answer in the linked dupe for the essence of the matter. – Bill Dubuque Dec 26 '20 at 20:37

3 Answers3

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The first line of (iii) establishes the implication "if $e$ has the given property, then $e=-1$." But maybe no number has the given property—this if-then statement hasn't established that $-1$ has the given property, but only that it's the lone candidate. The "Check" actually establishes the implication "$-1$ has the given property".

This is actually a common nuance in high-school algebra in general, not just in this context. Personally I might label the two steps "Explore" and "Proof" instead of "Solve" and "Check" to clarify the difference.

Similar remarks apply to (iv): the first sentence is giving necessary conditions on the possible answer, but the "Check" is what establishes that the proposed answer is actually correct.

Greg Martin
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  • Thanks. I see that this things are linear so are checks pretty much useless (I know they are needed to "do the logic") in this particular case? Can you give counterexample, In other words can you please give example where check really prove that our guess/exploration is wrong) – 1b3b Dec 26 '20 at 20:34
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    @1b3b You are lucky in this case, as (as you say) you can see the equivalence directly and don't need to go in the roundabout way (prove the implication, then check). The author of the solutions went down the slower but safer route. SIde note: as a part of an unrelated question, I've tried to spill some wisdom at some earlier point about how, in general, we can perform calculations with some confidence that we won't make a mistake: https://math.stackexchange.com/a/3780331/700480 . That was about solving (in)equalities but it partially applies here too. –  Dec 26 '20 at 20:50
  • @1b3b If you use unidirectional arrows (noninvertible operations) like squaring then you can get extraneous roots, so you need to check that the possible roots are actual roots. Similarly if you multiply by a noninvertible element, e.g. when solving congruences. It's best to view this in arrow language as I explain in my answer. – Bill Dubuque Dec 26 '20 at 21:20
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This is not an answer to the Question. It is apparently a response to OP's underlying question, which has been asked repeatedly in comments to the other answers.

Let $x \ast y = x^y$ for $x,y > 0$. Explore: If $x \ast \varepsilon = x^\varepsilon = x$ for all $x$, then $\varepsilon = 1$. Check: $x \ast 1 = x^1 = x$ is true, but $1 \ast x = 1^x = x$ is false (for all $x \neq 1$). (This also demonstrates a failure of commutativity.)

What we have shown is that $1$ is a genuine identity on the right, but fails to be an identity on the left.

Let $x \ast y = xy^2$ for real numbers $x$ and $y$. Explore: If $x \ast \varepsilon = x\varepsilon^2 = x$ for all $x$, then $\varepsilon^2 = 1$, for instance $\varepsilon = -1$. Check : $-1 \ast y = (-1)y^2 \neq y$ (for all $y \neq -1$). So this fails. Alternatively, $\varepsilon = 1$ ... (works as a right-inverse for all $x$ and fails as a left-inverse for $y \neq 1$).

Finally, an example that is commutative and associative, but does not have an identity (on either side) or inverses.

Let $x \ast y = x + y$ for $x,y$ positive integers. Explore: If $x \ast \varepsilon = x + \varepsilon = x$, then $\varepsilon = 0$, which is not a positive integer.

Eric Towers
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Checking is needed only due to (needless) use of $\rm\color{#c00}{unidirectional}$ (vs. $\rm\color{#0a0}{bidirectional})$ arrows, i.e.

$$\begin{align} &\overbrace{x+e+1}^{\textstyle x*e} = x\ \color{#c00}\Longrightarrow\, e+1 = 0\ \color{#c00}\Longrightarrow\, e = -1\\[.2em] {\rm Better}\!\!:\ \ &x+e+1 = x\!\color{#0a0}\iff\! e+1 = 0\!\color{#0a0}\iff\! e = -1\end{align}\qquad\qquad$$

The $\rm\color{#c00}{unidirectional}$ inferences yield only that $\,e=-1\,$ is necessary for the identity to hold, not that it is sufficient. For sufficiency (reverse direction) we can either use the $\rm\color{#0a0}{bidirectional}$ arrows, or else explicitly check that the identity holds for $\,e = -1$.

As in the above inferences (which cancel $\,x\,$ then add $\,-1$) many equational inferences are naturally bidirectional, i.e. they yield equivalent equations, so there is no need to restrict their power by using them only unidirectionally.

Remark $ $ These painstaking verifications can be eliminated if we simply note that the essence of the matter boils down to the fact that the operation arises simply via transport of structure.

Bill Dubuque
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  • So, if we have some equation involving $e$ and if that equations have $n$ solutions, $e_1, e_2, ..., e_n$ then there are $n$ neutral elements? But it seems weird to me that number of $e$'s is greater then, well, $1.$ – 1b3b Dec 26 '20 at 21:26
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    @1b3b Hypothetically yes, but that can't occur here because in this theory neutral elements are unique; indeed if $,e',$ and $e$ are neutrals then $, e' = e*e' = e\ \ $ – Bill Dubuque Dec 26 '20 at 21:32
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    Take for example $xy = |x - y|.$ If* $y$ is indeed neutral element ($y := e$), then it must be $|x - e| = x.$ So, only two candidates are $e_1 = 0$ and $e_2 = 2x.$ (We can immediately eliminate $e_2$ from our candidate's list because $e$ cannot depend on $x$, of course.) Now what that check in the book exactly does? Because obviously if we plug that two values in the equation we will get two valid equations. I am confused... – 1b3b Dec 26 '20 at 21:39
  • In other words, if we take two elements, $x$ and double of it, $2x$, our operation will indeed spit out $x$ back every time, but that's not what we want for neutral element in general. Because we want same $e$ for everybody (every $x$ we choose) and is that conclusion "check" in a sense? – 1b3b Dec 26 '20 at 21:45
  • @1b3b Yes, a neutral element is an element of the underlying set of the algebraic structure, not a function on that set. – Bill Dubuque Dec 26 '20 at 21:49
  • Yes, then that checking work for $e$ is easier when comparing to $x'$ because $x'$ depends on $x$ we have chosen. In original example, $x*y = x + y + 1$, we get that if $y = x'$ then it must be $x + x' + 1 = e = - 1 \iff x' = - x - 2$... Can you please give counterexample, example when check is useful? – 1b3b Dec 26 '20 at 21:54
  • @1b3b Yes, a neutral element is a constant (or nullary operation, i.e a map that takes no arguments and returns an element), but the inverse map is a unary operation (i.e. a map that takes one element to another). Btw, you may find helpful my emphasis of the power of bidrectional inferences in various posts. – Bill Dubuque Dec 26 '20 at 22:01
  • Thanks. I would highly appreciate one small and simple example when just solving the equation isn't enough, for $e$ and/or $x'$ :) – 1b3b Dec 26 '20 at 22:09
  • @1b3b Some examples: you probably already know that the noninvertible operation of squaring can introduce extraneous roots. Similarly for multiplying by noninvertibles, e.g. here and here. – Bill Dubuque Dec 26 '20 at 22:21
  • Also, why Pinter checks both $xx'$ and $x'x$ (and he does the same for $e$)? He proved before that $xy = x + y + 1$ is commutative. – 1b3b Dec 26 '20 at 22:22
  • @1b3b Yes, that check is unneeded given commutativity is already known. – Bill Dubuque Dec 26 '20 at 22:26
  • Suppose all our elements for which $xy = x + y + 1$ is defined are in $A.$ Obviously, for $xy$ to work in $A$ we have to define operation $x + y$ (because we calculate using ordinary addition) on $A$ first and we did it. This checks are logically useful (if and only if) but there is no need to repeat them twice because we got $e = - 1$ and it's obvious that $- 1$ must work because it's only solution of the equation. Anyway, thanks and sorry for taking your time. – 1b3b Dec 26 '20 at 22:34
  • @1b3b I'm not sure what you are trying to say in your prior comment. These points are addressed in my answer. Did you read it? Is is clear? – Bill Dubuque Dec 26 '20 at 22:36
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    Yes, it's clear but that checks made me confused at the first. Please forgive me, but I see that it's just traditional task to torture students... I perfectly understand iff statements but this checks taked my shoes off. Take care and happy new 2021! – 1b3b Dec 26 '20 at 22:46