Is checking really needed?

Question

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:

Answer 1

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.

Answer 2

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.

Answer 3

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.