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:

Eric Towers · Accepted Answer

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 $x ast varepsilon = xvarepsilon^2 = x$ 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.

Bill Dubuque · Answer

Checking is needed only due to (needless) use of $rmcolor{#c00}{unidirectional}$ (vs. $rmcolor{#0a0}{bidirectional})$ 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 = -1end{align}qquadqquad$$
The $rmcolor{#c00}{unidirectional}$ 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 $rmcolor{#0a0}{bidirectional}$ 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.

Greg Martin · Answer

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.

Is checking really needed?

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