# Vlorbik's Diner

## Classes, Laws, and Associations: Not JAWOPPA Anymore!

Posted by vlorbik on November 12, 2009

Let “Algebra 101” denote a generic
“remedial” algebra course for college
students.

For historical reasons, one typically finds
“introductory” material in Algebra 101
on Number Systems
(the Natural Numbers, the Integers,
and so on) and on such algebraic
“laws” as “The Associative Law
of Multiplication”… which is of course
the statement that for any
$x$, $y$, and $z$ (in any of our systems),
$x(yz) = (xy)z$.

For practical purposes however
this material is actually much
more {\it advanced} than most of the
rest of the course. However
{\it logically} prior these ideas
may be to {\it formal} discussions
we will be duty-bound to count
ourselves lucky if we can see
some precise {\it calculations}
from our students, and insisting
on correct use of vocabulary
courts disaster (since, for example,
many instructors are themselves
incapable of work at this level).

So we throw the facts out there in a big pile
knowing it’ll baffle ’em; it’s easy enough
to be a whole lot easier to understand
than the textbook and indeed several
students in any decent-size class
will have prepared minds (this
amounts to a certain “cultural
literacy” consisting mainly of
a “college level” of ordinary
words-on-paper literacy) and
get quite a bit out of this work
in a pretty short time…
so it’s a long way from a total loss
classwide. But for many,
maybe most, it’s just blah blah blah.

And this is even more of a shame than usual.
I’ve taken the time… stolen it even…
from a few of these Algebra 101 classes
to work carefully with {\it one little bit}
of this “high-level” stuff: associativity.
Enough to know that one can easily
bring {\it the whole class} (or the subset
that actually shows up and works;
never forget these are college students)
to work out some pretty convincing
exercises.

{\bf 1.} Prove that $((ab)(cd))e = (a(bc))(de)$ showing
each application of the associative law as a separate “step”.

$\langle(ab)[cd]\rangle e=$
$(ab)\langle[cd]e\rangle=$
$(ab)\langle c[de]\rangle=$
$\langle (ab)c\rangle[de]=$
$\langle a(bc)\rangle[de]$

I’m going to start yelling now in the certainty that
I won’t be heard. This looks trivial and is;
that’s why it’s important. If it {\it isn’t} trivial,
you don’t have any idea what
associativity even means
in any way that matters
outside classrooms. It takes
about a week to get a class to do this
though and there’s just no time.

I’m claiming here that there’s no point
in continuing to present lists of algebraic laws
at this level if we actually seek
student understanding. I now go on
to claim that this is perfectly well-known.
These are weeder courses and
this is a way to make flowering young minds
into the weeds they’ve unwittingly
signed up to be.

1. ### cut & paste | the livingston reviewsaid

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2. ### Vlorbik On Math Ed | the livingston reviewsaid

[…] 11/09/09 notes for chapter zero blahblahs 11/12/09 classes, laws, and associations associativity 11/12/09 And Left Me In Reputeless Banishment “A” and “a”; […]