# Difference between revisions of "Associative implies generalized associative"

## Statement

Let $S$ be a set and $*$ be a binary operation on $S$. Suppose that $*$ is an associative binary operation; in other words:

$a * (b * c) = (a * b) * c \ \forall \ a,b,c \in S$

Then, for any positive integer $n$, every possible parenthesization of the expression:

$a_1 * a_2 * \dots * a_n$

is equivalent.

## Proof

The technique of proof here is a strong form of induction: we assume the result is true for every $m < n$, and we prove it for $n$. Note that the cases $n = 1,2$ are tautological. For the case $n = 3$, there are precisely two possible parenthesizations, and associativity tells us that they are both equal.

Thus, assume $n > 3$, and that the result holds for all $m < n$. We will show that any way of parenthesization of:

$a_1 * a_2 * \dots * a_n$

is equal to the left-associated expression:

$((\dots ( a_1 * a_2) * a_3) * \dots ) * a_n)$

Let's prove this. For any method of parenthesization, there is an outermost multiplication, and in terms of this outermost multiplication, we can view the parenthesization as:

$A * B$

where $A = a_1 * \dots * a_m$ and $B = a_{m+1} * \dots * a_n$, both parenthesized in some unknown way, with $0 < m < n$. Applying the induction assumption, $A$ equals the left-associated expression for $a_1 * a_2 * \dots * a_m$ and $B$ equals the left-associated expression for $a_{m+1} * a_{m+2} * \dots * a_n$.

Now, in the case that $B$ is a single letter (i.e., $m = n - 1$), we are already done: $A * B$ equals the left-associated expression. If not, we can write $A * B$ as:

$A * (C * a_n)$

where $C$ is the left-associated expression for $a_{m+1} * \dots * a_{n-1}$. Applying associativity, we get that the original expression equals:

$(A * C) * a_n$

which is precisely the left-associated expression for $a_1 * a_2 * \dots * a_n$.

## References

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 18-19, Section 1.1, Proposition 1 (statement on page 18, proof on page 19)