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### Proving the Associative Property

```
Date: 02/24/2001 at 15:15:07
From: Cesar Valverde
Subject: Associative properties

In a binary operation, for example:

# | a  b  c
---+---------
a | b  c  a
b | c  b  b
c | a  b  a

I understand that I can prove it is commutative (i.e. a#b = b#a)
because there's symmetry along the diagonal (the one that goes "\").
But how can I prove that it is associative, i.e. a#(b#c) = (a#b)#c?
```

```
Date: 02/26/2001 at 17:24:54
From: Doctor Achilles
Subject: Re: Associative properties

Hi Cesar,

Thanks for writing to Dr. Math.

You're right that you can prove the commutative property holds because
of the diagonal. However, you actually are very close to a proof that
the associative property DOES NOT hold.

The associative property of an operation says that for all values of
x, y and z, (x*(y*z)) = ((x*y)*z). To prove it, you need an argument
that shows that it always holds. To disprove it, you need one
counterexample, i.e. one time when (x*(y*z)) = ((x*y)*z) is false.

(a*(b*c)) in your operation can be evaluated like this. Write the
expression to evaluate:

(a*(b*c))

Just like anything else, we have to do what's inside the parentheses
first:

(a*b)

Now we can evaluate the answer:

c

Try the same thing for ((a*b)*c). Remember to evaluate what's inside
the parentheses first. See what you come out with. If it's not c,
then you have a counterexample. You can then disprove the associative
property by simply saying that if it did hold, then the two
expressions would have the same value, since they have different
values, it doesn't hold.

other math topics, please write back.

- Doctor Achilles, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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