A set has the commutative property under a particular operation if the result of the operation is the same, even if you switch the order of the elements that are being acted on by the operation.
More formally, if x and y are variables that represent any 2 arbitrary elements in the set we are looking at (let’s call the set we are looking at A), and the symbol # represents our operation, then the commutative property , for A with the operation # would be:
x#y = y#x.
This means that the commutative property only holds for the set A and the operation # if, no matter what elements we take from A and put in place of x and y, x#y will always give us the same result as y#x.
It is much easier to understand a property by looking at examples than it is by simply talking about it in an abstract way, so let's move on to looking at examples so that you can see exactly what we are talking about when we say that a set has the commutative property:
First let’s look at a few infinite sets with operations that are already familiar to us:
a) The set of natural numbers is commutative under the operation of addition, because it is true that for any two natural numbers x and y, x+y = y+x.
b) The set of integers is not commutative under the operation of division, because for any two integers x and y, there are many cases where x ÷ y ≠ y ÷ x!
For example:
2 ÷ 5 = 2/5 or 0.4
5 ÷ 2 = 5/2 or 2.5
0.4 ≠ 2.5!
So 2 ÷ 5 ≠ 5 ÷ 2
So x ÷ y = y ÷ x is not true for all integers x and y!
to see some more examples!
If all we have is an operation TABLE, it can be a little bit more difficult to tell whether or not a set is commutative under a particular operation. To figure out if there is a simple way to tell if a set with an operation table is commutative under the given operation,
e) Here is an operation table for the set {a,b,c} and the operation *:
* |
a |
b |
c |
a |
a |
b |
c |
b |
b |
a |
c |
c |
c |
c |
a |
From the table we can see that:
a*a=aa*b=b b*a=b So a*b = b*a
a*c=c c*a=c So a*c = c*a
b*b=a
b*c=c c*b=c So b*c = c*b
c*c=a
So if the variables x and y represent any two arbitrary elements in the set {a,b,c}, then x*y = y*x. So the set {a,b,c} under the operation * outlined in the operation TABLE above is commutative!
f) Here is an operation table for the set {a,b,c} and the operation ~:
~ |
a |
b |
c |
a |
a |
c |
c |
b |
c |
a |
b |
c |
b |
c |
a |
From the table we can see that:
a~a=aa~b=c b~a=c So a~b = b~a
a~c=c c~a=b So a~c ≠ c~a
b~b=a
b~c=b c~b=c So b~c ≠ c~b
c~c=a
So if the variables x and y represent any two arbitrary elements in the set {a,b,c}, then it will not be true in all cases that x~y = y~x. So the set {a,b,c} under the operation ~ outlined in the operation TABLE above is not commutative!
Do we notice a pattern here in the tables when a set is commutative under a certain operation? If you look at the tables, you’ll see that the area where the results are located is symmetrical across the diagonal (the diagonal is the purple shaded area) when it is commutative and the area where the results are is asymmetrical (not symmetrical) across the diagonal if it is not commutative!
The set {a,b,c} with the operation * as defined by this table was commutative:
* |
a |
b |
c |
a |
a |
b |
c |
b |
b |
a |
c |
c |
c |
c |
a |
Notice that for this table, each result is exactly the same as the one that mirrors it on the other side of the diagonal; if we were to fold this table in half right on the diagonal (the purple line) then the two folded sides would be exactly the same: the b would fold right on top of the b on the other side of the diagonal, the c would fold right on top of the c right on the other side of the diagonal and the c would fold right on top of the c on the other side of the diagonal! The two halves of results on either side of the diagonal mirror one another, so this operation table is symmetric across the diagonal!
The set {a,b,c} with the operation ~ as defined by this table was not commutative:
* |
a |
b |
c |
a |
a |
c |
c |
b |
c |
a |
b |
c |
b |
c |
a |
Notice that for this table, not every result is exactly the same as the one that mirrors it on the other side of the diagonal; if we were to fold this table in half right on the diagonal (the purple line) then the two folded sides would be different: the c would fold right on top of the c on the other side of the diagonal, but the c would fold right on top of the b on the other side of the diagonal and the b would fold right on top of the c on the other side of the diagonal! The two halves of results on either side of the diagonal do not mirror one another, so this operation table is asymmetric across the diagonal!
g) Let’s consider the operation @ acting on the set { β, γ, δ}given by the following GROUP TABLE:
@ |
β |
γ |
δ |
β |
β |
δ |
δ |
γ |
δ |
β |
γ |
δ |
γ |
δ |
β |
This set is not commutative under the operation of @ because the operation table is asymmetrical across the diagonal! δ and γ are not the same and γ and δ are not the same!
to see another example.
Now return to Blackboard to answer Group Lecture Questions 3: Commutativity!