Question

For all \[a,b \in R\], we define \[a * b = \left| {a - b} \right|\]. Show that, \[ * \] is commutative but not associative.

Answer 1

Hint: We are given two variables like a and b. we should know that commutative and associative are the two properties. Commutative is like \[a * b = b * a\] and associative is like \[\left( {x * y} \right) * z = x * \left( {y * z} \right)\]. This is generally used in solving the equations. But here we are given one more condition that is \[ * \] symbol. It shows that, \[a * b = \left| {a - b} \right|\] So we will take some values for a, b, x, y and z, and get the proof.

Complete step-by-step answer:
Given that, \[a,b \in R\]
Also they define the situation as \[a * b = \left| {a - b} \right|\]
Now we have to prove that, \[ * \] is commutative but not associative. That is \[a * b = b * a\] is true or holds and \[\left( {x * y} \right) * z = x * \left( {y * z} \right)\] is false.
1.commutative property:
let a=2 and b=3
so we can write,
\[a * b = \left| {a - b} \right|\]
Putting the values,
\[2 * 3 = \left| {2 - 3} \right|\]
On subtraction it gives,
\[ = \left| { - 1} \right|\]
On taking the modulus,
\[ = 1\]
Now,
\[b * a = \left| {b - a} \right|\]
Putting the values,
\[3 * 2 = \left| {3 - 2} \right|\]
On subtraction we get,
\[ = \left| 1 \right|\]
On taking the modulus,
\[ = 1\]
We observe that the commutative nature holds for the condition above that is,
\[a * b = b * a = 1\]
Thus first statement, \[ * \] is commutative is true.

2.Associative property:
Let x=2, y=3 and z=4
So we can write,
\[\left( {x * y} \right) * z\]
Putting the values,
\[ = \left( {2 * 3} \right) * 4\]
Taking the property \[a * b = \left| {a - b} \right|\] mentioned in the question for the bracket,
\[ = \left| {2 - 3} \right| * 4\]
On subtracting we get,
\[ = \left| { - 1} \right| * 4\]
Taking the value of modulus,
\[ = 1 * 4\]
Now again using, \[a * b = \left| {a - b} \right|\]
\[ = \left| {1 - 4} \right|\]
On subtracting and taking the modulus out,
\[ = \left| { - 3} \right|\]
\[ = 3\]
This is one side.

\[x * \left( {y * z} \right)\]
Putting the values,
\[2 * \left( {3 * 4} \right)\]
Taking the property \[a * b = \left| {a - b} \right|\] mentioned in the question for the bracket,
\[ = 2 * \left| {3 - 4} \right|\]
On subtracting we get,
\[ = 2 * \left| { - 1} \right|\]
Taking the value of modulus,
\[ = 2 * 1\]
Now again using, \[a * b = \left| {a - b} \right|\]
\[ = \left| {2 - 1} \right|\]
On subtracting and taking the modulus out,
\[
   = \left| 1 \right| \\
   = 1 \\
\]
This is the other side.

Now we can clearly see that the LHS and RHS are not equal.
\[\left( {x * y} \right) * z \ne x * \left( {y * z} \right)\]
 So the sign \[ * \] is not applicable for associative property.
Thus proved both the properties.

Note: Note that, we should not get confused between the properties. That is the biggest part where students get confused. Also note that there is no such restriction on using the variables and values also. But remember the condition will not change that . Though we used x, y and z it will be applied the way it is given.