Chapter 1, Problem 2RE

To determine

To describe: The property of the real numbers being used, $\left(a+b\right)\left(a-b\right)=\left(a-b\right)\left(a+b\right)$ .

Answer to Problem 2RE

Commutative property of multiplicity is used in the expression.

Explanation of Solution

Given information:

The expression, $\left(a+b\right)\left(a-b\right)=\left(a-b\right)\left(a+b\right)$

The commutative property is a property in which, $a\cdot b=b\cdot a$ .it means when we multiply two numbers as, , $a\cdot b$ ,and then we add $b\cdot a$ and equate these both are equal by this property.

It is like two constants are multiplyed and they are equal the the commutative property is there.

As the two numbers are multiply, order doesn’t matter.

For example: $\begin{array}{c}4\cdot 3=3\cdot 4\\ 12=12\end{array}$

Apply the property , $a\cdot b=b\cdot a$

Rewrite the expression as,

  $\begin{array}{c}\left(a+b\right)\left(a-b\right)=\left(a-b\right)\left(a+b\right)\\ a^2-ab+ba-b^2=a^2+ab-ba-b^2\\ 0=0\\ \end{array}$ .

The above expression is satisfied to the commutative property.