###### What is the Distributive Property?

The distributive property (aka distributive law) is a property of real numbers which says that the product of a number say a and the sum (or difference) of two numbers say b and c, is equal to the sum (or difference) of the product of a and b and a and c. In symbol, we have:

The Distributive Property of Real Numbers

I’m sure you have used this property to work with numbers. A typical question that involves the distributive property would be: Find the product of x and (x-7). Using the distributive property, you will get $x(x-7)=x^2 - 7x$. This means that if x = 10, one way to calculate the product of 10 and (10-7) is 10×3 or 10×10-70. Both of these will give you 30. This property is true for all real numbers.

Here’s a more interesting problem where you can apply the distributive property of real numbers.

###### Problem

Warning: You do not need your calculator for this.

###### Solution

I’m not going to give the solution for each. Instead I’ll use a general form. I say general because it has the same structure as the three problems. The x represents any real numbers.

$(1-x)(1+x+x^2+x^3+x^4+x^5)$

= $1(1-x)+x(1-x)+x^2(1-x)+x^3(1-x)+x^4(1-x)+x^5(1-x)$

= $1-x+x-x^2+x^2-x^3+x^3-x^4+x^4-x^5+x^5-x^6$

= $1-x^6$

This in fact works for n number of terms. You can try proving the statement below to show that it is true for n terms in any x.

or

### 2 Responses to “Application of Distributive Property”

1. Bravo Erlina! Encouragements to pursuit!

Jean-Yves Rollin, Skype scolaire75

2. This can also be used to create a fake-proof that 1+1+1…+1 = 1.As a exercice you could ask your students to show what’s wrong with the of the following argument:

Consider the sum:
1+x+x²+x³+…+x^n = 1

We want to find a number x such that the above equation is solved. Then, f we multiply both sides by 1-x we have:
(x-1)(1+x+x²+x³+…+x^n) = x-1

Then, by the above formula:
x*x^n – 1 = x – 1

Cutting out the -1:
x*x^n = x

So we cut out a x in both sides:
x^n=1

Therefore, x = 1. But if we put this value in the original sum:
1+1+1²+1³+…+1^n = 1

By the way, that’s a cool way of looking into the geometric progression sum formula!