1. Applications of Expansion and Factorisation
Slideshow 17
Mathematics
Mr. Richard Sasaki

2. Objectives
Use the difference of two squares to make certain numerical calculations
Be able to prove certain simple numerical facts

3. The Difference of Two Squares
What is the difference of two squares?
The law states that if an expression 𝑎+𝑏 is multiplied by its conjugate, we get the difference of 𝑎 2 and 𝑏 2 .
𝑎+𝑏 𝑎−𝑏 ≡ 𝑎 2 − 𝑏 2 for ∀ 𝑎, 𝑏∈ℝ.
This rule has many uses to help make calculations easier.

4. The Difference of Two Squares
𝑎+𝑏 𝑎−𝑏 ≡ 𝑎 2 − 𝑏 2 for ∀ 𝑎, 𝑏∈ℝ.
Let’s use this identity to help us make some numerical calculations.
Example
Calculate 51×49.
Consider the above for 𝑎=50, 𝑏=1.
51×49= −1
= 50 2 − 1 2
=2500−1
=2499

5. The Difference of Two Squares
We also need to consider the opposite principle.
𝑎 2 − 𝑏 2 ≡ 𝑎+𝑏 𝑎−𝑏 for ∀ 𝑎, 𝑏∈ℝ.
Example
Calcuate − 5 2 .
Consider the above for 𝑎=45, 𝑏=5.
45 2 − 5 2 = −5
=50×40
=2000

6. 6399
396
2491
2451
8019
3536
999900
8096
48.91
If 𝑥 refers to the girl…
If 𝑥 refers to the boy…
𝑥+2 2 − 𝑥 2 =32
𝑥 2 − 𝑥−2 2 =32
𝑥 2 +4𝑥+4− 𝑥 2 =32
𝑥 2 − 𝑥 2 +4𝑥−4=32
4𝑥+4=32⇒𝑥=7
4𝑥−4=32⇒𝑥=9
The boy is 7 and his sister is 9.
600
1800
6600
900000
−9000

7. Numerical Proofs
Before we look at proofs, we should recall some vocabulary.
Odd -
A number in the form 2𝑛−1 where 𝑛∈ ℤ + . (1, 3, 5, 7, …)
Even -
A number in the form 2𝑛 where 𝑛∈ ℤ + . (2, 4, 6, 8, …)
Consecutive Integers -
A pair of integers 𝑛, 𝑛+1 where 𝑛∈ℤ.
(for example 15 and 16.)

8. Proofs (Format) When you prove something, there should be three steps:
A statement declaring any variables you need for your proof. (eg: Let an even number be a number be in the form 2𝑛 where 𝑛∈ ℤ + .)
Step 2 -
Perform the necessary calculation (manipulation). Write what you need to in the correct form.
Step 3 -
State that as it’s now in the correct form, the proof for the original expression is now complete.

9. Numerical Proofs Let’s try a simple proof and divide the steps up.
Example
Prove that an odd number squared is odd.
Step 1
Let an odd number 𝑎=2𝑛−1 where 𝑛∈ ℤ + .
Step 2
If 𝑎=2𝑛−1, 𝑎 2 = 2𝑛−1 2 =4 𝑛 2 −4𝑛+1
=2(2 𝑛 2 −2𝑛)+1
Step 3
As 𝑎 2 =2 2 𝑛 2 −2𝑛 +1 is in the form 2𝐴+1 where 𝐴∈ ℤ + , 𝑎 2 must be odd. ∴ An odd number squared is odd.

10. Numerical Proofs Example If two integers are odd, their sum is even.
Step 1
Consider two integers 𝑎, 𝑏 where 𝑎=2𝑚−1, 𝑏=2𝑛−1 and 𝑚, 𝑛∈ ℤ + .
Step 2
The sum of 𝑎 and 𝑏, 𝑎+𝑏= 2𝑚−1 +(2𝑛−1)
=2𝑚+2𝑛−2=2(𝑚+𝑛−1).
Step 3
As 𝑎+𝑏 is in the form 2𝐴 where 𝐴∈ ℤ + , 𝑎+𝑏 is even. ∴ If two integers are odd, their sum is even.

11. Answers – Easy (Top)
Let an odd number be in the form 2𝑛−1 where 𝑛∈ ℤ + . 7=2∙4−1 where 4∈ ℤ + . As 7 can be written in the form 2𝑛−1 where 𝑛∈ ℤ + , 7 is odd.
Let an even number be in the form 2𝑛 where 𝑛∈ ℤ =2∙6 where 6∈ ℤ + . As 12 can be written in the form 2𝑛 where 𝑛∈ ℤ + , 12 is even.
Let an even number 𝑎=2𝑛 where 𝑛∈ ℤ + . If 𝑎=2𝑛, 𝑎 2 = 2𝑛 2 =4 𝑛 2 =2 2 𝑛 As 𝑎 2 is in the form 2𝐴 where 𝐴∈ ℤ + , 𝑎 2 is even. ∴ An even number squared is even.

12. Let an even number 𝑎=2𝑛 where 𝑛∈ ℤ +
Let an even number 𝑎=2𝑛 where 𝑛∈ ℤ + . If 𝑎=2𝑛, 𝑎 2 =𝑛⇒ 𝑎 2 ∈ ℤ + as 𝑛∈ ℤ + .∴ Half of an even number is an integer.
Consider two integers 𝑎, 𝑏 where 𝑎=2𝑚, 𝑏=2𝑛 and 𝑚, 𝑛∈ ℤ + . The sum of 𝑎 and 𝑏, 𝑎+𝑏= 2𝑚+2𝑛=2 𝑚+𝑛 . As 𝑎+𝑏 is in the form 2𝐴 where 𝐴∈ ℤ + , 𝑎+𝑏 is even. ∴ If two integers are even, their sum is even.
Let 𝑎 be an odd number where 𝑎=2𝑛−1,𝑛∈ ℤ + . As 𝑎 2 = 2𝑛−1 2 , 𝑎 2 =4 𝑛 2 −4𝑛+1=2 2 𝑛 2 −2𝑛 2 𝑛 2 −2𝑛 +1. As 𝑎 2 is in the form 2𝐴+1 where 𝐴∈ ℤ + , 𝑎 2 is odd. Also, its root 𝑎=2𝑛−1 is odd. ∴ The positive root of a square odd number is odd.

13. Let a pair of consecutive odd numbers 𝑎=2𝑛−1, 𝑏=2𝑛+1, 𝑛∈ ℤ +
Let a pair of consecutive odd numbers 𝑎=2𝑛−1, 𝑏=2𝑛+1, 𝑛∈ ℤ + . 𝑎𝑏+1= 2𝑛−1 2𝑛+1 + 1=4 𝑛 2 . As 𝑎𝑏+1 is in the form 4𝐴 where 𝐴∈ ℤ + , the product of two consecutive odd numbers plus 1 is a multiple of 4.
Consider two integers 𝑎, 𝑏 where 𝑎=2𝑚−1, 𝑏=2𝑛 and 𝑚, 𝑛∈ ℤ + . The sum of 𝑎 and 𝑏, 𝑎+𝑏=2𝑚− 1+2𝑛=2 𝑚+𝑛 −1. As 𝑎+𝑏 is in the form 2𝐴− 1where 𝐴∈ ℤ + , 𝑎+𝑏 is odd. ∴ The sum of an odd and even number is odd.
Consider two integers 𝑎, 𝑏 where 𝑎=2𝑚−1, 𝑏= 2𝑛−1 and 𝑚, 𝑛∈ ℤ + . The product of 𝑎 and 𝑏, 𝑎𝑏= 2𝑚−1 2𝑛−1 =2 2𝑚𝑛−𝑚−𝑛 +1. As 𝑎𝑏 is in the form 2𝐴+1 , where 𝐴∈ ℤ + , the expression is odd. ∴ The product of two odd numbers is odd.

14. Answers – Hard (Bottom)
Consider two integers 𝑎, 𝑏 where 𝑎=2𝑚−1, 𝑏= 2𝑛 and 𝑚, 𝑛∈ ℤ + . The product of 𝑎 and 𝑏, 𝑎𝑏= 2𝑛 2𝑚−1 =2(2𝑚𝑛−𝑛). As 𝑎𝑏 is in the form 2𝐴 where 𝐴∈ ℤ + , the expression is even. ∴ The product of an odd and even number is even.
Let an odd number 𝑛 be in the form 𝑛=2𝑎−1 where 𝑎∈ ℤ + . By substitution, 3𝑛+7=3 2𝑎−1 +7=6𝑎+ 4=2 3𝑎+2 . As 3𝑛+7 is in the form 2𝐴 where 𝐴∈ ℤ + , the expression is even. ∴ If 𝑛 is odd, 3𝑛+7 is even.