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

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

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.

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= 50+1 50−1 = 50 2 − 1 2 =2500−1 =2499

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 45 2 − 5 2 . Consider the above for 𝑎=45, 𝑏=5. 45 2 − 5 2 = 45+5 45−5 =50×40 =2000

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 3800000 −9000

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.)

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.

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.

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.

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 + . 12=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 2 . As 𝑎 2 is in the form 2𝐴 where 𝐴∈ ℤ + , 𝑎 2 is even. ∴ An even number squared is even.

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.

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.

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.