Geometry Honors: Unit 2 Day 2 Algebraic Proof

Properties of Real Numbers Addition Property of Equality –

!"!! = !, !ℎ!"!! + ! = ! + ! Ex.

4 = 2 ∗ 2 = 2!

4+ 6 = 2! + 6 =

Subtraction Property of Equality – !"!! = !, !ℎ!"!! − ! = ! − !

Ex. 4 = 2 ∗ 2 = 2!

4− 1 = 2! − 1 =

Multiplication Property of Equality –

!"!! = !, !ℎ!"!!" = !" Ex.

4 = 2 ∗ 2 = 2!

4 ∗ 6 = 2! ∗ 6 =

Division Property of Equality –

!"!! = !, !ℎ!"!!! =!!

Ex. 4 = 2 ∗ 2 = 2!

!! = !

!

! =

Reflexive Property of Equality –

! = ! Ex.

9 = 9!, ! =

Symmetric Property of Equality – !"!! = !, !ℎ!"!! = !

Ex.

4 = 2! ⟺ 2! = 4

Transitive Property of Equality – !"!! = !!!"#!! = !, !ℎ!"!! = !

Ex. 4 = 2 ∗ 2!!"#!2 ∗ 2 = 2!

4 = 2!

Substitution Property of Equality – !"!! = !, !ℎ!"!!!!"#!!"!!"#$%&"'!!"!!

Ex. 34 =

68

!! +

!! !!!!!=!!!!!!!!!!!!!!!

Distributive Property–

! ! + ! = !" + !" Ex.

2(6+ 4) 2(10) = 20 2 6+ 4 = 2 6 + 2 4 = 12+ 8 = 20

An algebraic proof is one that is made up of a series of algebraic statements.

Example 1: Prove that if −5 ! + 4 = 70, then ! = −18. Given – −5 ! + 4 = 70

Proposition – – ! = −!"

< What we know, what we want >

Definitions –

Properties –

Postulates –

< How we get there >

Algebra −5 ! + 4 = 70

! = −!"

Property (Reason)

Substitution Property of Equality Example 2: Which postulate can be used to justify each statement?

a. If 4+ −5 = 1, then!! + 4+ −5 = ! − 1

b. If 5 = !, then ! = 5

Example 3: Prove that if 2! − 3 = 5, then ! = 4.

Geometric Properties Reflexive Property – Segments Angles

!!" = !".!!!!!!!!!!!!!!!!!∠1 = !∠1.

Symmetric Property – Segments

!!"!!" = !",!" = !".! Angles

If !∠1 = !∠2, then !∠2 = !∠1.

Transitive Property – Segments If !" = !"!!"#!!" = !",!" = !".! Angles If !∠1 = !∠2, and !∠2 = !∠3!then !∠1 = !∠3.

Example 4: Prove that if ∠!"# ≅ ∠!"# and ∠!"# ≅ ∠!"#, then ! = 6. Given – ∠!"# ≅ ∠!"# ∠!"# ≅ ∠!"# Definitions – ≅ Properties – Transitive Property Postulates – None (for now) Proposition – ! = 6