7.8: Induction and Inequalities
This is the third in a series of lessons on mathematical proofs. In this lesson we continue to focus mainly on proof by induction, this time of inequalities, and other kinds of proofs such as proof by geometry.
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James Sousa: Mathematical Induction
Guidance
The Transitive Property of Inequality
Below, we will prove several statements about inequalities that rely on the transitive property of inequality:
If a < b and b < c , then a < c .
Note that we could also make such a statement by turning around the relationships (i.e., using “greater than” statements) or by making inclusive statements, such as a ≥ b .
It is also important to note that this property of integers is a postulate , or a statement that we assume to be true. This means that we need not prove the transitive property of inequality.
You encountered other useful properties of inequalities in earlier algebra courses:
 Addition property: if a > b , then a + c > b + c .
 Multiplication property: if a > b , and c > 0 then ac > bc .
Example A
Prove that for .
Solution:
 1. The base case is n = 4: 4! = 24, 2 ^{ 4 } = 16. 24 ≥ 16 so the base case is true.
 2. Assume that k ! ≥ 2 ^{ k } for some value of k such that k ≥ 4
 3. Show that ( k +1)! ≥ 2 ^{ k+1 }


( k +1)! = k !( k +1) Rewrite ( k +1)! in terms of k ! ≥ 2 ^{ k } ( k +1) Use step 2 and the multiplication property. ≥ 2 ^{ k } (2) k +1 ≥ 5 >2, so we can use the multiplication property again. = 2 ^{ k+1 }

Therefore n ! ≥ 2 ^{ n } for n ≥ 4.
Example B
For what values of x is the inequality x > x ^{ 2 } true?
Solution:
The inequality is true if x is a number between 1 and 1 but not 0.
Example C
Prove that 9 ^{ n }  1 is divisible by 8 for all positive integers n .
Solution:
1. Base case:  If n = 1, 9 ^{ n }  1 = 91 = 8 = 8(1) 

2. Inductive hypothesis:  Assume that 9 ^{ k }  1 is divisible by 8. 
3. Inductive step:  Show that 9 ^{ k+1 }  1 is divisible by 8. 

9 ^{ k }  1 divisible by 8 8 W = (9 ^{ k } 1) for some integer W 9 ^{ k+1 }  1 = 9(9 ^{ k }  1) + 8 = 9(8W) + 8,which is divisible by 8
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Guided Practice
1) Prove that for all positive integers n where .
2) Prove that for all integers .
3) Prove that for all integers .
Answers
1) Use the three steps of proof by induction

Step 1) Base Case:
 ..... This checks out
 Step 2) Assumption:
 Step 3) Induction Step: starting with prove
 ..... If then this is true
 ... Multiply both sides by
 for all positive integers n where
2) Use the three steps of proof by induction
 Step 1) Base Case: (n = 1) or, if you prefer, (n = 2)
 Step 2) Assumption:
 Step 3) Induction Step: starting with prove
 and ..... assuming as specified in the question
 ..... combine the two statements above
 ..... add to both sides
 ..... from above
 for all integers
3) Prove that 2 n + 1 < 2 ^{ n } for all integers n > 3
 Step 1) Base case: If n = 3, 2(3) + 1 = 7, 2 ^{ 3 } = 8 : 7 < 8, so the base case is true.
 Step 2) Inductive hypothesis: Assume that 2 k + 1 < 2 ^{ k } for k > 3
 Step 3) Inductive step: Show that 2( k + 1) + 1 < 2 ^{ k + 1 }
 2( k + 1) + 1 = 2 k + 2 + 1 = (2k + 1) + 2 < 2 ^{ k } + 2 < 2 ^{ k } + 2 ^{ k } = 2 (2 ^{ k } ) = 2 ^{ k + 1 }
Explore More
Prove the following inequalities.
 For what values of x is the inequality x > x ^{ 2 } true?
 Prove that 3 ^{ n } > n ^{ 2 } for all positive integers n .
Prove the following inequalities.
 for
 Given: are positive numbers, prove the following:
 for
Complete the following geometric induction proofs.
 Prove that side length of a quadrilateral is less than the sum of all its other side lengths.
 Prove that side length of a pentagon is less than the sum of all its other side lengths.
 Prove that it is possible to color all regions of a plane divided by several lines with two different colors, so that any two neighbor regions contain a different color.
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Description
Learning Objectives
Here you will learn about applying the concepts of mathematical induction to inequalities.