<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

Induction and Inequalities

Transitive, addition, and multiplication properties of inequalities used in inductive proofs.

Atoms Practice
%
Progress
 
 
 
MEMORY METER
This indicates how strong in your memory this concept is
Practice
Progress
%
Practice Now
Turn In
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.

Induction and Inequalities

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 ab.

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.

Examples

Example 1

Prove that \begin{align*}n!\geq 2^n\end{align*} for \begin{align*}n\geq 4\end{align*}.

Step 1) The base case is n = 4: 4! = 24, 24 = 16. 24 ≥ 16 so the base case is true.

Step 2) Assume that k! ≥ 2k for some value of k such that k ≥ 4

Step 3) Show that (k+1)! ≥ 2k+1

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

Therefore n! ≥ 2n for n ≥ 4.

Example 2

For what values of x is the inequality x > x2 true?

The inequality is true if x is a number between -1 and 1 but not 0.

Example 3

Prove that 9n - 1 is divisible by 8 for all positive integers n.

1. Base case: If n = 1, 9n - 1 = 9-1 = 8 = 8(1)
2. Inductive hypothesis: Assume that 9k - 1 is divisible by 8.
3. Inductive step: Show that 9k+1 - 1 is divisible by 8.
9k - 1 divisible by 8 \begin{align*}\Rightarrow\end{align*} 8W = (9k-1) for some integer W
9k+1 - 1 = 9(9k - 1) + 8 = 9(8W) + 8,which is divisible by 8

Example 4

Prove that \begin{align*}2^n < n!\end{align*} for all positive integers n where \begin{align*}n \geq 4\end{align*}.

Use the three steps of proof by induction:

Step 1) Base Case: \begin{align*}2^4 < 4!\end{align*}

\begin{align*}2 \cdot 2 \cdot 2 \cdot 2 < 1 \cdot 2 \cdot 3 \cdot 4\end{align*}

\begin{align*}16 < 24\end{align*} ..... This checks out

Step 2) Assumption: \begin{align*}2^k < k!\end{align*}

Step 3) Induction Step: starting with \begin{align*}2^k < k!\end{align*} prove \begin{align*}2^k(k + 1) < k!(k +1)\end{align*}

\begin{align*}2^k(k + 1) < (k + 1)!\end{align*}

\begin{align*}2 < k + 1\end{align*} ..... If \begin{align*}k \geq 4\end{align*} then this is true

\begin{align*}2^k \cdot 2 < 2^k (k + 1)\end{align*} ... Multiply both sides by \begin{align*}2^k\end{align*}

\begin{align*}2^{k + 1} < 2^k (k + 1)\end{align*}

\begin{align*}2^{k + 1} < (k + 1)!\end{align*}

\begin{align*}\therefore 2^n < n!\end{align*} for all positive integers n where \begin{align*}n \geq 4\end{align*}

Example 5

Prove that \begin{align*}n^2 < 3^n\end{align*} for all integers \begin{align*}n > 2\end{align*}.

Use the three steps of proof by induction:

Step 1) Base Case: (n = 1) \begin{align*}1^2 < 3^1\end{align*} or, if you prefer, (n = 2) \begin{align*}2^2 < 3^2\end{align*}

Step 2) Assumption: \begin{align*}k^2 < 3^k\end{align*}

Step 3) Induction Step: starting with \begin{align*}k^2 < 3^k\end{align*} prove \begin{align*}(k + 1)^2 < 3^{k + 1}\end{align*}

\begin{align*}k^2 \cdot 3 < 3^k \cdot 3\end{align*}

\begin{align*}2k < k^2\end{align*} and \begin{align*}1 < k^2\end{align*} ..... assuming \begin{align*}2 < k\end{align*} as specified in the question

\begin{align*}2k + 1 < 2k^2\end{align*} ..... combine the two statements above

\begin{align*}k^2 + 2k + 1< 3k^2\end{align*} ..... add \begin{align*}k^2\end{align*} to both sides

\begin{align*}(k + 1)^2 < 3k^2\end{align*}

\begin{align*}(k + 1)^2 < 3 \cdot 3^k\end{align*} ..... from above

\begin{align*}(k + 1)^2 < 3^{k + 1}\end{align*}

\begin{align*}\therefore n^2 < 3^n\end{align*} for all integers \begin{align*}n > 2\end{align*}

Example 6

Prove that \begin{align*}2n+1 < 2^n \end{align*} for all integers \begin{align*}n>3\end{align*}.

Use the three steps of proof by induction:

Step 1) Base case: If n = 3, 2(3) + 1 = 7, 23 = 8 : 7 < 8, so the base case is true.

Step 2) Inductive hypothesis: Assume that 2k + 1 < 2k for k > 3

Step 3) Inductive step: Show that 2(k + 1) + 1 < 2k + 1

2(k + 1) + 1 = 2k + 2 + 1 = (2k + 1) + 2 < 2k + 2 < 2k + 2k = 2 (2k) = 2k + 1

  

Review

Prove the following inequalities.

  1. \begin{align*}5^k<(k+5)!\end{align*}
  2. \begin{align*}1^k < (k + 1)!\end{align*}
  3. \begin{align*}4^k< (k + 4)!\end{align*}
  4. \begin{align*}2^k<(k + 2)!\end{align*}
  5. For what values of x is the inequality x > x2 true?
  6. Prove that 3n > n2 for all positive integers n.

Prove the following inequalities.

  1. \begin{align*}\frac{1}{n + 1} + \frac{1}{n+2} + \frac{1}{n + 3} + ... + \frac{1}{2n}> \frac{13}{24}(n > 1)\end{align*}
  2. \begin{align*}2^n \geq n^2\end{align*} for \begin{align*}n = 4, 5, 6,...\end{align*}
  3. \begin{align*}\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ... +\frac{1}{n^2}\end{align*}
  4. Given:\begin{align*}x_1, ..., x_n\end{align*} are positive numbers, prove the following:\begin{align*}\frac{(x_1+ ...+ x_n)}{n} \geq (x_1 \cdot ...\cdot x_n)^{\frac{1}{n}}\end{align*}
  5. \begin{align*}n! \geq 3^n\end{align*} for \begin{align*}n = 7, 8, 9, ....\end{align*}

Complete the following geometric induction proofs.

  1. Prove that side length of a quadrilateral is less than the sum of all its other side lengths.
  2. Prove that side length of a pentagon is less than the sum of all its other side lengths.
  3. 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.

Review (Answers)

To see the Review answers, open this PDF file and look for section 7.8. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Vocabulary

!

The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol "!" denotes factorial. n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n .

factorial

The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol "!" denotes factorial. n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n .

induction

Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.

inequality

An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are <, >, \le, \ge and \ne.

Integer

The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...

postulate

A postulate is a statement that is accepted as true without proof.

proof

A proof is a series of true statements leading to the acceptance of truth of a more complex statement.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Induction and Inequalities.
Please wait...
Please wait...