Applications and harder induction
Worked examples
Inequality proof (Bernoulli's inequality)
Straightforward
Problem
Answer
Recursive sequence
Moderate
Problem
Answer
Inequality with a non-standard base case
Challenging
Problem
Answer
Practise
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
LHS . RHS . ✓
**Inductive hypothesis:** Assume is true for some integer :
**Inductive step:** Show is true, i.e., .
Starting from the LHS of :
This is exactly the RHS of . ✓
**Conclusion:** By the principle of mathematical induction, is true for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
RHS . ✓
**Inductive hypothesis:** Assume is true: .
**Inductive step:** Show is true, i.e., .
Using the recurrence relation and the inductive hypothesis:
This equals , which is the formula for . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
. ✓
**Inductive hypothesis:** Assume is true for some integer : .
**Inductive step:** Show is true, i.e., .
Since , we have , so:
Therefore . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
. Divisible by 7. ✓
**Inductive hypothesis:** Assume is true: for some integer , i.e., .
**Inductive step:** Show is true, i.e., .
Since is an integer, is divisible by 7. ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
LHS . RHS . ✓
**Inductive hypothesis:** Assume is true:
**Inductive step:** Show is true.
This equals , the RHS of . ✓
**Conclusion:** By the principle of mathematical induction, is true for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
. So . ✓
**Inductive hypothesis:** Assume is true for some : .
**Inductive step:** Show is true, i.e., .
From the inductive hypothesis, multiplying both sides by 2 (both positive):
It now suffices to show , i.e., .
Completing the square: .
Since , we have , so , giving . ✓
Therefore . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
. Divisible by 9. ✓
(Check : . ✓)
**Inductive hypothesis:** Assume is true: for some integer , i.e., .
**Inductive step:** Show .
Since is an integer, . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
LHS . RHS . ✓
**Inductive hypothesis:** Assume is true:
**Inductive step:** Show is true.
Factoring the numerator: .
This is the RHS of . ✓
**Conclusion:** By the principle of mathematical induction, is true for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
RHS . ✓
(Check: . ✓)
**Inductive hypothesis:** Assume is true for some : .
**Inductive step:** Show is true, i.e., .
Using the recurrence relation and the inductive hypothesis:
This is the formula for . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
. ✓ (equality holds)
**Inductive hypothesis:** Assume is true for some : for all .
**Inductive step:** Show is true, i.e., .
Since , we have . Multiplying both sides of the inductive hypothesis by preserves the inequality:
Expanding the right side:
Since for all real :
This is the RHS of . ✓
**Conclusion:** By the principle of mathematical induction, for all integers and all .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
and . Since , is true. ✓
**Inductive hypothesis:** Assume is true for some : .
**Inductive step:** Show is true, i.e., .
Since , we have . Using the inductive hypothesis:
Since :
Therefore . ✓
**Conclusion:** By the principle of mathematical induction, for all integers .
> **Note:** The result fails for small : , , . That is why the base case is , not .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Base case** ():
LHS . RHS . So LHS RHS. ✓
**Inductive hypothesis:** Assume is true: .
**Inductive step:** Show is true, i.e., .
It remains to show , i.e., .
Rationalising the right side:
Since , we have , so:
Therefore , which gives . ✓
**Conclusion:** By the principle of mathematical induction, is true for all integers .
Open Math
Applications and harder induction
Proof · ME-12-01
Worked solutions and answers at openmath.au/year-12/extension-1/proof-by-mathematical-induction/applications-and-harder-induction