Proof by Induction (DP IB Analysis & Approaches (AA)): Revision Note

Dan Finlay

Written by: Dan Finlay

Reviewed by: Mark Curtis

Updated on

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Proof by Induction

What is proof by induction?

  • Proof by induction is a way of proving a result is true for a set of integers

    • by showing that if it is true for one integer

      • then it is true for the next integer

  • It can be thought of as falling dominoes

    • Assuming some domino falls

      • the assumption step

    • you need to show it causes (induces) the next domino to fall

      • the induction step

    • If you can also show that the first domino falls

      • the basic step

    • then putting together the above mean that all dominoes will fall!

      • the conclusion step

What are the steps for proof by induction with series?

  • For example: prove that sum from r equals 1 to n of r squared equals 1 over 6 n left parenthesis n plus 1 right parenthesis left parenthesis 2 n plus 1 right parenthesis for all n element of straight integer numbers to the power of plus

    • It has a left-hand side, L H S equals sum from r equals 1 to n of r squared

    • and a right-hand side, R H S equals 1 over 6 n left parenthesis n plus 1 right parenthesis left parenthesis 2 n plus 1 right parenthesis

  • STEP 1

    The basic step: Show result is true for bold italic n bold equals bold 1

    • Substitute n equals 1 into both sides individually and show they are equal

      L H S equals space sum from r equals 1 to 1 of r squared equals 1 squared equals 1
R H S equals 1 over 6 cross times 1 cross times left parenthesis 1 plus 1 right parenthesis left parenthesis 2 cross times 1 plus 1 right parenthesis equals 1 over 6 cross times 2 cross times 3 equals 1

  • STEP 2

    The assumption step: Assume the LHS and RHS are equal for some bold italic n bold equals bold italic k (where k is some integer)

    • Replace n with k in the statement

    • Write: "Assume sum from r equals 1 to k of r squared equals 1 over 6 k left parenthesis k plus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis is true"

  • STEP 3

    The inductive step: You need to prove that the LHS and RHS are equal for n equals k plus 1

    • Start with the LHS for n equals k plus 1

      • It helps to take the last term out: sum from r equals 1 to k plus 1 of r to the power of 2 space end exponent equals sum from r equals 1 to k of r squared space plus open parentheses k plus 1 close parentheses squared

      • Now substitute in the assumption (from step 2) :

      • sum from r equals 1 to k plus 1 of r to the power of 2 space end exponent equals 1 over 6 k left parenthesis k plus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis space plus open parentheses k plus 1 close parentheses squared

      • Expand and factorise:
        1 over 6 open parentheses k plus 1 close parentheses left square bracket k left parenthesis 2 k plus 1 right parenthesis plus 6 open parentheses k plus 1 close parentheses right square bracket
equals 1 over 6 open parentheses k plus 1 close parentheses left square bracket 2 k squared plus 7 k plus 6 right square bracket
equals 1 over 6 open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses 2 k plus 3 close parentheses

    • Check if it is the same as the RHS for n equals k plus 1

      table row cell R H S end cell equals cell 1 over 6 open parentheses k plus 1 close parentheses open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses open parentheses 2 open parentheses k plus 1 close parentheses plus 1 close parentheses end cell row blank equals cell 1 over 6 open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses 2 k plus 3 close parentheses end cell end table

    • So LHS = RHS for n equals k plus 1 (as long as the assumption is true)

  • STEP 4

    The conclusion step: Explain in words how the above steps make the result true for all positive integers, n element of straight integer numbers to the power of plus, using the following two sentences:

    • "If it is true for n equals k, then it is true for n equals k plus 1."

    • "As it is true for bold italic n bold equals bold 1, the statement is true for all n element of straight integer numbers to the power of plus."

What are the steps for proof by induction with divisibility?

  • For example: prove that f left parenthesis n right parenthesis equals 4 to the power of n minus 1 is divisible by 3 for all positive integers n

    • Being divisible by 3 is the same as being a multiple of 3

  • STEP 1

    The basic step: Show result is true for bold italic n bold equals bold 1

    • Substitute n equals 1 into the function

    • f open parentheses 1 close parentheses equals 4 to the power of 1 minus 1 equals 3which is divisible by 3

  • STEP 2

    The assumption step: Assume f left parenthesis k right parenthesis equals 4 to the power of k minus 1 is divisible by 3 for some bold italic n bold equals bold italic k where k is some integer

    • Replace n with k in the statement

    • Write "divisible by 3" as "equals 3 m" where m is a positive integer

    • Sof open parentheses k close parentheses equals 4 to the power of k minus 1 equals 3 m where m element of straight integer numbers to the power of plus 

    • It helps to make 4 to the power of k the subject: 4 to the power of k equals 1 plus 3 m 

  • STEP 3

    The inductive step (method 1): You need to show that the n equals k plus 1 case is divisible by 3

    • Substitute n equals k plus 1 into the function

      • f left parenthesis k plus 1 right parenthesis equals 4 to the power of k plus 1 end exponent minus 1

    • Use index laws to substitute in the assumption (step 2) 4 to the power of k equals 1 plus 3 m 

      • f left parenthesis k plus 1 right parenthesis equals 4 open parentheses 4 to the power of k close parentheses minus 1 equals 4 open parentheses 1 plus 3 m close parentheses minus 1 equals 3 open parentheses 4 m plus 1 close parentheseswhich is a multiple of 3

    • So f left parenthesis k plus 1 right parenthesisis divisible by 3 (as long as the assumption is true)

  • STEP 3

    The inductive step (method 2): An alternative method is to show that the difference f left parenthesis k plus 1 right parenthesis minus f left parenthesis k right parenthesis is a multiple of 3

    • This is because f left parenthesis k plus 1 right parenthesis minus f left parenthesis k right parenthesis equals 3 cross times... rearranges to f left parenthesis k plus 1 right parenthesis equals f open parentheses k close parentheses plus 3 cross times...

    • So f left parenthesis k plus 1 right parenthesis is divisible by 3 (as long as the assumption is true, i.e. that f left parenthesis k right parenthesis is divisible by 3)

    • This method is not always that practical

      • When it is, a hint will be given

  • STEP 4

    The conclusion step: Explain in words how the above steps make the result true for all integers, using the following two sentences:

    • "If it is true for n equals k, then it is true for n equals k plus 1."

    • "As it is true for bold italic n bold equals bold 1, the statement is true for all n element of straight integer numbers to the power of plus."

What types of statements might I be asked to prove by induction?

  • You may be asked to use proof by induction in the following contexts

    • Series (see above)

      • Some useful tricks are

        • writing sum from r equals 1 to k plus 1 of f open parentheses r close parentheses equals f open parentheses k plus 1 close parentheses plus sum from r equals 1 to k of f open parentheses r close parentheses for the inductive step

        • writing open parentheses k plus 1 close parentheses factorial equals open parentheses k plus 1 close parentheses cross times open parentheses k factorial close parentheses if factorials are involved

    • Divisibility (see above)

      • A useful trick for the inductive step is using a to the power of k plus 1 end exponent equals a cross times a to the power of k

    • Complex numbers

      • You can prove de Moivre’s theorem by induction

    • Derivatives

      • These may involve the chain, product and quotient rule

      • A useful trick for the inductive step is using f to the power of open parentheses k plus 1 close parentheses end exponent open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction open parentheses f to the power of open parentheses k close parentheses end exponent open parentheses x close parentheses close parentheses

Examiner Tips and Tricks

Learn all the steps for proof by induction and ensure your conclusion makes mathematical sense!

Worked Example

Prove by induction that sum from r equals 1 to n of r open parentheses r minus 3 close parentheses equals 1 third n open parentheses n minus 4 close parentheses open parentheses n plus 1 close parentheses for n element of straight integer numbers to the power of plus.

1-5-1-ib-aa-hl-proof-by-induction-we-solution

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Dan Finlay

Author: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.