Proof by Induction

Proof by Induction is used when you have a proposition that is expressed for a infinite set of numbers and it cannot be easily solved using a form of direct proof. An example would be the statement 7 divides (32n-2n) for all integers n≥1. This cannot be easily proven since it states for an infinite number of integers; contradiction may work but for that your saying there exists an integer that doesn't satisfy the proposition which would be a bit tricky to prove.

The are 5 simple steps that can be taken to make an induction proof:

  1. Propositional Function: Express the proposition as a function 𝚸(n) which depends on n.
  2. Base Case: Prove 𝚸(1) is true. (Note the 𝚸(1) just means the start of n which could equate to say 𝚸(53))
  3. Induction Hypothesis: Assume there exists an n for which 𝚸(n) is true, and fix such an n. This can be done because in the previous step it was shown that there is such an n.
  4. Induction Step: Show how the truth of 𝚸(n) implies that 𝚸(n+1) is also true.
  5. Conclusion: Conclude by mathematical induction that 𝚸(n) is true for all n.

Dr. Scott Beaver

As one can see by showing 𝚸(1) is true and showing that if 𝚸(n) is true then 𝚸(n+1) is true, the proof chains down the line to infinity. If 𝚸(1) is true then 𝚸(2) is true which then 𝚸(3) is true and so on.

Example:

7 divides (32n-2n) for all integers n≥1.

Proof:

Whenever induction might be needed its always good to test it for the first few numbers and maybe a few random ones just in case it fails.

n (32n-2n) (32n-2n) / 7
1 7 1
2 77 11
3 721 103
5 59017 8431
7 4782841 683263
9 387419977 55345711

As we expected it worked for a couple numbers so we proceed by induction.

  1. Propositional Function: 𝚸(n)="7 divides 32n-2n"
  2. Base Case: 𝚸(1)="7 divides 32*1-21=7" which 7 divides 7 thus 𝚸(1) is true
  3. Induction Hypothesis: Let n be greater then 0 and satisfy "7 divides 32n-2n"
  4. Induction Step: Consider
    • 32(n+1)-2n+1 =
    • 32n+2-2n+1 =
    • 9*32n-2*2n =
    • (7+2)*32n-2*2n =
    • 7*32n+2*(32n-2n)
    Our induction hypothesis states 7 divides 32n-2n and thus 7 divides 2*(32n-2n). We know that 7 divides 7*32n since 7 divides 7 thus 7 divides 7*32n+2*(32n-2n). Therefore, if 𝚸(n) is true, 𝚸(n+1) is also true.
  5. Conclusion: Thus, by mathematical induction, for all integers n≥1, 7 divides 32n-2n.