1 / 14

Proving Square Tiling Using Mathematical Induction

This document explores the concept of mathematical induction through a specific statement, denoted as S(n). The statement claims that any n squares can be cut into pieces and reassembled into a single square, applicable for n ≥ 2. The proof begins by verifying S(2), establishes its validity for S(3) through S(5), and demonstrates that S(n) holds true for all integers n ≥ 2. Additionally, the paper discusses the principle of mathematical induction, providing a fundamental basis for proving properties about positive integers, illustrated with sum formulas.

alfonso-lindsay
Télécharger la présentation

Proving Square Tiling Using Mathematical Induction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mathematical Induction

  2. Consider the following statement • Any n squares can be cut into pieces and then pasted into one single square. • Let us denote this statement by S(n). • Of course n  2. • First of all, let us consider S(2):

  3. Conclusion • Hence S(2) is true, that is, any 2 squares can be cut into pieces and then pasted into one single square. • But how about other statements? • So, let’s go on to S(3).

  4. By S(2) By S(2)  S(3) is true.

  5. By S(2) By S(3)  S(4) is true.

  6. By S(2) By S(4)  S(5) is true.

  7. S(3) S(4) S(5) Conclusion • Hence, we can see that S(n) is true for all integers n  2 because: 1. S(2) is true, 2. If S(k) is assumed to be true, then S(k+1) is true. S(2)

  8. Principle of Mathematical Induction • Let S(n) be a statement about a positive integer n  1. Then S(n) is always true if: 1. S(1) is true. 2. If S(k) is assumed to be true, then S(k + 1) is true.

  9. Example To prove: For all n  1, 1 + 2 + 3 + … + n = Proof: When n = 1, L.H.S. = 1 R.H.S. = L.H.S. = R.H.S. when n = 1.

  10. Next, assume that LHS = RHS when n = k, that is, 1 + 2 + 3 + … + k = Then, when n = k + 1, LHS = 1 + 2 + 3 + … + k + (k + 1)

  11. Conclusion Hence, by principle of M.I., LHS = RHS for all n  1.

  12. Worksheet • Show that for any natural number n, • 12 + 22 + 32 + … + n2 =

More Related