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CSE 20 – Discrete Mathematics

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CSE 20 – Discrete Mathematics

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  1. Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.Based on a work at http://peerinstruction4cs.org.Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org. CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett

  2. Today’s Topics: • More review of Proof by Contradiction • Irrational numbers

  3. 1. Proof by Contradiction Following up on a very common source of errors on the previous midterm.

  4. Proof by Contradiction Steps • What are they? • 1. Assume what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of what you are proving (a contradiction). • 1. Assume the opposite of what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of your assumption (a contradiction). • 1. Assume the opposite of what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of some fact you already showed (a contradiction). • Other/none/more than one.

  5. Thm: 6 - 7 for some integer x for some integers a,b, b≠0 for some integer x for some integers a,b, b≠0 Other/none/more than one Proof (by contradiction): Assume not, that is, assume is rational. Then:

  6. Thm: 6 - 7 Proof (by contradiction): Assume not, that is, assume 6 - 7 is rational. Then 6 - 7 for some integers a,b, b≠0. //now isolate radical //isolated, but we can’t say much about sqrt(8) = = Because integers are closed under subtraction and multiplication, 6b-a is an integer, and 14b is an integer. Moreover, 14b≠0 because b≠0. So the left-hand side is rational. But sqrt(2) is irrational. A contradiction!

  7. Thm: 6 - 7 Proof (by contradiction): Assume not, that is, assume 6 - 7 is rational. Then 6 - 7 for some integers a,b, b≠0. //now isolate radical //isolated, but we can’t say much about sqrt(8) = = Because integers are closed under subtraction and multiplication, 6b-a is an integer, and 14b is an integer. Moreover, 14b≠0 because b≠0. So the left-hand side is rational. But sqrt(2) is irrational. A contradiction! Is there a shorter way that reaches a contradiction sooner?

  8. Thm: 6 - 7 Proof (by contradiction): Assume not, that is, assume is rational. Then 6 - 7 for some integers a,b, b≠0. Because integers are closed under subtraction and multiplication, would be an integer (and therefore a rational) if were also an integer. But is irrational, a contradiction! Is this proof valid? YES NO

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