1. Question: Why is it so difficult for a student to write a coherent argument?

2. I try to address two issues with my students: – Understanding Language – Parsing Logic

3. Rules for Beginners 1.Always phrase the thing you are proving as an “If…then” sentence. When you cannot, be content with one that starts, “There…exists.” 2.Always start a proof with the word “Assume” and then copy the hypothesis. Start with a new “Assume” every time you hit an “and.” 3.To prove, “There exists x such that…” Write down “x = “ even if you cannot finish the thought.

4. Facts about Beginners They will never do any of this.

5. The Concept of the “Minitheorem.” After every assumption you make, you have reduced your objective to a minitheorem. Ask yourself, “What do I need to prove now?” Revert back to the rules for beginners.

6. Example: Prove that if x and y are real numbers with x<y, then for all natural numbers n, there are real numbers z 1, z 2, z 3,… z n so that for all j=1,2,…n x < z 1 < z 2 < z 3 <… < z n <y

7. Assume x and y are real numbers. Assume x<y. Minitheorem: For all natural numbers n, there are real numbers z 1, z 2, z 3,… z n so that for all j=1,2,…n x < z 1 < z 2 < z 3 <… < z n <y

8. Assume x and y are real numbers. Assume x<y. Proof by Induction: Minitheorem: There are real numbers z 1, z 2, z 3,… z n so that for all j=1,2,…n x < z 1 < z 2 < z 3 <… < z n <y

9. Assume x and y are real numbers. Assume x<y. Proof by Induction: Base Step: For n=1, there are real numbers z 1, z 2, z 3,… z n, so that x < z 1 < z 2 < z 3 <… < z n <y. Inductive Step: If for n, there are real numbers z 1, z 2, z 3,… z n so that x < z 1 < z 2 < z 3 <… < z n <y, then for n+1, there are real numbers z 1, z 2, z 3,… z n+1 so that x < z 1 < z 2 < z 3 <… < z n+1 <y.

10. Assume x and y are real numbers. Assume x<y. Proof by Induction: Base Step: For n=1, there is a real numbers z 1 so that x < z 1 <y. Proof: Let z 1 =……..

11. Assume x and y are real numbers. Assume x<y. Inductive Step: If for n, there are real numbers z 1, z 2, z 3,… z n so that x < z 1 < z 2 < z 3 <… < z n <y, then for n+1, there are real numbers z 1, z 2, z 3,… z n+1 so that x < z 1 < z 2 < z 3 <… < z n+1 <y.

12. Assume x and y are real numbers. Assume x<y. Proof by Induction: Base Step Inductive step: Minitheorem: If there are real numbers z 1, z 2, z 3,… z n so that x < z 1 < z 2 < z 3 <… < z n <y, then for there are real numbers z 1, z 2, z 3,… z n+1 so that x < z 1 < z 2 < z 3 <… < z n+1 <y. Proof: Assume there are real numbers z 1, z 2, z 3,… z n so that x < z 1 < z 2 < z 3 <… < z n <y. Let z n+1 =