Math 99 (Notes)

Summary of What to Prove:

 To Prove: Do: P\implies Q "Assume P is true," prove Q is true, or "Assume Q is false," prove P is false, or "Assume P is true and Q is false", produce a contradiction. P\iff Q Prove (P\implies Q)\And(Q\implies P), or prove (P\implies Q)\And(\Not P\implies\Not Q), or prove (\Not Q\implies\Not P)\And(Q\implies P), or prove (\Not Q\implies\Not P)\And(\Not P\implies\Not Q) (\forall x)(P(x)) "Let x be an arbitrary \ldots" Prove P(x). (\exists x)(P(x)) "Take x = \ldots" Prove P(x) for this x. A \subseteq B Prove (\forall x\in A)(x\in B) i.e., if x\in A then x\in B. A = B Prove (A \subseteq B) \And (B \subseteq A). A = \emptyset Prove (\forall x)(x\notin A) (frequently best to use proof by contradiction). x\in A \cup B Prove (x\in A) \Or (x\in B). x\in A \cap B Prove (x\in A) \And (x\in B). x\in A - B Prove (x\in A) \And (x\notin B). \Not(P(x) \implies Q(x)) Prove (\exists x)(P(x)\And\Not Q(x)). \Not (P(x) \iff Q(x)) Prove (\exists x)(P(x)\And\Not Q(x))\Or(\exists x)(Q(x)\And\Not P(x)). \Not(\exists x)(P(x)) Prove (\forall x)(\Not P(x)). \Not(\forall x)(P(x)) Prove (\exists x)(\Not P(x)). A\not\subseteq B Prove (\exists x)(x\in A \And x\notin B). A \ne B Prove (A\not\subseteq B) \Or (B\not\subseteq A). ie, there is an x\in A where x\notin B or      there is an x\in B where x\notin A. A \ne \emptyset Prove (\exists x)(x\in A). x\notin A \cup B Prove (x\notin A) \And (x\notin B). x\notin A \cap B Prove (x\notin A) \Or (x\notin B). x\notin A - B Prove (x\notin A) \Or (x\in B).

 To prove P by contradiction: "Assume P is false" then show that you arrive at a contradition. To prove (\forall n\in\N)(P(n)) by Mathematical Induction, show the following: P(1) is true For all k\in\N, if P(k) is true then P(k+1) is true.

 Math 99 (Winter 2005) web pages Created: 01 Feb 2005 Last modified: Feb 1, 2005 8:38:22 PM Comments to: dpvc@union.edu