CMSC250

Recall the valid argument

\(\begin{array}{rll} (1) & p \Rightarrow \sim z &\\ (2) & z \land (p \lor r) &\\ (3) & \sim q &\\ (4) & z & \text{ Specialization (2)}\\ (5) & {\sim} p & \text{ Modus Tollens (1,4)}\\ (6) & {\sim} p \lor q & \text{ Generalization (5)}\\ \therefore & p \Rightarrow q & \text{Definition of Implication} \end{array}\)

Three things:

• Intro (Preamble)
• Body
• Conclusion

Three things:

• Intro (Preamble)
Things typically found here:
• Type of Proof
• Assumptions
• What it is you Want To Prove (WTP)
• Body
• Conclusion

Three things:

• Intro (Preamble)
• Body
Things typically found here:
• Use of Laws, Rules, operations to conclude more information
• Steps to obtain the WTP (must logically follow)
• Conclusion

Three things:

• Intro (Preamble)
• Body
• Conclusion
Things typically found here:
• QED, \(\blacksquare\)
• Explaination of your Conclusion

Prove: \((\forall x\in \mathbb{Z})[\text{ if } x \text{ is even, then } 3x+7 \text{ is odd}]\)

Prove: \((\forall x\in \mathbb{R})[\text{ if } x > 5 \text{ then } x^2 -x - 10 > 10]\)

Prove: \(\text{ if } a\vert b \text{ and } b\vert c \text{ then } a\vert c\)

Disprove: All even integers can only be written as a sum of two primes no more than one way.

Recall: Disproving \(\forall\) usually means proving \({\sim}\forall\)

Disprove: If \(ac \equiv bc\) (mod \(m\)) then \(a \equiv b\) (mod \(m\))

Direct Proof by Exhaustion (of a universal statement)

We show all possible cases are true

Note: must be finite possibilities

Prove: \((\forall x \in \{1,2,4,6\})[x^2 + 1\) is prime]

Sometimes proving \(p \Rightarrow q\) is hard

Prove: \((\forall a,b \in \mathbb{Z})\)[if \(a \nmid (2b)\), then \(a \nmid b\)]

Sometimes proving \(p \Rightarrow q\) is hard

Recall: \(p \Rightarrow q \equiv {\sim}q \Rightarrow {\sim}p\)

Proof via contrapositive: Instead of proving \(p \Rightarrow q\), directly prove \({\sim}q \Rightarrow {\sim}p\)

Prove: \((\forall a,b \in \mathbb{Z}]\)[if \(a \nmid (2b)\), then \(a \nmid b\)]

Prove: If the square of an integer is even, then that integer is even

Sometimes statements are not implications

Sometimes proving a statement is true is hard

Prove: There is no greatest integer

Proof via Contradiction: Assume the negation, and then find any contradiction

Note: This is when two facts disagree, or a definition breaks

Prove: \([\forall x \in \mathbb{R})[x \neq x + 1]\)

Prove: There is no greatest integer

Prove: \((\forall x \in \mathbb{Z})\)[if \(x\) is even, then \(x\) is not odd]

Recall \(\sim (p \Rightarrow q) \equiv (p \land {\sim} q)\)

Prove: \((\forall x \in \mathbb{Z}\))[if \(x \equiv 3\) (mod \(6\)), then \(x \not \equiv 2\) (mod \(3\))]