New in-class teams starting Monday. Enjoy your last day with your current team!
A2 is out, due Wednesday; on material through today.
Commutative: \[ \begin{align*} P \land Q \equiv Q \land P\\ P \lor Q \equiv Q \lor P\\ \end{align*} \]
Distributive: \[ \begin{align*} P \land (Q \lor R) \equiv (P \land Q) \lor (P \land R)\\ P \lor (Q \land R) \equiv (P \lor Q) \land (P \lor R)\\ \end{align*} \]
Associative: \[ \begin{align*} P \land (Q \land R) \equiv (P \land Q) \land R\\ P \lor (Q \lor R) \equiv (P \lor Q) \lor R\\ \end{align*} \]
Contrapositive: \[ P \Rightarrow Q \equiv (\neg Q) \Rightarrow (\neg P) \\ \]
DeMorgan’s Laws: \[ \begin{align*} \neg(P \land Q) \equiv (\neg P) \lor (\neg Q)\\ \neg(P \lor Q) \equiv (\neg P) \land (\neg Q) \end{align*} \]
Negating Quantifiers: \[ \begin{align*} \neg (\forall x \in X, P(x)) \equiv \exists x \in X, \neg P(x)\\ \neg (\exists x \in X, P(x)) \equiv \forall x \in X, \neg P(x) \end{align*} \] Negating conditionals: \[ \neg(P \Rightarrow Q) \equiv P \land \neg Q \]
Do Exercises Parts A-C
Proof is the process of demonstrating that a statement (often, a theorem or a proposition) is true. Generally, we start with things we already know to be true - either definitions or theorems that have been proved already. Then we show that given those things we know are true, the thing we wish to prove must also be true.
Many things we wish to prove are phrased as conditional statements, so we’re starting with techniques for proving those. For example, here’s a
Proposition: If \(x\) is odd, then \(x^2\) is odd.
Definition: An integer \(n\) is even if \(n = 2a\) for some integer \(a \in \mathbb{Z}\).
Observation: by (confusing) convention, definitions are usually stated as a conditional but meant to imply a biconditional. That is, we interpret the above definition to mean: \[ n \text{ is even} \Leftrightarrow \exists a \in Z, n = 2a \] What is the distinction? We are saying that \(n\) is not even if it does not satisfy the given property, where the unidirectional implication would be inconclusive about that case.
Definition: An integer \(n\) is odd if \(n = 2a + 1\) for some integer \(a \in \mathbb{Z}\).
Definition: Suppose \(a\) and \(b\) are integers. We say that \(a\) divides \(b\), written \(a | b\), if \(b = ac\) for some \(c \in \mathbb{Z}\). In this case we also say that \(a\) is a divisor of \(b\), and that \(b\) is a multiple of \(a\).
Definition: Two integers have the same parity if they are both even or they are both odd. Otherwise they have opposite parity.
Example:
Proposition: If \(x\) is odd, then \(x^2\) is odd.
Proof: Suppose \(x\) is odd.
\(\vdots\)
Therefore \(x^2\) is odd.
Our job is now to figure out what can replace \(\vdots\) such that “\(x^2\) is odd” logically follows from what comes before it. Here’s one natural way to do it:
Proposition: If \(x\) is odd, then \(x^2\) is odd.
Proof: Suppose \(x\) is odd.
Then \(x = 2a + 1\) for some \(a \in \mathbb{Z}\), by definition of an odd number.
Thus, \[ \begin{align*} x^2 &= (2a + 1)^2\\ &= 4a^2 + 4a + 1\\ &= 2(2a^2 + 2a) + 1 \end{align*} \] Therefore, \(x^2 = 2b + 1\), for \(b = 2a^2 + 2a\).
Thus \(x^2 = 2b + 1\) for some integer \(b\).
Therefore \(x^2\) is odd. \(\blacksquare\)
Note: \(\blacksquare\) is a common way to indicate that a proof is complete. It is read “QED”, standing for “quod erat demonstrandum”, which is latin for “which was to be demonstrated”.