New teams!
Week 2 Survey - complete by tonight
A2 due Wednesday
A3 (tentatively) out Wednesday, due a week later
Take two minutes to think about group work experiences you’ve had in the past (in this class or otherwise). Consider:
Take five minutes with your team to:
Write your norms on the Team Norms sheet.
I’ll collect these after class and redistribute them with the Exercise/Problem sheets each day.
We’ll occasionally revisit norms and make any adjustments deemed necessary.
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”.
Sometimes it’s easiest to break a proof down into multiple cases. For example:
Proposition: If \(x\) is an integer, then \(x^2\) has the same parity as \(x\).
Proof:
Even case: Suppose \(x\) is an even integer. Then \(x = 2a\) for some integer \(a\).
\(x^2 = 4a^2 = 2(2a^2)\), so \(x^2\) can be written as \(x^2 = 2b\) where \(b = 2a^2\). Thus \(x^2\) is even.
Odd case: Suppose \(x\) is an odd integer. Then \(x = 2a + 1\) for some integer \(a\).
\(x^2 = 2a^2 + 4a + 1 = 2(a^2 +2a) + 1\), so \(x^2\) can be written as \(x^2 = 2b + 1\) where \(b = a^2 + 2a\). Thus \(x^2\) is odd.
We have shown that even numbers have even squres, and odd numbers have odd squares; this completes the proof that \(x^2\) has the same parity as \(x\). \(\blacksquare\)
If two cases are symmetric or similar enough that proving one generalizes to the other case(s), you can prove one and use “without loss of generality” to indicate that the proof is different in only trivial ways from the case you’ve handled.
For example:
Proposition: If two integers have opposite parity, then their sum is odd.
Let \(x\) and \(y\) be two integers with opposite parity, and suppose without loss of generality that \(a\) is odd and \(b\) is even.
(…)
We can then proceed to prove this case, knowing that the case where \(a\) is even and \(b\) is odd will be identical except with \(a\) and \(b\) swapped in all steps of the proof.