A3 is due tonight
Themes from Week 3 Survey:
+: exercises, group work, getting the hang of LaTeX(!)
Labs are challenging. Some general tips:
Proofs are challenging
In-class exercises are helpful!
The Fibonacci sequence is defined as follows: \[ \begin{align*} F_0 &= 1\\ F_1 &= 1\\ F_n &= F_{n-1} + F_{n-2} \end{align*} \]
The standard inductive step is to show \(S_k \Rightarrow S_{k+1}\). But if we need to, we can prove $S_{k+1} using the fact that not only has the prior domino fallen, but all dominoes before it have fallen as well. In other words, a valid induction step is also to show that: \[ (S_0 \land S_1 \land S_2 \land \ldots \land S_k) \Rightarrow S_{k+1} \] This is called strong induction.
Example: Chocolate bars! See handwritten notes.
Proposition: All horses are the same color.
Proof: by induction.
Base case: Suppose we have a group of \(n = 1\) horses. Since there is only one, all horses in this group are the same color.
Inductive step: Suppose we have a group of \(k\) horses. We wish to show that a group of \(k+1\) horses are all the same color.
Consider the group of \(k+1\) horses, and select one horse to exclude. The remaining \(k\) horses are all the same color by the inductive hypothesis.
Now, choose a different horse to exclude. The remaining \(k\) horses, which includes the first one we chose to exclude above, are all the same color. Hence, all \(n+1\) horses are the same color.
What’s wrong with this proof?
This is a combination of proof by induction and proof by contradiction. The basic idea is to show the base case holds, then prove the inductive hypothesis by contradiction. You can do this by supposing that it does not always hold that \(S_{k-1} \Rightarrow S_{k}\). We let \(k\) be the smallest example for which this is the case, and thus suppose \(S_{k-1}\) is true but \(S_k\) is not; then show that this leads to a contradiction.