Solve each of the following exercises.
a) Let us show that for every natural number $n$, $\Sigma_{i=0}^n 2^i = 2^{n+1} -1$. For $n=0$, $\Sigma_{i=0}^0 2^i = 2^0 = 1 = 2^1-1$. Now suppose that $\Sigma_{i=0}^k 2^i = 2^{k+1} -1$. Then $\Sigma_{i=0}^{k+1} 2^i = (\Sigma_{i=0}^k 2^i) + 2^{k+1} = (2^{k+1} -1) + 2^{k+1} = 2^{k+2}-1$. Therefore, $\forall n \in \mathbb{N}$ $\Sigma_{i=0}^n 2^i = 2^{n+1} -1$.
b) DeMorgan's law can be generalized to a union (similarly, intersection) of any $n$ sets from some common universe: $\overline{A_1 \cup \dots \cup A_n} = \overline{A_1} \cap \dots \cap \overline{A_n}.$Indeed, the cases of $n=0$ and $n=1$ are trivial, and for $n=2$ it is our standard DeMorgan's law, $\overline{A_1 \cup A_2} = \overline{A_1} \cap \overline{A_2}$.
Let $\overline{A_1 \cup \dots \cup A_i} = \overline{A_1} \cap \dots \cap \overline{A_i}$ for every $i$ such that $2 \leq i < k$, and any sets $A_1, \dots, A_i$.
As $k>2$, for some $j$ such that $1 \leq j < k$, $\overline{A_1 \cup \dots \cup A_k} = \overline{A_1 \cup \dots \cup A_j \cup A_{j+1} \cup \dots \cup A_k}.$
Then $$\overline{A_1 \cup \dots \cup A_k}= \overline{A_1 \cup \dots \cup A_j \cup A_{j+1} \cup \dots \cup A_k} = $$ $$= \overline{(A_1 \cup \dots \cup A_j) \cup (A_{j+1} \cup \dots \cup A_k)} = $$ $$= \overline{(A_1 \cup \dots \cup A_j)} \cap \overline{(A_{j+1} \cup \dots \cup A_k)} =$$ $$= (\overline{A_1} \cap \dots \cap \overline{A_j}) \cap (\overline{A_{j+1}} \cap \dots \cap \overline{A_k})=$$ $$ =\overline{A_1} \cap \dots \cap \overline{A_j} \cap \overline{A_{j+1}} \cap \dots \cap \overline{A_k} =$$ $$= \overline{A_1} \cap \dots \cap \overline{A_k}.$$ Thus generalized DeMorgan's law holds for any $n$ and any sets $A_1, \dots, A_n$.
For this part of the lab, you will be working in groups of three. In the first part of the exercise, each of three people in the group picks one of the topics below, and solves the corresponding questions. Then, each of you will explain to two of your peers how you have solved these questions. Your explanation should be good enough that they can then solve similar questions on their own.
Imagine that you have a bar of chocolate which is a $n \times m$ grid of pieces ("squares"). To share it with your friends, you would like to break it up into individual squares, without ever breaking an individual square. How many steps would it take for you to do so, if at any step you can break a piece of chocolate in half horizontally or vertically? Does it depend on the order in which you break it, and if so, what is the best order?
Imagine $n$ people standing in a field, each holding a pie. At a signal, everybody throws a pie into their nearest neighbour (assume that distances between each pair of people are all different, so in particular everybody has one nearest neighbour).