🚧 Continuing through 1.4

chapter_1/1_4/investigate/investigate.md

@@ -0,0 +1,32 @@
+# Investigate!
+
+A **mini sudoku puzzle** is a 4 x 4 grid of squares, divided into four 2 x 2
+boxes. The goal is to fill each square with a digit from 1 to 4, such that no
+digit repeats in any row, any column, or any box.
+
+Here is a simple mini sudoku puzzle you can try to solve.
+
+![image 1_4_1_investigate_1](./1_4_1_investigate_1.png)
+
+You might notice that the solution to the above puzzle has its four outside
+corners all different, and its four middle squares all different.
+
+The goal of this _Investigate!_ question is to prove that this is not a
+coincidence: Suppose a mini sudoku puzzle has all different numbers in its four
+corners (marked with # below). Prove that the center four squares (marked with *
+below) must also contain different numbers.
+
+![image 1_4_1_investigate_2](./1_4_1_investigate_2.png)
+
+## Try it. 1.4.1
+
+Try placing numbers into an empty mini sudoku puzzle. See if you can break the
+statement we were asked to prove in the _Investigate!_ activity. What stops you?
+Briefly explain whether you think the statement is true or false, and why.
+
+A:
+
+I think the statement is true. When I tried to make two of the middle squares
+the same number, the sudoku rules forced a contradiction because the numbers
+would repeat in a row, column, or box. The different corners seem to force the
+middle squares to all be different too.

chapter_1/1_4/investigate/preview_activity.md

@@ -0,0 +1,96 @@
+# Preview Activity
+
+Consider the statement:
+
+If $ab$ is an even number, then $a$ or $b$ is even.
+
+Which of the proofs below appear to be valid proofs of this statement? Note: You
+can assume all the algebra below is correct (because it is).
+
+1.
+
+Suppose $a$ and $b$ are odd. That is, $a = 2k + 1$ and $b = 2m + 1$ for some
+integers $k$ and $m$. Then
+
+$$
+
+ab = (2k + 1)(2m + 1) \\
+
+\quad = 4km + 2k + 2m + 1 \\
+
+\quad = 2(2km + k + m) + 1.
+
+$$
+
+Therefore $ab$ is odd.
+
+2.
+
+Assume that $a$ or $b$ is even -- say it is $a$ (the case where $b$ is even will
+be identical). That is, $a = 2k$ for some integer $k$. Then
+
+$$
+
+ab = (2k)b \\
+
+\quad = 2(kb).
+
+$$
+
+Thus $ab$ is even.
+
+3.
+
+Suppose that $ab$ is even but $a$ and $b$ are both odd. Namely,
+$ab = 2n, a = 2k + 1$ and $b = 2j + 1$ for some integers $n$, $k$, and $j$. Then
+
+$$
+
+2n = (2k + 1)(2j + 1) \\
+
+2n = 4kj + 2k + 2j + 1 \\
+
+n = 2kj + k + j + \frac{1}{2}.
+
+$$
+
+But since $2kj + k + j$ is an integer, this says that the integer $n$ is equal
+to a non-integer, which is impossible.
+
+4.
+
+Let $ab$ be an even number, say $ab = 2n$, and $a$ be an odd number, say
+$a = 2k + 1$.
+
+$$
+
+ab = (2k + 1)b \\
+
+2n = 2kb + b \\
+
+2n - 2kb = b
+
+$$
+
+Therefore $b$ must be even.
+
+A:
+
+1. This one doesn't appear to be a valid proof at first glance to me. Although
+   the logic appears correct, we are not solving for if $ab$ is odd. If we
+   somehow proved that $ab$ was _not_ odd, then maybe this would prove that $ab$
+   is even, but the initial statement "If $ab$ is an even number, then $a$ or
+   $b$ is even.$ is contradicted by the following statement "Suppose $a$ and $b$
+   are odd."
+
+2. This proof appears to be valid for the given statement. Since $a = 2k$ for
+   all natural numbers, this is guaranteed to be even and multiplying it by $b$
+   does not change that this will result in an even number.
+
+3. I don't see the logic in this one. This does not appear to be a valid proof
+   of the statement, but I cannot say as to why.
+
+4. This proof appears to be valid for the given statement, but I am struggling
+   to say as to why.
+
+Note: Not changing my answers here, but I mostly got these all wrong.