**Puzzle 14**
Checking some concrete values, \\(2(1) \leq 3, 2(2) \not \leq 3, 2(2) \leq 5, 2(3) \not \leq 5\\). These suggest the function \\(g(b) = \lfloor b/2 \rfloor \\) is our maximum.
More formally, we want \\(g(b) = max\\){\\( a : 2a \leq b, a \in Z \\)}, We need to show it's in our set, and that any other element in our set is smaller.

First, \\(2\lfloor b / 2 \rfloor \leq b \\) so \\(g(b) \in \\){\\( a : 2a \leq b \\)}.
Second, division by 2 and flooring are both monotonic functions, so if a is in our set, we have
$$ 2a \leq b \Rightarrow a \leq b/2 \Rightarrow \lfloor a \rfloor \leq \lfloor b/2 \rfloor \Rightarrow a \leq \lfloor b/2 \rfloor $$
\\(\lfloor b/2 \rfloor\\) is the required maximum.

**Puzzle 15** This argument is analogous, except with \\(\lceil b / 2 \rceil \\). I would type it out, but I don't have time currently (famous last words).

**Puzzle 16** I'm going to give an observation, but my understanding on this isn't complete.

Given the definitions for adjunctions introduced in this lecture, it's clear they are unique (**Edit: This is true for the given example, but isn't true for every preorder, I shouldn't have said this was clear. And because the Galois connection definition is well defined for any preorder, my following suggestion won't generalize to a characterization for preorders by way of uniqueness!**). Which means the definition in puzzle 12 is equivalent to the max definition. We can therefore show we can prove properties from one version to the other. I'll give the direction I've currently figured out.

Suppose \\(g\\) is defined as in problem 14. Because all our functions are monotonic we have
$$f(a) \leq b \Rightarrow g(f(a)) \leq g(b) \Rightarrow a \leq g(b)$$
$$a \leq g(b) \Rightarrow f(a) \leq f(g(b)) \Rightarrow f(a) \leq b$$
Because \\(f(g(b)) \leq b\\) by definition of g. (It's the largest element x such that \\(f(x) \leq b\\)).

It should be possible to show these definitions are equivalent to maximizing in the sense defined in problem 14.