About the maximum likelihood, when we convert the maximization problem into minimization, why we take the negative? On page 12, we take $log$ on both side. 

$\max_{\boldsymbol{w}}L\boldsymbol({w})=\max_{w}\displaystyle\prod_{n=1}^Np(t^{(i)}|x^{(i)};\boldsymbol{w})$

$\ell(\boldsymbol{w})=-logL(\boldsymbol{w})$

$\ \ \ \ \ \ \ =-\displaystyle\sum_{i=1}^Nlog\ p(t^{(i)}|x^{(i)};\boldsymbol{w})$

The $log$ function is increasing as $\boldsymbol{w}$ increase. Why we have to take the negative?
It is common to define optimization problems as minimization problems instead of maximization. And by multiplying your target functions with $-1$ you can transform one into the other:

$$\max_{w} \log{L(w)} \Leftrightarrow \min_{w} -log{L(w)}$$

So to maximize the log-likelihood you minimize the negative log-likelihood. Basically it just comes down to conventions in optimization theory. 

Moreover, since $L(w) \in [0,1]$ its logarithm $\log{L(w)}$ will be less than or equal to $0$ (note that $log{0}$ is not defined). Accordingly $\max_{w} \log{L(w)}$ means to maximize a negative number which is, at least to me, less intuitive than minimizing a positive number.  

The more interesting part is actually the log-transformation which increases numerical stability of your calculations (since it "transforms" the multiplication to a sum and thereby reduces the risk of underflowing).