Q:

Rules for taking complements of probabilities

I'm seeing some examples in my book where they take complements of probabilities and say some probabilities are less/greater than or equal to others. I just wanted to clarify that I am understanding them correctly.
Consider the example
$P[|\hat{p} - p| \leq \epsilon] \geq 1 - \delta$
The book immediately "subtracts both sides from 1". My first question is, what's the purpose of this form in the first place, if all subsequent steps require the below form?
$1 - P[|\hat{p} - p| \leq \epsilon] \leq 1 - (1 - \delta)$
$P[|\hat{p} - p|  \gt \epsilon] \leq \delta$
I kind of have an understanding as to what is happening here. They just take the "opposite" of both sides, but what causes the sign change? I thought the only way the sign changes is if you divide both sides by a negative number.
They also claim that
$P(|\hat{p} - p| \gt \epsilon) \leq P(|\hat{p} - p| \geq \epsilon)$
I don't know how to start explaining the above statement is true.

A:

For your first question:
$$P[|\hat{p} - p| \leq \epsilon] \geq 1 - \delta$$
Multiply by $-1$:
$$-P[|\hat{p} - p| \leq \epsilon] \leq -(1 - \delta)$$
Add $1$:
$$1-P[|\hat{p} - p| \leq \epsilon] \leq 1-(1 - \delta)$$
which is equivalent to 
$$1-P[|\hat{p} - p| \leq \epsilon] \leq \delta$$
Hence 
$$P[|\hat{p} - p| > \epsilon] \leq \delta$$
For your second question:
It is due to 
$|\hat{p}-p| > \epsilon \implies |\hat{p}-p| \geq \epsilon$.
In general, if $A \implies B$, then $P(A) \leq P(B)$

