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 $P(A) \leq P(B)$
A: Without loss of generality, let $X= |\hat{p} - p|$. 
Assuming $P(X \leq \epsilon) \geq 1 - \delta$, we obtain 
$$
\delta \geq 1-P(X\leq \epsilon)
$$
by adding $\delta$ to both sides and subtracting $P(X\leq \epsilon)$ from both sides. 
Since $P(X\leq \epsilon)+ P(X>\epsilon)=1$, we have 
$1-P(X\leq \epsilon)=P(X>\epsilon)$. 
We thus conclude that 
$$
\delta \geq P(X>\epsilon). 
$$
Finally, 
the inequality $P(X>\epsilon)\leq P(X\geq \epsilon)$ holds because: 
$$
\begin{align*}
P(X\geq \epsilon) &= P(X>\epsilon) + P(X=\epsilon) \\
&\geq P(X>\epsilon).  
\end{align*}
$$ 
