SG. 4 Given that $x$ and $y$ are real numbers and $\left|x-\frac{1}{2}\right|+\sqrt{y^{2}-1}=0$.
Let $p=|x|+|y|$, find the value of $p$.
Reference: 2006 FI4.2 $\ldots y^{2}+4 y+4+\sqrt{x+y+k}=0$. If $r=|x y|, \ldots$
Both $\left|x-\frac{1}{2}\right|$ and $\sqrt{y^{2}-1}$ are non-negative numbers.
The sum of two non-negative numbers $=0$ means each of them is zero
$$
x=\frac{1}{2}, y= \pm 1 ; p=\frac{1}{2}+1=\frac{3}{2}
$$