Question 1102473
N) {{{highlight(sometimes)}}} true.
{{{N}}} = the set of natural numbers, or counting numbers, {1, 2, 2, ...}
For {{{q=4}}} , there is {{{p=2}}} such that {{{2^2=4}}} ,
but for {{{q=2}}} , if {{{p^2=2}}} , then {{{p =" " +- sqrt(2)}}} does not belong to set N.
 
Z) {{{highlight(sometimes)}}} true.
{{{Z}}}= the set of integers
For {{{q=4}}} , there is {{{p=2}}} such that {{{2^2=4}}} ,
but for {{{q=2}}} , if {{{p^2=2}}} , then {{{p =" " +- sqrt(2)}}} does not belong to set Z,
and for integer {{{q=-4}}} there is neither integer, nor rational, nor real number {{{p}}} such that {{{p^2=-4}}} .
 
Q) {{{highlight(sometimes)}}} true.
{{{Q}}}= the set of rational numbers.
All the examples above still work the same way,
so it is still sometimes true, and sometimes not true.
No need to look for more examples.
 
R) {{{highlight(sometimes)}}} true.
{{{R}}}= the set of real numbers.
For any non-negative real number {{{q>=0}}} ,
there are always two real numbers {{{p=" "+- sqrt(q)}}} , such that {{{p^2=q}}} . 
However, for negative numbers, that is not true.
For {{{q=-1}}} , there is no real number {{{p}}} such that {{{p^2=-1}}} .
 
C) {{{highlight(always)}}} true.
{{{C}}}= the set of complex numbers.
For every complex number {{{q}}} there are always two complex numbers {{{p}}} and {{{-p}}} , such that {{{p^2=(-p)^2=q}}} .
Any complex number can be written as {{{q=r(cos(theta)+i*sin(theta))}}} ,
with {{{r}}} being a non-negative real number, {{{r>=0}}} ,
and {{{theta}}} being an angle with {{{0<=theta<2pi}}} ,
For complex number {{{p=sqrt(r)(cos(theta/2)+i*sin(theta/2))}}} , {{{p^2=q}}} .
For {{{-p=sqrt(r)(cos(2pi-theta/2)+i*sin(2pi-theta/2))}}} , {{{p^2=q}}} too.
and {{{p=