How can I prove every positive real number has square root? Does it ask that we can express any positive real number as square root of something? like 4 is equal to square root of 16?
 A: Hint: Consider the set $\{x\in \Bbb{R}_{\geq 0} \colon x^2<y\}$. By the completeness axiom, there is a least upper bound of this set, call it $x_0$. 
If $x_0^2<y$, then by density of rationals, there is some rational number $q$ in $(x_0^2,y)$. Does this make sense? 
If $x_0^2>y$. What happens?
So there's only one choice left.
A: On the opposite, it is asking if any positive number is the square of something i.e. if
$$x=\left(\sqrt x\right)^2$$ is always possible.
A: $f(x)=x^2$ from $[0,\infty)\to [0,\infty)$ is a strictly increasing continuous function,since $f(0)=0$ and $\lim_{x\to\infty}f(x)=\infty$ the function is bijective,hence it's inverse is also bijective since $\sqrt{x}$ is the inverse that implies it's also surjective.This is true if the definition of square root is such that it is the inverse of the square function.
A: Your title says that the problem is:


*

*(a) Prove that every positive number has a square root, that is, for any $y > 0$, prove that there exists $x$ such that $x^2 = y$.


Your text says:


*

*(b) Prove that every positive number is the square root of another number, that is, for any $z > 0$, there exists $w = z^2$.


(b) is obviously true, and does not entail (a).
How to prove (a), now? This depends on how you have introduced real numbers. I guess you did it as the closure of the set of rational numbers. 
Then if you show that $x$ in (a) can be written as the limit of a sequence $x_n = f(x_{n-1})$ with $f$ defined with the basic arithmetic operations, you'll be able to conclude.
A: If we cheat a bit (by presupposing our claim) we can give the following justification:
Assume that $x_0\gt0$ is not a square root of any real number. Then, we can define the continuous function:
$$
f(x)=\frac{1}{x^2-x_0}
$$
for all real $x$. In particular, it is continuous in the segment $[0,1+x_0]$, and so must obtain its maximum there. However, the function is not bounded in that segement - a contradiction.
