# Why is $\sqrt{x}$ a function? [duplicate]

How $\sqrt{x}$ can be a function when $\sqrt{4}$ is equal to $-2$ and $2$?

• To point out the obvious, nothing can equal two different things. – Git Gud Nov 22 '14 at 12:54
• @GitGud: On this site, when encountering an expression like $i^i$, many people will immediately and silently assume that it is infinitely many things at once, and use that as a justification for saying that it is a particular one of them. – Marc van Leeuwen Nov 22 '14 at 13:00
• $\sqrt{4}$ is only equal to $2$. $\pm\sqrt{4}$ on the other hand, is equal to both $2$ and $-2$. – Frank Vel Nov 22 '14 at 13:02
• @fvel No, $\pm \sqrt 4$ doesn't equal anything because it's not an object. When one writes $a=\pm b$, this is short for the disjunction $a=b\lor a=-b$, the symbol $\pm b$ does not denote an object and if it did, it would denote only one object. – Git Gud Nov 22 '14 at 13:05
• @omidh Usually I am against 'real world' examples because you can only take the discussions so far before needing to formalize it, but OK. In that sentence the verb to 'to be' is not identity (=), but rather a predicate. Equality has no business being there. – Git Gud Nov 22 '14 at 13:41

The notation $\sqrt{x}$ refers to the principal square root of $x$. The principal square root is the non-negative square root. The notation $-\sqrt{x}$ refers to the negative square root of $x$.

Hence,

\begin{align*} \sqrt{4} & = 2\\ -\sqrt{4} & = -2 \end{align*}

The function $f(x) = \sqrt{x}$ has domain $[0, \infty)$ and range $[0, \infty)$. Its graph is the upper half of the parabola $x = y^2$.

The function $g(x) = -\sqrt{x}$ also has domain $[0, \infty)$, but its range is $(-\infty, 0]$. Its graph is the lower half of the parabola $x = y^2$.

See the diagram below. • So, why we don't behave equations like we behave functions? Why $\sqrt{x} doesn't mean principal square of x in equations? – omidh Nov 22 '14 at 13:15 • The symbol$\sqrt{x}$does mean the principal square root in equations. If you have$\sqrt{x} = 5$, then you solve it by squaring both sides to obtain$x = 5^2 = 25$. Check:$\sqrt{25} = 5$. On the other hand,$\sqrt{x} = -5$has no solution since squaring both sides yields$x = 25$, but$\sqrt{25} = 5 \neq -5$. If you have$x^2 = 25$, then taking square roots yields$x = \pm \sqrt{25} = \pm 5$. Check:$(\pm 5)^2 = 25$. Notice that the solutions of the equation$x^2 = 25$include both the principal square root,$\sqrt{25} = 5$, and negative square root,$-\sqrt{25} = -5$, of$25$. – N. F. Taussig Nov 22 '14 at 13:27 Let$f$be a function from$[0,\infty]\to [0,\infty]$by$f(x)=x^2$. You can see that$f$is invertable as it is$1-1$and onto.$f^{-1}(x)=g(x)=\sqrt x$defined from$[0,\infty]\to [0,\infty]$as an invverse of it. Thus$\sqrt 4=2\$.

• Note that The things that which determine function is not just an equation but also domain and range. When you forgot domain and range, you will confuse. – mesel Nov 22 '14 at 13:05