Does $x^{0.5}$ have a negative solution?

Question

A typical graph of $f(x) = n^x$ shows only positive solutions:

But it seems like, for some values of $x$, there are negative solutions. For example, $2^{1/2}$ is $\sqrt{2}$, which has a negative solution.

Some say $\sqrt{2}$ is defined to be just the positive root. But I'm not sure this applies to $2^{1/2}$. From the rule $x^a x^b = x^{a+b}$, we get $2^{1/2} = 2^{1/2}$, and can conclude that $2^{1/2} = \pm\sqrt{2}$, with a negative solution.

Is there a reason to define $n^x$ as only having positive solutions?

Answer 1

There is some confusion here. The number $\sqrt2$ is just that: a number. And numbers don't have solutions. When $\lambda\geqslant0$, the expression $\sqrt\lambda$ denotes the only non-negative square root of $\lambda$.

Besides, if $a>0$ and $b\in\Bbb R$, $a^b$ is $e^{b\log a}>0$. Also, if $b=\frac mn$, with $m\in\Bbb Z$ and $n\in\Bbb N$, then$$a^b=a^{m/n}=\sqrt[n]a^{\,m}>0.$$

Answer 2

The square root function is defined to be positive (you can as well consider the negative square root function, but this is rarely done). Remember that a function must have a unique value.

Now the exponential function of equation $y=a^x$ (where $a>0$) is always taken to be positive, otherwise changes of sign at odd fractions would cause unacceptable discontinuities.

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Do not confuse a function that you want to plot and the solutions to an equation such as $x^2=2.$

Answer 3

In mathematics, when we generalize a concept we either specify rules that uniquely determine the generalization, or else have an umbrella term for multiple generalizations. As an example of each, in that order:

• Starting from $x^0:=1,\,x^{n+1}:=x^nx$ as a definition of $x^n$ for non-negative integers $n$, we generalize to rational $n$ by requiring $\color{red}{x^ax^b=x^{a+b}}$ and $\color{blue}{x>0\implies x^n>0}$, and to real $n$ by demanding $n$-continuity. This cannot define $x^n$ for $x<0$ unless $n$ is rational, with odd denominator in its lowest terms. It also cannot define $0^n$ for any $n<0$. Even the case $0^0$ is complicated.
• The axioms of group theory generalize certain facts about the addition of integers, but don't uniquely determine whether we're talking about the addition of real numbers, which are one example of a group. (The previous sentence isn't intended as an accurate summary of the history behind group theory; indeed, it could have just as easily started by considering real numbers first.) Instead, there are lots of "groups" with different properties.

Your question essentially amounts to, "why do we uniquely define $2^{1/2}$ as the positive root of $x^2=2$ as per the first bullet point, when we could (as per the second bullet point) consider all functions satisfying the red part but not necessarily the blue part"? Such a red-only approach wouldn't be as interesting here as with groups, since once you understand the non-negative definition of a square root you know all the available generalizations. Instead, it's more convenient to give the function $\sqrt{x}$ a unique definition on the non-negative reals.

Answer 4

I am a little bit confused by the use of "solution" here, do you mean $n^x=0$ or a "value" of this function for a given $n,x$.

Either way, if $n\in\mathbb{N},x\in\mathbb{R}$ we can say that $n^x>0$. If we were to extend this function to negative $n$ then it would equal negative values, but for the most part would be complex-valued.