Must square root of $e$ be positive?

Question

I have always thought that there is two solutions to the square root of a real number, one being positive and the other being negative. However, in Penrose's book, A Road to Reality, he seems to claim that $e^{1/2}$ will always give a positive answer, since $e^n$ is defined as $1+ \frac{n}{1!} +\frac{n^2}{2!} + \ldots$, so substituting $\frac{1}{2}$ into the equation will give us a positive number. And so logarithm defined with base $e$ will be unambiguous since there is only one answer for every $e^n$. However, I find it quite puzzling because $-e^{1/2}$ will also give me $e$ when squared, doesn't it?

Answer 1

An old chestnut (and I'm sure there are many other posts on this, but here's another short attempt):

For real numbers $x$ which are zero or positive, the square root of $x$ is defined to be the real number $y \geq 0$ such that $y^2 = x$. It's easy to see that $y$ is unique and we denote it by the symbols $\sqrt{x}$. We make the square root unique in part so that the function $f(x) = \sqrt{x}$ is well defined.

If you then ask the question: what are the solutions of $x^2 = c$, for a positive, real number $c$, then there are two: $x = \sqrt{c}$ and $x = -\sqrt{c}$.

Similarly, the reason why the formula for solving the generic quadratic equation over the reals of $ax^2 + bx + c = 0$ has a $\pm$ symbol is because the expression $\sqrt{b^2 - 4ac}$ has a unique value when it exists.

So what Penrose is doing is using the established convention for what $\sqrt{x}$ means for a positive real number $x$.

Answer 2

I think the confusion arises because you are reading $e^{1/2}$ as "the set of solutions to $x^2=e$", and Penrose is defining $e^{1/2}$ as $\sum_{k\ge0}\frac{(1/2)^k}{k!}$. He is not using $e^{1/2}$ the way $y^{1/2}$ is defined in complex analysis, but rather as a notation for a function given by a convergent series. It would be more clear if he defined $$ \exp(n)=\sum_{k\ge0}\frac{n^k}{k!} $$ and then said that $\exp(1/2)$ is unambiguously defined.

Answer 3

If the solution to $x^2-e=0$ doesn't have the two roots $\pm\sqrt e$ some fundamental theorems of algebra has to be revised. But of course, also the exponential function must be well defined for $\displaystyle x=\frac{1}{2}$.

I guess it is not perfect to claim $\displaystyle a^{\frac{1}{2}}=\pm\sqrt a$.