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So, is $4^{1/2} =$ just $+2$ , or is it $+2$ and $-2$ ? In high school I learnt that it was just $+2$.

i see already in the 1st answers to my questions that this is a startlingly difficult question to answer! normally in maths, there is no such ambiguity, and i would like to understand what sort of symbology should be used to remove the ambiguity of this.

Note that i came to this question originally from having found issues in this matter in complex numbers.

there i came across the expression "Principle Root", perhaps for strictly clear math symbology, a little plus suscript for that might be

  • what is needed to indicate that root?
  • And if you expect 2 answers (is ALL the roots, rather than the one with the largest real component), what symbology would you use then to make that clear?

i am looking for a general rule here, the power of a half is a simple case. what if it was power of some other fraction? how should the question be expressed so as to indicate if the principle root is requested, or all roots?

Randor
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  • It is $\pm 2$. Let's simply the following: $2^2=4$ but also $(-2)^2=4$! – pawel Feb 20 '21 at 08:55
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    That +2 that you learned in highschool is called the principal square root as far as I know. – Kevin.S Feb 20 '21 at 09:03
  • According to wikipedia: "https://en.wikipedia.org/wiki/Square_root" The principal square root can be written in exponent notation as you did (see the third paragraph of the post) – Kevin.S Feb 20 '21 at 09:05
  • So to indicate which root you want, usually, we can use $\sqrt{x}$ to represent the nonnegative one and $-\sqrt{x}$ the negative one. Again, wikipedia also mentions this. – Kevin.S Feb 20 '21 at 09:07
  • OK. when you want two roots to be included, just use $\pm\sqrt{x}$, put a plus-minus sign...... – Kevin.S Feb 20 '21 at 09:12
  • and what symbology if x is complex and its not square root but 10th root? – Randor Feb 20 '21 at 09:16
  • i want to know what is the general rule here. normally in my experience, maths symbology is meant to be clear and unambiguous. it seems we have something ambiguous here, and i want to know how to express the mathematical question in an unambiguous way. And then , i will be wanting to understand the proof of why it is not true that i^2=+1 , i guess that will go to a seperate post – Randor Feb 20 '21 at 09:21
  • @Randor In general, I'm afraid that there is no standard notation for things like 10th roots. However, in complex numbers, we often use the polar form, i.e., $re^{i\theta}$. So for instance, if you want to solve $x^3=1$, then you can specify which root you want by clearly stating "in this case, we need e.g., $e^{i\frac{2}{3}\pi}$". Also, the $\pm$ sign only works in $\Bbb{R}_+$ because there is no "positive complex number"... – Kevin.S Feb 20 '21 at 10:50
  • you can write the 10th root as either x^(1/10) or as 10th root of x (ie with a 10 inside the v of the radical).
    to indicate you want the principal root, how should that be indicated?
    – Randor Feb 20 '21 at 10:54

4 Answers4

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By definition (when $a> 0$), $a^{1/2}$ and $\sqrt a$ both mean the positive square root of $x$. If you wanted to include both numbers which square to give $a$, you would need to write $\pm a^{1/2}$ or $\pm\sqrt{a}$.

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$4^{1/2}$ is evaluated to be $2$.

Fundamentally, the function $f(x)=x^{1/2}$ has a range of $[0,\infty)$.

In comparison, if we were looking for the solutions to the equation $$x^2=4$$ instead, then in solving the quadratic equation we would get $x=\pm2$.

Andrew Chin
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An expression or function evaluation has only one value (or "answer") - or perhaps none in case it is not defined. Even though the equation $x^2=9$ has two real solutions $x=3$ and $x=-3$, the expression $\sqrt 9$ has only one value $3$.

What "is" $a^b$, anyway? In order to know what is meant by $a^b$, we need to define the expression $a^b$ for suitable $a$ and $b$. Unfortunately, we don't do this all at once, but first only for a few cases and then extend the definition. And even more unfortunately, sometimes we encounter different routes to extend along and have to keep an eye on consistency. Here's an overview.

For

non-negative integers $a,b$,

i.e., when these can be interpreted as the cardinalities of finite sets, we can readily define $a^b$ as the number of maps from a set $B$ of cardinality $b$ to a set $A$ of cardinality $a$. For example, simple counting then gives us $3^2=9$, $2^3=8$, $0^0=1$, and so on. One can even show that this implies $$\tag1 a^0=1,\quad a^{n+1}=a\cdot a^n$$ and more general relations such as $$\tag2 a^{b+c}=a^ba^b,\quad a^{bc}={(a^b)}^c,\quad (ab)^c=a^cb^c.$$ We can use $(1)$ to extend our definition (namely, by using recursion as definition) to the case

$a$ an arbitrary number, $b$ a non-negative integer.

When $a\ne0$, we can even use the recursion in $(1)$ to go backwards and thereby extend to

$a\ne0$, $b\in\Bbb Z$.

Note that we cannot nicely define $0^b$ when $b$ is a negative integer. Well, theoretically we can define anything - but a definition should be useful. For example, we want to be able to make use of properties $(2)$ without having to check for applicability. Thus any extension of our definition should make sure that $(2)$ keeps holding. This is called the Principle of Permanence. Therefore any useful definition of $0^{-n}$ should make $0=0^{-n}\cdot 0=0^{-n}\cdot 0^n=0^0=1$, which is absurd.

But can we extend to rational exponents $b=\frac nm$, $n\in \Bbb Z$, $m\in \Bbb Z_{>0}$ (respecting the Principle of Permanence)? That is, $x=a^{\frac nm}$ should be a number with the property $x^m=a^n$. Sometimes there is no (real) number with this property, (namely when $a<0$, $m$ even, $n$ odd; or when $a=0$ and $\frac nm<0$). Sometimes there is exactly one real solution (namely when $m$ is odd; also when $a=0$ and $\frac nm>0$). And sometimes there are several solutions (namely when $a>0$, $m$ even, $n$ odd). In the latter case we have to remember that we want to define the value of $a^b$ as a function of $a$ and $b$. The convention is to take the positive value, which has a huge advantage when we extend even further. But first let's summarize that we managed to deal with

$b\in\Bbb Q$, provided $a>0$, or $a=0$ and $b\ge0$, or $a<0$ and denominator of $b$ is odd.

Apparently, things are already becoming somewhat convoluted.

Thanks to our choice odf positive solution above, we have a new property: If $a>0$, then the map $\Bbb Q\to\Bbb R$, $b\mapsto a^b$ is continuous! This allows us to extend to

$a>0$, $b\in\Bbb R$

simply by continuity! This cannot work for negative $a$ because of the gaps at rationals with even denominator. And it almost works for $a=0$ by defining $0^b=0$ for all $b>0$, but neither can we define this for negative exponents, nor can we avoid the discontinuity at $b=0$, i.e., $0^0=1\ne 0=\lim_{b\to0^+}0^b$.

Once we know enough calculus, we can use a different way to extend our definition by setting $a^b=\exp(b\ln a)$ using the exponential function and natural logarithm. This agrees with the above as long as $a>0$ (in particular, we still have $4^{\frac12}=2$, period). It cannot deal with the case of $a=0$ where the logarithm is not defined and also has serious trouble when $a<0$. On the other hand, it allows us to use complex exponents $b$. The goal now seems to be to extend the definition of the $\ln$ function. There are nice was to do so for any open subset $U$ of $\Bbb C$ that is a simply connected neighbourhood of $1$. While there is a somewhat natural choice $U=\Bbb C\setminus (-\infty,0]$ for a maximal such $U$, different choices may sometimes be more useful (and do ultimately lead to different values for some $a^b$!). Note that this standard choice (the "principal branch") ignores the cases with $a<0$, and just combining the two methods is also no good idea regarding Permanence.

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By definition, taking square root (aka arithmetic root) results in only non-negitive answer. You simply ignore the fact that $(-2)^2 = 4$.

But if you want to get all possible "answers", you should solve an equation like this one:

$$ x^2 = 4 $$

Both left and right parts of this equation are positive so you can take square root on both of them. Since $\sqrt{x^2} = |x|$ we get the following equation:

$$ |x| = 2 $$

The absolute value $|x|$ tells us that both $-2$ and $2$ are roots, not only $2$.

Deepak
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