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I am aware that 1/0 is undefined for two reasons:

  1. If you would have to give an answer to this it is infinity which is not a number but a concept;
  2. The limit of 1/x for $x \to 0$ is either positive or negative infinity and since it has two limits we cannot state what its exact value is

How can we use this for square roots? It has multiple answers so why do we pick the positive one? if $x^2 = 16 \implies x = \sqrt{16} $ or $x = -\sqrt{16}$ for respectively the positive and negative solution. This implies that the square root function has a single answer and we must negate its answer to obtain the second solution. I understand this is extremely practical but it feels like 'cheating' since we know that there are 2 solutions, but in the function we simply ignore one of them. Is this correct?

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    $\begingroup$ There are two square roots of $x$, but the symbol $\sqrt x$ is defined to be the positive one. $\endgroup$
    – Akiva Weinberger
    Feb 29, 2016 at 22:35
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    $\begingroup$ We aren't ignoring the second root. We are being rigorous in naming the two roots: the positive one is denoted $\sqrt{x}$, and the negative one is denoted $-\sqrt{x}$. That way, there is no ambiguity in using the symbol. $\endgroup$
    – MPW
    Feb 29, 2016 at 22:52
  • $\begingroup$ Your overall question seems to be the definition of the square root function $f(x)=\sqrt{x}$. As for roots of quadratic equations, there will always be two (complex roots). $\endgroup$
    – Lutz Lehmann
    Feb 29, 2016 at 22:55
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    $\begingroup$ Here is a related question. $\endgroup$
    – N. F. Taussig
    Mar 1, 2016 at 11:46

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Take it this way: $a = \sqrt{16}$ is just a symbol denoting something.
It denotes (by definition) the non-negative root of $x^2 = 16$.
Since $16 = a^2 = (-a)^2$, it means the other solution of this equation is $-a = -\sqrt{16}$

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  • $\begingroup$ So if I understand correctly there is such a thing as the square root function which always returns the non-negative root of a number; and then there's the mathematical concept of a square root for which there exists at least two solutions (positive and negative ones) for which the square root function will return only the non-negative one? $\endgroup$
    – Héctor van den Boorn
    Feb 29, 2016 at 23:04
  • $\begingroup$ @HéctorvandenBoorn Well, sort of but ... no, the sole fact that $b^2=x$ does not mean that $b$ is the square root of x. So in my notation above the $(-a)$ is not a square root of $x$. I would not make any distinction between the square root function and the math concept of square root. $\endgroup$
    – peter.petrov
    Mar 1, 2016 at 9:23
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Deciding whether $1/0$ is positive or negative is not really a problem. The real problem is there is no number $x$ such that $0 \times x = 1$. In contrast, there are exactly two distinct numbers $x_+$ and $x_-{}$ such that $(x_+)^2 = (x_-)^2 = 16$.

It's useful to be able to say sometimes which of the possible square roots we mean in a certain expression. If you want the negative root, write $-\sqrt{16}$. If you want to leave open the possibility of using either square root, write $\pm\sqrt{16}$ as is done in most presentations of the formula for the roots of a quadratic equation.

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In general, if $x$ is a positive number, there are $n$ complex numbers whose $n$th power is $x$. But there is only one real, positive number. That is defined to be $x^{1/n}$.

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    $\begingroup$ This isn't true. While, yes, some authors do adopt this convention, it isn't standard in the same way as $\sqrt{x}$. There are, for instance, plenty of authors (especially those covering topics involving complex numbers) who reserve $\sqrt[n] x$ for the positive real $n$-th root of a number, leaving $x^{1/n}$ either undefined, to represent a "generic" $n$-th root in computations, or as denoting the set of all $n$-th roots of $x$. $\endgroup$
    – Dan
    Feb 29, 2016 at 22:45
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This has to be taken into account when solving equations.

If you have (amongst other constraints) : $$x^2 = 16$$

then at first take you must consider:

Not:

$$x = \sqrt{16} $$

But:

$$x = \pm \sqrt{16}$$

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  • $\begingroup$ No, this is only valid for the solutions of the quadratic equation $x^2-16=0$. $\endgroup$
    – Lutz Lehmann
    Feb 29, 2016 at 22:47
  • $\begingroup$ @LutzL Yes, you're right of course. Edited answer accordingly. $\endgroup$
    – Paul Evans
    Feb 29, 2016 at 22:50
  • $\begingroup$ But the question was about the square root function, even if it is not explicit. $\endgroup$
    – Lutz Lehmann
    Feb 29, 2016 at 22:52
  • $\begingroup$ @LutzL It is also specifically about $x^2 = 16$. Edited answer again to be clearer. $\endgroup$
    – Paul Evans
    Feb 29, 2016 at 22:55
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If you want to solve the equation $x^2=16$ you can rewrite it $$0=x^2-16=(x+4)(x-4)$$One of the factors must be zero and you get $x=-4$ or $x=4$. A quadratic equation typically has two roots (occasionally a double root, and sometimes the roots are complex numbers).

The interpretation of the expression $\sqrt {16}$, as a matter of convention, is to set the value equal to the positive square root.

Quite often with expressions which can take multiple values, convention (or choice) will assign a principal value, so that the expression can be used with confidence.

When working in more general contexts it is not always true that principal values so chosen will always respect the normal laws of arithmetic, so care needs to be taken with expressions which involve multiple choices of principal value. Various common "paradoxes" in the manipulation of complex numbers can be attributed to this.

For example define $i=\sqrt {-1}$ then a "paradox" arises from $$i=\sqrt {-1}=\sqrt {\frac 1{-1}}=\frac {\sqrt 1}{\sqrt {-1}}=\frac 1i=\frac i{i^2}=-i$$

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