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I found some inconsistencies of notations in my calculus textbook: for example, whenever it uses $\sqrt4$, it assumes that $\sqrt4 = 2$ instead of $\pm2$. But whenever it solves an equation such as $x^2 = 4$, it gives me the answer that $x = \pm2$. I think that square root and square are inverse operations. What I mean is that: if $x^2 = y$ then $\sqrt{y} = x$. Here one of the appropriate solutions would be either $x = \pm2$ and $y = 4$ or $x = 2$ and $y = 4$.

an4s
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OptatootatpO
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3 Answers3

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There is no inconsistency. The square root is a function, hence has a single value. Unless otherwise specified, the positive root is taken.

The solutions of

$$x^2=4$$ are $$x=\pm\sqrt4.$$

2

From here,

Every non-negative real number $a$ has a unique non-negative square root, called the principal square root, which is denoted by $\sqrt a$, where $\sqrt{\,\,}$ is called the radical sign or radix. For example, the principal square root of $9$ is $3$, which is denoted by $\sqrt9 = 3$, because $3^2 = 3\cdot3 = 9$ and $3$ is nonnegative.

and

Every positive number $a$ has two square roots: $\sqrt a$, which is positive, and $−\sqrt a$, which is negative. Together, these two roots are denoted as $\pm\sqrt a$. Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root.

an4s
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Your book is correct.

The equation $$x^2=4$$ has two solutions and they are $$x =\pm 2$$

One of them is positive and the other one is negative.

The positive one is denoted by $\sqrt 4 $ and the negative one is denoted by $- \sqrt 4$ That is $$\sqrt 4 =2\\ -\sqrt 4 =-2 $$

We like $$\sqrt x : [0, \infty ) \to [0, \infty ) $$ to be a function and functions do not have more than one value at a single point.

Similarly $\sqrt 9 =3$ and $- \sqrt 9 =-3$

Mohammad Riazi-Kermani
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