Why is equating these two symbols an error?

[EngWiPy]

Hello,

My calculus book says that readers who are writting $$\sqrt{9}$$ as $$\pm3$$ must stop doing that, because it is incorrect. The question is: why is it incorrect?

Regards

 

[arildno]

Because the square root of a number A is DEFINED to be the unique, non-negative number whose square equals A.

 

[EngWiPy]

Because the square root of a number A is DEFINED to be the unique, non-negative number whose square equals A.

I didn't understand. For any real number a there are two square roots: a positive square root, and a negative square root. How is the square root is unique?

 

[fleem]

When you encounter a square root in an equation, use the +/- thing. If somebody asks you, "What is the square root of four", say "two". Its just what mathematicians have decided we will mean, when "square root" is used in each of those contexts.

 

[arildno]

I didn't understand. For any real number a there are two square roots: a positive square root, and a negative square root. How is the square root is unique?

Incorrect.

For any non-negative number "a", the equation:
$$x^{2}=a$$
has two SOLUTIONS:
$$x_{1}=\sqrt{a},x_{2}=-\sqrt{a}$$

The $\sqrt{a}$ is a non-negative number.

 

[HallsofIvy]

To expand on arildno's point: S David, would you say that the solution to $x^2= 5$ is $\sqrt{5}$ or $\pm \sqrt{5}$? I suspect you will say the latter and the point is that the whole reason we need the "$\pm$" is because $\sqrt{5}$ itself only gives one of them: the positive root.

 

[EngWiPy]

Incorrect.

For any non-negative number "a", the equation:
$$x^{2}=a$$
has two SOLUTIONS:
$$x_{1}=\sqrt{a},x_{2}=-\sqrt{a}$$

The $\sqrt{a}$ is a non-negative number.

Referreing to the book whose name is: Calculus (7th ed), for Anton, Bivens, and Davis, Appendix B at the bottom of the page, it says the following:

Recall from algebra that a number is called a square root of A if its square is A. Recall also that every positive real number has two square roots, one positive and one negative; the positive square root is $$\sqrt{A}$$ and the negative square root is $$-\sqrt{A}$$. For example, the positive square root of 9 is $$\sqrt{9}=3$$ and the negative square root of 9 is $$-\sqrt{9}=-3$$

After this review, it says the thing I started with. Is this differ from what I said in post #3 in this thread?

 

[arildno]

The SYMBOL $\sqrt{a}$ always signifies a non-negative number.

Therefore, $-\sqrt{a}$ is always a non-positive number.

Colloquially, we call this "the negative square root of a", whereas if we want to be über-precise, we ought to call it "the (additive) negative OF the square root of a"

(alternatively, "minus square root of a", in complete agreement of calling -2 for "minus two")

 

[slider142]

The symbol $\sqrt{9}$ is shorthand for "the principal square root of 9" (not simply "a square root of 9") where the principal square root is a function. A function has only a single output for each input, therefore equating it to the symbols $\pm 9$ which is shorthand for the set {9, -9} is an error.