How do you read $\pm ?$ Why does $|x|=3\implies x=\pm 3 ?$

Question

Question 1

How is the symbol '±' read? "Plus AND Minus" or Plus OR Minus"? All this time I've been reading it as Plus-Minus. My math teacher says that it's the former; but then in $$x^2 = 9 \implies x= ± 3, $$ how does $x$ equal two values? $x$ should equal $+3$ OR $-3,$ right?

Question 2

My teacher says that $$|x| ≠ ± x\tag A$$ but that $$|x| = 5 \implies x = ±5.\tag B$$ Wait... what? What is going on here?

He also says that $$\sqrt{x^2}= |x|.$$ But this means that $$|3|=\sqrt{9} =3,$$ which, by $(\text B),$ implies that $$ |3| = ± 3;$$ doesn't this contradict $(\text A) ?$

Answer 1

1) Think of $x^2=9$ as having two possible solutions: $x=3$ or $x=-3$. You could also say that $x=3$ and $x=-3$ are valid solutions. This is more a matter of wording I think.

2) When the value of $x$ is known, we can either have $|x|=x$ or $|x|=-x$ but not both. Consider $x=-2$. Then $|-2| = 2 = -(-2)$. So here, we do have $|x|=-x$. But $|-2|=2 \neq -2$, i.e. $|x| \neq x$. For $x=2$ we have the cases switched around.

3) With regards to your edit, see this and this.

Answer 2

The notation $±$ should be read plus or minus (at least, this is how it goes in French).

I don't know why your maths teacher would say otherwise.

Generally speaking, this is a shortcut to avoid writing two different expressions that are identic up to a sign. You can avoid it by writing both, and connecting them with an "or".

(NB: there is absolutely no maths in there, only notational convention, or even notational abuse.)

Answer 3

The symbol $\pm$ should be read plus or minus.

Background

By convention, if $x \geq 0$, then $\sqrt{x}$ is the principal square root of $x$, meaning the nonnegative square root of $x$. Hence, $\sqrt{9} = 3$. If we want to denote the negative square root, we write $-\sqrt{x}$. Thus, the negative square root of $9$ is expressed in the form $-\sqrt{9} = -3$. Notice that the principal square root of a number is uniquely defined.

The absolute value of a real number $x$, denoted $|x|$, is its distance from $0$ on the real number line. Since distances are nonnegative, the absolute value of a number is never negative. If $x \geq 0$, then $|x| = x$. For instance, $|5| = 5$ since $5$ is five units from $0$ and $|0| = 0$ since $0$ is zero units from $0$. If $x < 0$, then $|x| = -x$. For instance, $|-4| = 4$ since $-4$ is four units from $0$. Consequently, we can write a piecewise definition of the absolute value as follows: $$|x| = \begin{cases} x & \text{if $x \geq 0$}\\ -x & \text{if $x < 0$} \end{cases}$$ Observe that \begin{align*} \sqrt{5^2} & = \sqrt{25} = 5 = |5|\\ \sqrt{0^2} & = \sqrt{0} = 0 = |0|\\ \sqrt{(-4)^2} & = \sqrt{16} = 4 = -(-4) = |-4| \end{align*} Observe that if $x \geq 0$, then $\sqrt{x^2} = x = |x|$. If $x < 0$, then $\sqrt{x^2} = -x = |x|$. Therefore, we obtain the alternative definition of absolute value: $$\sqrt{x^2} = |x|$$

Solve the equation $x^2 = 9$.

\begin{align*} x^2 & = 9\\ \sqrt{x^2} & = \sqrt{9} && \text{take the principal square root of each side of the equation}\\ |x| & = 3 && \text{evaluate the principal square roots} \end{align*} The equation $|x| = 3$ means that $x$ is a real number whose distance from $0$ on the real number line is $3$. There are two such numbers. They are $3$ and $-3$. However, $x$ cannot simultaneously be both $3$ and $-3$. Hence, we say that $x = 3$ or $-3$. Our solution set is $S = \{3, -3\}$. Thus, this equation has two distinct solutions that happen to be opposite in sign. When this occurs, the solution is sometimes expressed in the form $$x = \pm 3$$ to indicate that if $x$ is a solution of the equation, then $x = 3$ or $x = -3$. To put is another way, either $3$ or $-3$ could be substituted for $x$ to make the equation $x^2 = 9$ true.

Clearly, it does not make sense to say $x = 3$ and $x = -3$ since $3 \neq -3$. Consequently, it also does not make sense to read the symbol $\pm$ as plus and minus.

The symbol $\pm$ indicates that the symbol may be substituted by either the positive or negative values of the variable (see this definition). Therefore, it does not make sense to write $$|x| = \pm x$$ since there is only a single value which may be substituted for $|x|$. If $x \geq 0$, that value is $x$; if $x < 0$, that value is $-x$.

Answer 4

This answer is concerned with real numbers only.

1. How is the symbol '±' read?

   The notational shorthand $$y=\pm x,$$ that is, $$\text{“$y$ equals plus }\textit{or}\text{ minus $x$”,}$$ is definitionally equivalent to $$y\in\{x,-x\}.$$

2. $|x| = 5 \implies x = ±5$

   True; in fact, $$|x| = 5 \iff x = ±5,\tag0$$ since for each $x,$ each side of the biconditional means precisely that $x$ is an element of $\{-5,5\}.$

   $ |3| = ± 3$

   True; indeed, $|3|\in\{-3,3\}.$

   $|x| ≠ ± x$

   False; $$\text{for each }x,\:\:|x|=\pm x.\quad\quad\Large✓\tag1$$

3. The ± notational shorthand makes statement $(1)$ look like a mathematical identity, $|x|\equiv\pm x,$ and misleadingly suggest that $|x|$ and $\pm x$ are mutually substitutable, when in fact $$|x|\not\equiv\pm x.\quad\quad\Large✓\tag2$$ To be clear: $\boldsymbol{|x|=\pm x}$ is not a genuine equation, and statement $(1)$ means: $$\text{for each }x,\:\:|x|\in \{x,-x\}\tag1$$ $$\text{for each }x,\;\big(\,|x|=x\quad\textit{or}\quad |x|=-x\,\big).\tag1$$

4. Observe that $$\boxed{\;y=|x|\quad \iff\quad y=\pm x\:\:\:\textit{and}\:\:\:y\ge0\;};$$ this result entails statements $(0)$ and $(2).$

   Hence, \begin{align}& x\in\{-4,-2,0\}\\\iff {}& x^2+3x=\pm x \\\kern.6em\not\kern-.6em\implies{}& \color{brown}{x^2+3x=|x|}\\\iff{}& x^2+3x=\pm x \:\:\:\textit{and}\:\:\:x^2+3x \ge0 \\\iff{}& x\in\{-4,0\}.\end{align}

Answer 5

• $ \pm $ is pronounced as plus or minus

• $x=|4| \implies x=4,-4$ because both $|4|$ and $|-4|$ are equal to $4$