Skip to main content

Section 10.1 So Many Symbols!

As we continue forward into higher levels of mathematics, we will encounter more mathematical notation. Students sometimes start to struggle because they've never developed a set of tools to help them interpret those symbols. It's like trying to read a language when you don't have a sense of the grammar. You can sometimes piece things together and follow what's happening, but most of the time you're completely lost.

We will begin with familiar territory. At some point in their lives, most students learned some version of the order of operations (often called PEMDAS). This is a set of instructions for the correct way to perform a complex calculation by prioritizing certain calculations. Here is how the order of operations is usually presented:

  • Parentheses and other grouping symbols.

  • Exponents

  • Multiplication and division (performed left-to-right)

  • Addition and subtraction (performed left-to-right)

This is fine as far as the basics are concerned. And when working with numbers, students are usually able to do this correctly. But it's not long after we start introducing variables that students can feel overwhelmed by symbols.

The best analogy for understanding this comes from thinking about a complex sentence in English: "The man carrying a large bag containing three apples and two pears bought a blue shirt with six silver buttons from the store on the corner of Main Street and West Park Avenue." There are a lot of details in that sentence, but there's also the "big picture" of the sentence. What is actually happening in the sentence? Some man bought a shirt from a store. From there we can go into specific details, such as what the man was holding, the color of the shirt, and the location of the store. But those details add information to the big picture without changing the core of the picture.

The focus of this section is not really on algebra, but on helping you familiarize yourself with the language of mathematics. Fortunately, mathematicians have created an entire notation to help you do this. Consider the following polynomial:

\begin{equation*} x^3 - 3x^2 - 7x + 11 \end{equation*}

At this point, your experience should lead you to interpret this as the sum of four terms. You should be able to visualize the plus and minus signs as being separators for the different terms. Compare this to how we would have to write it if we had to write everything out:

\begin{equation*} x \cdot x \cdot x + (-3) \cdot x \cdot x + (-7) \cdot x + 11 \end{equation*}

It is much harder for our brains to process this because we have to do a lot of extra work. It's harder to locate the plus signs from among all of the symbols, and we have to count out all of the products to figure out how many factors of there are.

The two key features here are the use of implied multiplication and exponents. These two bits of notation help us to visually condense the information and make it easier to read. The exponents also mean that we don't have to count out the products. We also have the convention that subtraction means addition of the opposite, which lets us write the negative coefficients as subtraction.

All of this notation helps us to identify that the "big picture" of this is that we have a sum of four terms. From there, we can choose to look at the details of those terms to see (for example) if there are any like terms that we might want to combine.

Another mathematical notation that was developed is the use of fractions for division. Interestingly, it's very rare in the modern mathematical world to use the \(\div\) symbol. It is not as rare, but still uncommon to use \(/\) to represent division as well. These tend to cause more confusion than clarity, which is why modern mathematics prefers the use of a fraction. This way, we can clearly distinguish between the numerator and denominator of a fraction.

For example, consider the following expression: \(x^2 + 3/x + 2\text{.}\) The strict (and proper) application of the order of operations is that this result is the same as \(x^2 + \frac{3}{x} + 2\text{.}\) But students will write that to mean \(\frac{x^2 + 3}{x + 2}\text{.}\) If you insist on writing your fractions with a diagonal slash, you will need to use parentheses to separate out the numerator and the denominator: \((x^2 + 3)/(x + 2)\text{.}\)

For this reason, students are strongly urged to not use the diagonal slash for division, but write their division using a long horizontal bar (long enough to span the width of all the terms in the numerator and the denominator) so that it is clear what is in the numerator and what is in the denominator.

Similarly, when using the square root, it is important that the bar of the square root is long enough (and only long enough) to cover the parts inside the square root. For example, \(\sqrt{x + 3}\) is not the same as \(\sqrt{x} + 3\text{,}\) and \(\sqrt{x \,+ \, } \, 3\) doesn't even make sense. So please be careful with what you're writing.

Activity 10.1.1. Identifying the "Big Picture" of an Expression.

A basic skill when learning to read mathematical notation is to mentally "group together" things to be able to process what's happening. For example, the polynomial from before can be seen as the sum of four terms.

\begin{equation*} x^3 - 3x^2 - 7x + 11 \longrightarrow \boxed{x^3} + \boxed{(-3x^2)} + \boxed{ (-7x) } + \boxed{11} \longrightarrow \boxed{A} + \boxed{B} + \boxed{C} + \boxed{D} \end{equation*}

Similarly, we can think of fractions as being basically a numerator divided by a denominator.

\begin{equation*} \frac{x^2 + 3}{x + 2} \longrightarrow \frac{\boxed{x^2 + 3}}{\boxed{ x + 2} } \longrightarrow \frac{ \boxed{A} }{\boxed{B}} \end{equation*}

Try it!

Consider the expression \(\frac{x + 3}{x - 2} + 3x\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.

Solution.
\begin{equation*} \boxed{ \frac{ x + 3 }{ x - 2 } } + \boxed{ 3x } \end{equation*}

This is the sum of a fraction and a monomial.

After seeing the "big picture" we can then break things down further. Once we're inside of the numerator or the denominator of the fraction, and that may consist of multiple terms.

\begin{equation*} \frac{x^2 + 3}{x + 2} \longrightarrow \frac{\boxed{x^2 + 3}}{\boxed{ x + 2} } \longrightarrow \frac{\boxed{ \boxed{x^2} + \boxed{3} }} {\boxed{ \boxed{x} + \boxed{2} } } \longrightarrow \frac{ \boxed{\boxed{A} + \boxed{B} }}{\boxed{\boxed{C} + \boxed{D}}} \end{equation*}

And we can go deeper into the individual terms and break them up into the coefficient (which is sometimes an unwritten and the variable part. And the variable part can be broken down into individual variables. But at this point, it's like focusing on the number of apples in the bag and you've lost sight of the big picture.

But so far, we've only focused on the basic arithmetic operations. The next large piece of language you will encounter are functions. We aren't going to go into the details of what functions are here. We're just going to use the notation to help you understand how to think about it when you get to them.

The basic shape of a function is \(f(x)\text{.}\) This represents a number just like the variable \(y\) would represent a number. The only difference is that with a function, there is a "rule" that tells you how to calculate the number. When we write \(f(x)\) we are saying to use the rule associated with \(f\) to do the calculation. In addition to the letters \(f\) and \(g\) (which are used for generic functions), there are also a number of special functions that have names. Here are a few examples:

Name Notation
sine \(\sin()\)
cosine \(\cos()\)
tangent \(\tan()\)
common logarithm \(\log()\)
natural logarithm \(\ln()\)
exponential \(\exp()\)

The empty parentheses indicate that there is a value to "plug into" the function. So there would normally be some expression inside of those parentheses, which can be a number (such as with \(\ln(2)\)), a variable (such as with \(\sin(x)\)) or a variable expression (such as with \(\exp(-x^2 + 1)\)).

At this point, it doesn't matter what these functions represent, other than they represent a method for calculating some specific number. So it doesn't matter that \(\ln(2) \approx 0.693\text{,}\) only that \(\ln(2)\) is some number that your calculator can calculate for you.

At least in terms of algebra, functions behave very similarly to variables. Here are some examples:

\begin{equation*} \begin{aligned} 2f(x) + 3f(x) \amp = 5f(x) \\ x \ln(3) - x \amp = (\ln(3) - 1)x \end{aligned} \end{equation*}

The important note is that the function name cannot be separated from its argument (the parentheses and everything inside of the parentheses). All of that notation should be seen as a single object. That is, you cannot factor out the \(f\) in the first example:

\begin{equation*} 2f(x) + 3f(x) \overset{\times}{=} f(2(x) + 3(x)) \end{equation*}

It may help for you to put a box around the function and its argument.

\begin{equation*} 2f(x) + 3f(x) = 2 \, \boxed{ f(x) } + 3 \, \boxed{ f(x) } \end{equation*}

The error of breaking apart the function from its argument is similar to other errors where students "distribute" incorrectly:

\begin{equation*} \begin{aligned} \frac{1}{a + b} \amp \overset{\times}{=} \frac{1}{a} + \frac{1}{b} \\ (x + y)^2 \amp \overset{\times}{=} x^2 + y^2 \\ \sqrt{n + m} \amp \overset{\times}{=} \sqrt{n} + \sqrt{m} \\ \sqrt{n^2 + m^2} \amp \overset{\times}{=} n + m \\ \sin(\theta + \phi) \amp \overset{\times}{=} \sin(\theta) + \sin(\phi) \end{aligned} \end{equation*}

It all comes down to not understanding what the symbols mean or how they behave.

Activity 10.1.2. Hiding Information in Boxes.

Students often feel intimidated by the notation. One trick is to think of the the function as being hidden inside of a box. This will "hide" the details so that you can focus on the big picture.

\begin{equation*} \begin{aligned} x \ln(3) + 4 \amp = \ln(6) \\ x \cdot \boxed{ A } + 4 \amp = \boxed{ B } \amp \eqnspacer \amp \text{Substitute} \\ x \cdot \boxed{ A } \amp = \boxed{ B } - 4 \amp \amp \text{Subtract $4$ from both sides} \\ x \amp = \frac{ \boxed{ B } - 4}{\boxed{A}} \amp \amp \text{Divide both sides by $\boxed{A}$} \\ x \amp = \frac{ \ln(6) - 4}{\ln(3)} \amp \amp \text{Substitute} \end{aligned} \end{equation*}

Try it!

Solve the equation \(x \sin(1) + 5 = 3x - \cos(2)\text{.}\) Do it once using a substitution similar to the example above, and then do it without that substitution. Use a complete presentation both times.

Solution.
\begin{equation*} \begin{aligned} x \sin(1) + 5 \amp = 3x - \cos(2) \\ x \cdot \boxed{ A } + 5 \amp = 3x - \boxed{ B } \amp \amp \text{Substitute} \\ x \cdot \boxed{ A } \amp = 3x - \boxed{ B } - 5 \amp \amp \text{Subtract $5$ from both sides} \\ x \cdot \boxed{ A } - 3x \amp = - \boxed{ B } - 5 \amp \amp \text{Subtract $3x$ from both sides} \\ \left( \boxed{ A } - 3 \right) x \amp = - \boxed{ B } - 5 \amp \amp \text{Factor out the $x$} \\ x \amp = \frac{ - \boxed{B} - 5}{\boxed{A} - 3} \amp \amp \text{Divide by $\boxed{A} - 3$} \\ x \amp = \frac{ - \cos(2) - 5}{\sin(1) - 3} \amp \amp \text{Substitute} \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} x \sin(1) + 5 \amp = 3x - \cos(2) \\ x \sin(1) \amp = 3x - \cos(2) - 5 \amp \amp \text{Subtract $5$ from both sides} \\ x \sin(1) - 3x \amp = - \cos(2) - 5 \amp \amp \text{Subtract $3x$ from both sides} \\ \left( \sin(1) - 3 \right) x \amp = - \cos(2) - 5 \amp \amp \text{Factor out the $x$} \\ x \amp = \frac{ - \cos(2) - 5}{\sin(1) - 3} \amp \amp \text{Divide by $\sin(1) - 3$} \end{aligned} \end{equation*}

Activity 10.1.3. Solving for an Expression.

When we solve equations, sometimes we're not solving for the variable, but some expression involving the variable. This is often the case when the variable is wrapped up in a function. The same type of process can be done by making a substitution similar to the ones above. It is important to remember to substitute back to the original variable if you do this.

\begin{equation*} \begin{aligned} 2 \sin(x) + \sqrt{2} \amp = 0 \\ 2 y + \sqrt{2} \amp = 0 \amp \eqnspacer \amp \text{Substitute $y = \sin(x)$} \\ 2 y \amp = -\sqrt{2} \amp \amp \text{Subtract $\sqrt{2}$ from both sides} \\ y \amp = - \frac{\sqrt{2}}{2} \amp \amp \text{Divide both sides by $2$} \\ \sin(x) \amp = - \frac{\sqrt{2}}{2} \amp \amp \text{Substitute back to the original variable} \end{aligned} \end{equation*}

Try it!

Solve the equation \(4 \exp(x) + 6 = \ln(4)\) for Do it once using a substitution similar to the example above, and then do it without that substitution. Use a complete presentation both times.

Solution.
\begin{equation*} \begin{aligned} 4 \exp(x) + 6 \amp = \ln(4) \\ 4 y + 6 \amp = \ln(4) \amp \amp \text{Substitute $y = \exp(x)$} \\ 4 y \amp = \ln(4) - 6 \amp \amp \text{Subtract $6$ from both sides} \\ y \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Divide both sides by $4$} \\ \exp(x) \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Substitute back to the original variable} \\ \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} 4 \exp(x) + 6 \amp = \ln(4) \\ 4 \exp(x) \amp = \ln(4) - 6 \amp \amp \text{Subtract $6$ from both sides} \\ \exp(x) \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Divide both sides by $4$} \end{aligned} \end{equation*}