Reading and writing mathematics

Reading mathematics is an active process. A short line such as

$$f(x)=\frac{3x-1}{x+2},\qquad x\ne-2$$

contains an instruction, a structure and a restriction. It defines the output of $f$ from an input $x$, says that the entire expression $3x-1$ is divided by $x+2$, and excludes $x=-2$ because that value makes the denominator zero.

Writing mathematics reverses the process. You arrange symbols and words so that another reader can recover your meaning and verify every important step. Correct answers matter, but a solution should also show why the answer follows.

Prerequisites

You should be able to:

• perform arithmetic with fractions, negative numbers, powers and roots;
• substitute into simple expressions;
• solve a linear equation;
• recognise basic algebraic notation.

Review mathematical language and notation whenever a symbol is unfamiliar. Review algebraic fluency if manipulating expressions is still difficult.

Read in layers

Do not try to absorb a complicated line in one glance. Read it in layers.

1. Identify the object. Is it an expression, equation, function, inequality, definition or conclusion?
2. Find the main operation or relation. In an equation this is often $=$. In an expression it may be an outer fraction, power or function.
3. Group the parts. Fractions, brackets, roots and function arguments create units of meaning.
4. State restrictions. Look for denominators, square roots, logarithms, domains and contextual constraints.
5. Translate into words. A correct verbal reading exposes many silent mistakes.

Worked example 1: uncover the structure

Read

$$y=2-\sqrt{\frac{x+3}{5}},\qquad x\geq-3.$$

Start at the outside. The main relation is $y=$. On the right, $2$ has a square root subtracted from it. The root contains the fraction $(x+3)/5$.

A precise reading is:

$y$ equals two minus the principal square root of the quotient of $x+3$ and $5$, for $x$ at least $-3$.

The order of operations is therefore

$$x\longmapsto x+3 \longmapsto \frac{x+3}{5} \longmapsto \sqrt{\frac{x+3}{5}} \longmapsto 2-\sqrt{\frac{x+3}{5}}.$$

For real values, the expression inside the square root must be non-negative:

$$\frac{x+3}{5}\geq0 \quad\Longrightarrow\quad x\geq-3.$$

This restriction is part of the mathematics, not an optional note.

The fraction bar groups expressions

The expression $\dfrac{3x+1}{2x-5}$ means $(3x+1)\div(2x-5)$. It does not mean $3x+\dfrac1{2x}-5$. When writing on one line, use brackets: (3x+1)/(2x-5).

Worked example 2: distinguish similar notation

Let $f(x)=x^2+1$. Compare

$$f(3),\qquad f(x+3),\qquad f(x)+3.$$

The brackets after $f$ contain its input:

$$f(3)=3^2+1=10.$$

For $f(x+3)$, replace every input $x$ in the rule by $x+3$:

$$f(x+3)=(x+3)^2+1=x^2+6x+10.$$

By contrast,

$$f(x)+3=(x^2+1)+3=x^2+4.$$

Thus $f(x+3)\ne f(x)+3$ in general. The first changes the input; the second changes the output. See functions for a fuller treatment.

Parse before calculating

An operation can act on one term or on a whole expression. Mark the scope before doing algebra.

Written form	Structure	Meaning
$-x^2$	$-(x^2)$	square first, then negate
$(-x)^2$	$(-x)(-x)$	square the negative quantity
$\sin^2x$	$(\sin x)^2$	square the sine value
$\sin(x^2)$	sine applied to $x^2$	square the input first
$\sqrt{x+4}$	root of the whole sum	both terms lie under the root
$\sqrt{x}+4$	root of $x$, then add $4$	only $x$ lies under the root

Worked example 3: evaluate exactly

Evaluate

$$-x^2+3(x-1)$$

when $x=-2$.

Substitute using brackets:

$$\begin{aligned} -x^2+3(x-1) &=-(-2)^2+3((-2)-1)\\ &=-4+3(-3)\\ &=-13. \end{aligned}$$

The first term is $-((-2)^2)=-4$. It is not $(-(-2))^2=4$. Brackets preserve the structure during substitution.

Self-check 1

Let $g(t)=3t-2$. Find:

1. $g(-4)$;
2. $g(2u)$;
3. $2g(u)$;
4. the real values of $x$ for which $\sqrt{7-2x}$ is defined.

Answers

1. $g(-4)=3(-4)-2=-14$.
2. $g(2u)=3(2u)-2=6u-2$.
3. $2g(u)=2(3u-2)=6u-4$.
4. Require $7-2x\geq0$. Hence $-2x\geq-7$, so $x\leq7/2$. The inequality reverses when dividing by $-2$.

Read definitions as tests

A mathematical definition gives necessary and sufficient conditions for using a term. Read both directions.

For example, an integer $n$ is even if there exists an integer $k$ such that

$$n=2k.$$

This says:

• if $n$ is even, it can be written as twice an integer;
• if $n$ can be written as twice an integer, it is even.

The phrase ”$k$ is an integer” matters. Without it, every real number could be written as $2k$ for some real $k$.

Worked example 4: use a definition

Show that the sum of two odd integers is even.

Let the odd integers be $2r+1$ and $2s+1$, where $r,s\in\mathbb Z$. Their sum is

$$\begin{aligned} (2r+1)+(2s+1) &=2r+2s+2\\ &=2(r+s+1). \end{aligned}$$

Since integers are closed under addition, $r+s+1\in\mathbb Z$. The sum has the form $2k$ for an integer $k$, so it is even.

Merely checking $3+5=8$ gives supporting evidence, but it does not establish the claim for all odd integers. Study deduction and exhaustion to develop proof structure.

Follow the logical direction

Each new line must follow from information already established. The symbols used between lines should state the relationship accurately.

• $A=B$ says two quantities have the same value.
• $A\Rightarrow B$ says if statement $A$ is true, then statement $B$ follows.
• $A\Longleftrightarrow B$ says each statement implies the other.
• $A\approx B$ says the values are approximately equal.

Worked example 5: reversible and irreversible steps

Solve

$$\sqrt{x+6}=x.$$

The principal square root is non-negative, so any solution must satisfy $x\geq0$. Squaring gives a necessary condition:

$$\begin{aligned} \sqrt{x+6}=x &\Rightarrow x+6=x^2\\ &\Rightarrow x^2-x-6=0\\ &\Rightarrow (x-3)(x+2)=0. \end{aligned}$$

The candidates are $x=3$ and $x=-2$. Check them in the original equation:

$$x=3:\quad \sqrt9=3 \quad\checkmark$$ $$x=-2:\quad \sqrt4\ne-2 \quad\times$$

Therefore

$$\boxed{x=3}.$$

Squaring is not reversible over the reals because $a^2=b^2$ allows $a=b$ or $a=-b$. Writing $\Longleftrightarrow$ at the squaring step would incorrectly claim equivalence and conceal the extraneous candidate.

Read connecting words mathematically

Words such as therefore, because, if, only if, for all and there exists carry logical information. Do not skim them. “If $P$, then $Q$” does not automatically mean “if $Q$, then $P$”.

Read displays, tables and diagrams critically

Mathematics is not only prose and algebra.

For a graph, identify:

• the variables and axis scales;
• whether the axes start at zero;
• units and permitted domain;
• intercepts, turning points, discontinuities and asymptotes;
• whether values are exact or estimated from the picture.

For a table, read headings and units before the entries. Decide whether rows are raw observations, frequencies, probabilities or calculated values. For a diagram, use only stated facts and standard markings. A shape that looks like a right angle is not necessarily one unless it is marked or follows from other information.

Worked example 6: read a rate from units

Suppose $V$ is measured in cubic centimetres and $t$ in seconds, with

$$\frac{dV}{dt}=-12.$$

The derivative is read “the rate of change of volume with respect to time”. Its units are

$$\frac{\text{cm}^3}{\text{s}}=\text{cm}^3\text{ s}^{-1}.$$

The negative sign says the volume is decreasing. It does not mean that the volume itself is negative. A complete interpretation is:

At that instant, the volume is decreasing at $12\text{ cm}^3\text{ s}^{-1}$.

See connected rates of change for problems in which several rates interact.

Write a solution that can be audited

A strong solution usually contains four kinds of information:

1. Definitions: state what introduced symbols represent.
2. Method: show the equation, formula or principle being used.
3. Transformation: give enough intermediate algebra to locate an error.
4. Conclusion: answer the actual question, with restrictions, units and suitable accuracy.

Worked example 7: turn scratch work into a solution

Solve

$$\frac{2x-1}{3}-\frac{x+2}{2}=1.$$

A readable solution clears both denominators using their lowest common multiple, $6$:

$$\begin{aligned} 2(2x-1)-3(x+2)&=6\\ 4x-2-3x-6&=6\\ x-8&=6\\ x&=14. \end{aligned}$$

Check in the original equation:

$$\frac{2(14)-1}{3}-\frac{14+2}{2} =\frac{27}{3}-\frac{16}{2} =9-8 =1.$$

Therefore $x=14$. The first line shows that the whole numerator $x+2$ is multiplied by $3$. Omitting brackets would produce the common error $-3x+2$ instead of $-3x-6$.

Worked example 8: communicate a contextual answer

A rectangle has length $x+4$ metres and width $x-1$ metres. Its area is $40\text{ m}^2$. Find its dimensions.

Form the equation:

$$(x+4)(x-1)=40.$$

Then

$$\begin{aligned} x^2+3x-4&=40\\ x^2+3x-44&=0\\ (x+11)(x-4)&=0. \end{aligned}$$

Thus $x=-11$ or $x=4$. However, lengths must be positive. If $x=-11$, both $x+4$ and $x-1$ are negative, so this algebraic solution is invalid in context. For $x=4$,

$$x+4=8,qquad x-1=3.$$

Hence the rectangle measures

$$\boxed{8\text{ m by }3\text{ m}}.$$

The boxed result answers the question. Boxing $x=4$ alone would leave the reader to infer the requested dimensions.

Use words where symbols alone are ambiguous

Equations should carry the calculation; words should carry decisions and meaning. Useful short statements include:

• “Since $x>0$, take the positive square root.”
• “Let $t$ be the time in seconds after release.”
• “The denominator is non-zero because $x\ne3$.”
• “This root is rejected because probability cannot exceed $1$.”
• “The model predicts a maximum height of $18.4$ m.”

Avoid chains that mix unrelated equalities. For example,

$$x^2=9=x=3$$

is false and also loses $x=-3$. Write

$$x^2=9 \quad\Longrightarrow\quad x=\pm3.$$

Similarly, do not write

$$A=\pi r^2=\pi(5)^2=78.5\text{ cm}^2$$

unless $A$ is actually the area throughout and the final value has been rounded appropriately. A clearer calculation is

$$\begin{aligned} A&=\pi r^2\\ &=\pi(5)^2\\ &=25\pi\text{ cm}^2\\ &\approx78.5\text{ cm}^2. \end{aligned}$$

This distinguishes exact equality from approximation.

Common weaknesses and repairs

Weak writing	Why it is weak	Repair
x = 3, 5	unclear and non-standard	write $x=3$ or $x=5$
$\sqrt9=\pm3$	the principal square root is $3$	write $\sqrt9=3$, but $x^2=9\Rightarrow x=\pm3$
a decimal in the middle of exact work	loses information	retain fractions, surds and $\pi$ until the end
unexplained calculator output	hides the model and method	show the equation or substitution first
no units	leaves the quantity uninterpreted	attach units to the final contextual answer
every algebra line joined by $=$	may claim false equality	use $=$ for values and $\Rightarrow$ for implications
rejecting a root without a reason	the reader cannot check the decision	cite the domain or contextual restriction

Self-check 2

1. Explain the difference between $\dfrac{1}{x}+2$ and $\dfrac{1}{x+2}$.
2. Solve $\sqrt{2x+3}=x$, checking all candidates.
3. A calculation gives $p=1.08$, where $p$ is a probability. What should you conclude?
4. Rewrite the claim ”$n^2$ is odd, so $n$ is odd” as an implication.

Answers

1. In $1/x+2$, only $1$ is divided by $x$. In $1/(x+2)$, $1$ is divided by the whole sum $x+2$. Their domains also differ: $x\ne0$ and $x\ne-2$, respectively.
2. A solution requires $x\geq0$. Squaring gives $2x+3=x^2$, so $x^2-2x-3=0$ and $(x-3)(x+1)=0$. The candidates are $3$ and $-1$. Only $x=3$ satisfies the original equation. Therefore $x=3$.
3. A probability must lie in $[0,1]$, so $1.08$ is impossible. Check the model, data entry and arithmetic rather than rounding it to $1$ without justification.
4. $n^2\text{ is odd}\Rightarrow n\text{ is odd}$. If a proof is required, prove the implication rather than only restating it.

A final reading and writing routine

Before starting a question:

1. identify what is given and what must be found;
2. define symbols, units and domains;
3. mark the structure of complicated expressions;
4. translate command words such as solve, show, prove, estimate and interpret into an action.

Before finishing:

1. check substitutions in the original statement;
2. check signs, scale, units and domain restrictions;
3. distinguish exact values from approximations;
4. state a conclusion that answers the question.

Next, study problem solving to choose and adapt methods, mathematical modelling to connect symbols with real situations, and communicating reasoning to develop examination solutions and proofs.