The principle of using mathematical terminology is basically choosing the right word for the right occasion. Here we split the examples according to the subject area.
Here we refer to things such as functions, polynomials and so on (as opposed to abstract algebra), which most secondary school students should know.
| Wrong | Correct | Comments |
|---|---|---|
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| The word ‘constant’ can be used as an adjective or a noun. As an adjective, it is used to describe the non-varying property of functions. As a noun, it is used to refer to a fixed and well-defined number and has a plural form ‘constants’. |
| Since $6=2\times 3$, so 6 divides 3. |
| A rule of thumb is that if $a$ divides $b$ (where $a$, $b$ are positive integers), then $a$ must be the smaller number. |
| We can move the graph of $y=e^x$ to the left by 1 unit to obtain the graph of $y=e^{x+1}$. | We can translate the graph of $y=e^x$ to the left by 1 unit to obtain the graph of $y=e^{x+1}$. | When referring to moving graphs by a fixed vector the verb ‘translate’ is used. |
| The multiplication of two positive numbers is positive. | The product of two positive numbers is positive. | The result of multiplication is ‘product’. |
| We count the number of integers between 1 and 100 which are not prime numbers. | We count the number of integers between 1 and 100 which are not prime. | There is redundant information in the wrong example (‘integers’ are ‘numbers’). The convention is to use only the adjective. |
| The thousand digit of 12345 is 2. | The thousands digit of 12345 is 2. | When referring to digits use the terms ‘unit digit’, ‘tens digit’, ‘hundreds digit’, etc. |
| The tenth digit of 12.34 is 1. | The tens digit of 12.34 is 1. | The tenth digit refers to the digit immediately to the right of the decimal point, i.e. 3 in this example. |
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| In mathematical proofs the convention is to use ‘if’ rather than ‘when’ for describing a condition. But in the second example, it is more popular to use ‘when’ since we are talking about a variable taking on a certain value. |
Plane geometry is another popular topic in the mathematics syllabus of secondary school.
| Wrong | Correct | Comments |
|---|---|---|
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Denote the centre of the circumcentre of $ABC$ by $O$. | Denote the circumcentre of $ABC$ by $O$. | ‘Circumcentre’ is the centre of the circumscribed circle. |
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| ‘Concyclic’ is used to describe points whereas ‘cyclic’ is used to describe polygons. |
| Let $ABCD$ form a square. |
| The correct examples are the mathematical conventions. |
| The length of the perimeter of the circle is $4\pi\,$cm. |
| In mathematics ‘circumference’ is the outer boundary of a circle, while ‘perimeter’ means ‘length of the boundary’. |
Mathematical induction is a widely used technique of proof in many branches of mathematics. Study the following proof. Can you point out all the problems in it?
Question. Show that $1+2+\cdots+n=\dfrac{n(n+1)}2$ for all positive integers $n$.
Solution.
Let $S(n)$ be the statement $1+2+\cdots+n=\dfrac{n(n+1)}2$ for all positive integers $n$.
Assume $S(k)$ is true for all positive integers $k$, s.t. $1+2+\cdots+k=\dfrac{k(k+1)}2$.
When $S(k+1)$, we have
\[\begin{align*} S(k+1) &= 1+2+\cdots+(k+1) \\ &= \frac{k(k+1)}2+(k+1) \\ &= \frac{k(k+1)+2(k+1)}2 \\ &= \frac{(k+1)(k+2)}2 \\ &= \frac{(k+1)[(k+1)+1]}2 \end{align*}\]Hence $n=k+1$ is also true.
By the principal of mathematical induction, $S(n)$ is true for all positive integers $n$.
There are close to ten errors in the proof. How many can you find?
| Wrong | Correct | Comments |
|---|---|---|
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Let $S(n)$ be the statement \[\boldsymbol{1+2+\cdots+n = \frac{n(n+1)}2}\]for all positive integers $\boldsymbol n$. | Let $S(n)$ be the statement \[\boldsymbol{1+2+\cdots+n = \frac{n(n+1)}2}.\] | $S(n)$ is a statement that depends on the value of $n$. We want to prove that $S(n)$ is true for all $n$. |
| $\boldsymbol{ 1+2+\cdots+1 }$ | $\boldsymbol1$ | We start adding from 1 and end at $n$. When $n=1$, there is only one term and we should not write ‘$+2$’ at all. |
| \[\begin{align*} \boldsymbol{1+2+\cdots+1} & \boldsymbol{ {}=\frac{1(1+1)}2} \\ \boldsymbol{1} & \boldsymbol{ {}=1} \end{align*}\] | \[\begin{align*} \textbf{LHS} & \boldsymbol{ {}=1} \\ \textbf{RHS} & \boldsymbol{ {}=\frac{1(1+1)}2 = 1} \\ \end{align*}\] | When proving an equality, we do not simplify both sides simultaneously. We may either start from one side and reach the other, or simplify both sides separately to obtain the same value or expression. |
| Assume $S(k)$ is true for all positive integers $k$. | Assume $S(k)$ is true for some positive integer $k$. | ‘$S(k)$ is true for all $k$’ is precisely the statement we need to prove, and therefore it does not make sense to assume $S(k)$ is true for all $k$. |
| s.t. $1+2+\cdots+k=\dfrac{k(k+1)}2$. | i.e. $1+2+\cdots+k=\dfrac{k(k+1)}2$. | Here we want to use the phrase ‘that is’ rather than ‘such that’, because we want to explain what we mean by $S(k)$ is true, rather than to talk about some consequences. |
| When $\boldsymbol{S(k+1)}$ | When $\boldsymbol{n=k+1}$ | When we say ‘when $n=\cdots$’, we are discussing what happens under that particular value of $n$. Here we need to prove that $S(k+1)$ is true, so we cannot say ‘when $S(k+1)$’. |
| $\boldsymbol{ S(k+1)=1+2+\cdots+(k+1) }$ | LHS $\boldsymbol{ {}=1+2+\cdots+(k+1)}$ | $S(k+1)$ is a statement and cannot be equal to $1+2+\cdots+(k+1)$. |
| Hence $\boldsymbol{n=k+1}$ is also true. | Hence $\boldsymbol{S(k+1)}$ is also true. | Only a statement can be true or false. $S(k+1)$ is a statement, but $n=k+1$ is not a statement since $n$ and $k$ are dummy variables and we cannot say $n=k+1$ is true or false. |
| By the principal of mathematical induction, $P(n)$ is true for all $n\geq 2$. | By the principle of mathematical induction, $P(n)$ is true for all $n\geq 2$. | ‘Principal’ is the head of a school and ‘principle’ is a general law or primary truth. |
These errors are specific to the use of terminology in the concepts of functions, limits, differentiation and integration.
| Wrong | Correct | Comments |
|---|---|---|
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$-1$ is not the domain of $\sqrt{x}$. |
| ‘Domain’ is a set and the intended meaning here is to clarify whether $-1$ is an element of this set. |
| The differentiation of $x^2$ is $2x$. | The derivative of $x^2$ is $2x$. | ‘Differentiation’ is the process of finding the derivative. |
| To solve for the points of intersection of the graphs, we suppose $x+1=2x+3$. | To solve for the points of intersection of the graphs, we set $x+1=2x+3$. | ‘Suppose’ is used for assumption. In this example there is no assumption; rather we assign two expressions to be equal to find $x$. |
| The function $f(x)=3x$ is strictly increasing in the interval $[0,1]$. | The function $f(x)=3x$ is strictly increasing on the interval $[0,1]$. | It is a mathematical convention to say ‘on an interval’ rather than ‘in an interval’. |
| Let $f(x)=\sin x$. When $\boldsymbol{ f(0) }$, we have $\boldsymbol{ \sin\, 0=0 }$. | Let $f(x)=\sin x$. When $\boldsymbol{ x=0 }$, we have $\boldsymbol{ f(0)=\sin\,0=0 }$. | The statement $\sin 0=0$ is true regardless of what happens to $f(0)$. Also, there must be a condition following the word ‘when’. For example, you could say ‘when $x$ is 2, the value of $f(x)$ is 5’. But in the example, $f(0)$ is not a condition. |
| Note that $\ln(\sin 0)$ has no solution. | Note that $\ln(\sin 0)$ is undefined. | We can only say that an equation (with some sort of variable) has no solution; but here $\ln(\sin 0)$ is a value (although undefined). |
| The $(k+1)$-th derivative of $\sin x$ is \[\boldsymbol{ \frac{d^kd}{dx^kx} } (\sin x) = \frac{d^k}{dx^k}(\cos x).\] | The $(k+1)$-th derivative of $\sin x$ is \[\boldsymbol{ \frac{d^k}{dx^k} } \left( \boldsymbol{ \frac{d}{dx} } (\sin x) \right) = \frac{d^k}{dx^k}(\cos x).\] | The notation in the first example is wrong. |
These errors are specific to the use of terminology in the theory of matrices and vector spaces.
| Wrong | Correct | Comments |
|---|---|---|
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The second entry of $A$ is 2. | The (1,2)-entry of $A$ is 2. | It is unclear what the ‘second’ entry of a matrix is — whether it is the second element in the first row or the first element in the second row. |
| Since $A$ is $3\times 4$ and $B$ is $4\times 5$, the matrix $\boldsymbol{A\times B}$ is $3\times 5$. | Since $A$ is $3\times 4$ and $B$ is $4\times 5$, the matrix $\boldsymbol{AB}$ is $3\times 5$. | Although not wrong, it is not a common practice to denote the multiplication of matrices using the ‘$\times$’ sign. Most of the time no symbol is used at all. |
| The matrix $A$ is positive. | The entries of the matrix $A$ are positive. | ‘Positive’ is used to describe real numbers. |
| For the $\boldsymbol{(2,3)}$-entry of $\boldsymbol A$ $= 2+3 = 5$. |
| ‘The $(2,3)$-entry of $A$’ can be equal to a number, but ‘for the $(2,3)$-entry of $A$’ is a phrase and cannot be equal to a number. |
| When $E$ is a Type III, … |
| ‘Type III’ is an adjective, so it should be followed by a noun, or we add the preposition ‘of’ beforehand. |
| If $A$ is rank $3$, then it must be invertible. |
| ‘$A$ is rank $3$’ is not a proper complete sentence. |
| We solve the characteristic polynomial of $A$ as follows. |
| We can ‘solve’ an equation but not a polynomial. |
| The matrix $A$ is linearly independent and so must be invertible. | The rows of the matrix $A$ are linearly independent and so must be invertible. | Linear independence refers to elements of a vector space (e.g.\ row/column vectors). The intended meaning here is that the row vectors of $A$ (multiple objects) are linearly independent, not the matrix $A$ itself (a single object) being linearly independent. |
| Hence $\boldsymbol{\mathbf{x},\mathbf{y}}$ is a basis. | Hence $\boldsymbol{ \lbrace \mathbf{x},\mathbf{y} \rbrace }$ is a basis. | A basis is a set of vectors in a vector space and therefore must be expressed as a set. |
| Since $A$ has rank 2, we have $\boldsymbol{ \textbf{null}\, A=3 }$. | Since $A$ has rank 2, we have $\boldsymbol{ \textbf{dim}\,(\textbf{null}\, A)=3 }$. | $\operatorname{null} A$ refers to the null space of $A$ and cannot be equal to a number. What is equal to 3 is the dimension of its null space (also called its nullity). |
| Since $A$ has rank 2, the rank of its null space is 3. | Since $A$ has rank 2, the dimension of its null space is 3. | ‘Null space’ is not a matrix and hence has no rank. |
| If a $3\times3$ matrix is invertible, its number of rank is 3. | If a $3\times3$ matrix is invertible, its rank is 3. | ‘Rank’ is already a number. |
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