The 2012 Annual National Assessments for Grade 9 include a question to select a particular type of number from a list. Each of these carries one mark only, and thus not much is expected from the learner except to circle their choice on their answer paper. This blog article will talk you through two worked examples.

The Exemplar paper has the following question:

1.1 Which of the following numbers is irrational?

$\begin{array}{ll} \text{A} & 0,\dot8 \\ \text{B} & 0,5 \\ \text{C} & -\sqrt3 \\ \text{D} & \sqrt{2\frac14} \end{array}$

The actual paper includes the following question:

1.2 Which of the following numbers is a rational number?

$\begin{array}{ll} \text{A} & \sqrt3 \\ \text{B} & \sqrt{16} \\ \text{C} & \sqrt{-9} \\ \text{D} & \sqrt{13} \end{array}$

Both of these questions require that you fully understand the differences between the types of numbers and how to determine if one number is of a particular type.

The types of numbers

type	description	examples
whole number	these are the countable numbers which start at one and continue forever. All of the whole numbers fall within the integers.	examples include 1, 2, 3, 12, 15, 134, 156743, 222333444555666
integers	The integers include the whole numbers, as well as the number zero (0), and all of the negative numbers. All integers fall within the set of rational numbers.	1, 2, 4, 5, -7, -45, 0
rational numbers	The rational numbers are numbers that can be expressed in terms of the ratio of two integers.	$\frac41, \frac14, \frac{12}{13}, \frac{-2}3$
irrational numbers	These numbers cannot be expressed as the ratio of two integers, and within the school curriculum the irrational numbers that you will encounter are those that are expressed as a square root of a number which is not a perfect square. The perfect squares are the squares of the integers, so that 3×3 = 9, and thus 9 is a perfect square. However, there is no integer which can multiply by itself to get to 3 or 5, or 8.	$\sqrt8, \sqrt5, \sqrt3$

However, note that $\sqrt9 = 3$ , since 3×3 = 9 and thus $\sqrt9$ is not irrational but is an integers, and this is also a rational number.

Reviewing the numbers from question 1.1

Let us look at each of the numbers presented in these two examples, and determine the type of each. Answering these questions requires that we are able to determine the type, and also to understand how these types of numbers are related to each other.

$\text{(A) }0,\dot8$

The dot about the 8 indicated that this is a recurring decimal, so that this number can be written as

$0,88888888888...$

where the digit 8 simply continues forever.

One part of your work in decimal numbers shows you how to convert a recurring decimal into a fraction, and thus this number is rational.

$\text{(B) }0,5$

You will know that 0,5 can be expressed as the fraction one half, or $\frac12$ . From this it is clear that 0,5 is also rational.

Since the first question asks which is irrational, then we still have two options to choose from, and they both look similar, both using a square root, but with different values.

You may be tempted to say that either of these two is thus irrational, using the (false) reasoning that anything with a square root is irrational. So let us see why there is only ONE answer to this question.

$\text{(C) }-\sqrt3$

This number looks irrational, since 3 is not a perfect square, and from what I said above this is thus a possible candidate for being irrational.

$\text{(D) }\sqrt{2\frac14}$

This number is a little more complex, since we have a mixed number within the square root sign, and you may be tempted to choose this one just because it LOOKS more complex.

Let us break this mixed number down into its parts:

$2\frac14 = \frac94$

and both 9 and 4 are perfect squares, since 3×3 = 9 and 2×2 = 4.

As a result, we can rewrite this as:

$\sqrt{\frac{3 \times 3}{2 \times 2}} = \sqrt{\frac32 \times \frac32} = \sqrt{\left({\frac32}\right)^2} = \frac32$

Reviewing the numbers from question 1.1

$\text{(A) } \sqrt3$

From the previous definitions it is evident that 3 is not a perfect square, and this this square root is irrational.

$\text{(B) } \sqrt{16}$

The number 16 is a perfect square, being $4 \times 4$ , and thus

$\sqrt{16} = \sqrt{4 \times 4} = \sqrt{4^2} = 4$

This is thus a whole number, which is also an integer, which is thus also a rational number.

$\text{(C) } \sqrt{-9}$

Whereas 9 is a perfect square, being $3 \times 3$ , -9 is not a perfect square, since no integer can be multiplied by itself to yield -9. This square root falls within the definition of imaginary numbers, which are numbers that concern the square roots of negative numbers.

This is thus clearly not rational.

$\text{(D) }\sqrt{13}$

Finally, we have the same situation as the answer choice for $\sqrt3$ in which 13 is not a perfect square, and this this is thus an irrational number.

Analysis of these questions

The analysis of this question is that again, like the previous question 1.1, the numbers given cannot be taken at face value and must be analysed first before the question that can be answered.

When answered this type of question you should always yourself whether any of the numbers presented can be simplified first, as though there has been a question such as…

Simplify $\sqrt{16}$

mathematics for South Africa

The types of numbers

Reviewing the numbers from question 1.1

Reviewing the numbers from question 1.1

Analysis of these questions