UNIT 2 – NOTATION, ORDER AND INEQUALITIES

 

Exercises:   [1]   [2]   [3]   [4]   [5]   [6]   [7]   [8]

 

INTRODUCTION

To communicate their ideas and results efficiently scientists use various symbols and shorthand forms such as CO2 for carbon dioxide, H2O or water or p for 3.14159… . The symbol S (sigma) meaning ‘add’ is another such abbreviation which is widely used, particularly in statistics.

When you read scientific articles or use instruments you will find that real numbers are usually written in a standard form called scientific notation. This is a convention for writing decimals using powers of 10.

So far you have worked with equalities and equations, but you do not have to look far to realise that this can be restrictive. If you are investigating rainfall, you may find that over the last 100years in only one year was there exactly 360 points (millimetres) or even more likely, that there was no year in which there was less than 360 points of rain. Or perhaps years in which the rainfall was between 700 and 1200 points. Problems of this sort lead to the study of inequalities, that is statements involving phrases such as greater than, less than and between.

 

OBJECTIVES

  1. To use S notation.

  2. To use scientific notation.

  3. To put real numbers in order using the inequality symbols ³ , , > and < .

  4. To represent inequalities as intervals on the real number line.

  5. To find the distance between two real numbers.

  6. To solve equations and inequalities involving |x|

 

ADVICE

You will need to use a calculator for some of the exercises in this unit.

 

 

S (SIGMA) NOTATION

This is a short way of writing ‘add the numbers…’. Suppose you have five measurements which you represent by the symbols x1, x2, x3, x4 and x5, then their sum

x1, x2, x3, x4 and x5 is abbreviated to .

Example 2a. if x1 = 1.0, x2 = 1.5, x3 = 0.9, and x4 = 1.2 then

= 1.0 + 1.5 + 0.9 + 1.2 = 4.6

= 1.02 + 1.52 +0.92 +1.22 = 5.5

= 2.6111… .

In general if there are n readings denoted by x1, x2, …, xn

then = x1 + x2 + …+ xn .

Exercises 2.1 Solutions

  1. If x1 = 1 , x2 = 2, x3 = 3 and x4 = 4, find

    2. (a) (b)

    (c) (d)

  2. If x1 = 2 , x2 = -3, x3 = 1, x4 = -2 and x5 = 0, find

    3. (a) (b)

    (c) (d)

  3. If x1 = x2 = … = xn = 2, find

 

DECIMAL NOTATION

When real numbers are written as decimals, it is customary to use only a certain number of figures.

e.g. 4/3 = 1.33333 correct to 5 decimal places,

p = 3.1416 correct to 4 decimal places,

100/3 = 33.33 correct to 4 significant figures,

2/61 = 0.0328 correct to 4 decimal places,

½ = 0.5 exactly.

So the decimal notation form of a number is often only an approximation to its actual value, with the accuracy of the approximation depending on the context in which the number is being used.

 

SCIENTIFIC NOTATION

In scientific work it is common to encounter very large and very small numbers. For such numbers decimal notation is cumbersome or impractical, for example

5/3126976 = 0.0000015990 correct to 10 decimal places,

= 0.00000 correct to 5 decimal places!

For uniformity and to retain accuracy, numbers are often written in a standard form with just one figure in front of the decimal point and multiplied by some power of ten.

e.g. 68,000 = 6.8 ´ 104

186,000 = 1.86 ´ 105

1.862 = 1.862 ´ 100

0.00068 = 6.8 ´ 10-4

e.g. A micron is a millionth of a metre, so that

1 micron (m ) = 1 ´ 10-6 m

This notation is commonly used by the manufacturers of electronic calculators; for example the value of is given by many calculators as

= 7.07107… ´ 10-1

 

Exercises 2.2 Solutions

  1. Write the following numbers in decimal notation

    2. (a) 1/8 (b) 13/5 (c) 22/7

  2. Write the following numbers in scientific notation

    3. (a) 246 (b) 5800 (c) 61.701

    (d) 0.000261 (e) 1/3

  3. What is the value of

    4. 1.02 ´ 102 + 2.5 ´ 103 ?

  4. Is 1.31 ´ 10-3 greater than or less than 13000.0 ´ 10-7 ?

  5. Are the following statements true or false?

    1. 1/103 = 10-3

    2. 1/10-3 = 103

  6. Rewrite ‘one millimetre is one thousandth of a metre’ in scientific notation.

 

REAL NUMBER LINE

A real number can be represented as a point on a line, as shown below in Figure 2.1. This line is called the real number line. The central point on it is zero. Numbers to the right of zero are positive and numbers to the left of zero are negative. 

______|______|______|______________|_____|________|_______|______|____
     -2  -1.436    -1              0   0.35       1      1½      2

Figure 2.1: Real number line

 

The integers are drawn as equally spaced points on the real number line. Between any pair of integers there is an infinite number of rational numbers, e.g. between 0 and 1 there would be also and so on. The rational numbers are said to be dense everywhere on the real line because between any two rational numbers (no matter how close) we can always find another one(for example their average). Also the positions of the irrational numbers can be marked on this line.

ORDER

A number a is said to be less than a number b , written a < b if a is to the left of b when they are drawn as points on the real line.

e.g. -2 < 0, 0.3567 < 0.591, -5 < -2.5.

Similarly a is greater than b, a > b if a is to the right of b on the real line.

e.g. 0.35 > 0, 2 > -1, p > 1, -5 > -10.

The symbol a b means a < b or a = b and similarly b ³ c means ‘b > c or b = c’.

 

Exercises 2.3 Solutions

Insert the correct symbol < or > between the following pairs of numbers:

1. -7 and 2 2. -1 and 3 3. and -2

4. 0 and -6 5. 3-1 and 3 6. -(-3) and 1

 

Note that , a > 0 means a is positive, and a < 0 means a is negative. Also a < b < c (Figure 2.2) means a < b and b< c (and therefore, of course, a < c ),

e.g. -2 < 1 < 5.

                  ___|____|____|___
                     a    b    c

Figure 2.2: a < b < c

INTERVALS, BOUNDS AND INEQUALITITES

The expression (or inequality) 2 < x < 5 (Figure 2.3) means that x may be any real number between 2 and 5 (but it cannot equal 2 or 5). 

              ___|_________|___
                 2<---x--->5

Figure 2.3: 2 < x < 5

 

Alternative expressions used to mean the same thing as 2 < x < 5 are:

  1. x is any real number in the open interval (2,5).

  2. x is (any real number0 in the open interval bounded by 2 and 5.

The term ‘open interval’ is used in the above example because the end-points (or bounds) of the interval (i.e. 2 and 5) are not possible values for the variable x.

If the end-points of the interval are possible values for the variable x, i.e. if a x b, x is said to be in the closed interval between a and b and the interval is written with square brackets as [a,b] (Figure 2.4).

 

[ x ]

a b

Figure 2.4: Closed interval [a,b]

 

Also if, for example, 0 < x 1 then x is said to be in the half-open (or half-closed) interval (0,1].

When an interval is specified by giving its bounds, the first number is the left end-point or the lower bound and the second the right end-point or upper bound.

e.g. 1 x 2 is written as [1,2] and not as [2,1]

(because 1 < 2).

The inequality x ³ 10 is expressed as x may be any real number greater than or equal to 10, no matter how large. It is represented on the number line by a ray (or infinite interval).

 

[ x

0 10

Figure 2.5: x ³ 10

 

To write x ³ 10 in interval notation, the symbol ¥ is used to represent infinity; ¥ is not a real number but it represent the idea of ‘something’ which is bigger than any number you can think of, no matter how large.

Using this notation, x ³ 10 is expressed as x is in the interval [10,¥ ]; the interval is bounded on the left by 10 but is said to be unbounded on the right.

Similarly, x -1 is represented on the real number line as shown in Figure 2.6.

 

x ]

-1 0

Figure 2.6: x -1

 

The interval corresponding to x -1 is (-¥ ,-1). It is unbounded on the left but bounded on the right by –1.

Sometimes it is possible to combine a pair of inequalities. For example, if x satisfies both the inequalities x < -1 and x > -2 , (as in Figure 2.7(a)), it must be in the interval (-2,1) so -2 < x 1

)

( x

-2 -1 0 1

Figure 2.7(a)

However, if x satisfies both the inequalities x < -2 and x > 1 it would be represented on the real number line by Figure 2.7(b).

 

x ) ( x

-2 0 1

Figure 2.7(b)

 

This is not a single interval so the inequalities cannot be combined.

 

Exercises 2.4 Solutions

  1. Write the following inequalities as intervals and represent them on the real number line:

    2. (a) 3 x < 5 (b) -1 x 1 (c) x > -2

  2. Represent on the real number line each of the following pairs of inequalities. Write them in interval notation. Where possible express the two inequalities as a single combined inequality

  1. x > 5 and x < 10

  2. x > 5 and x < 2

 

SIMPLIFYING INEQUALITIES

Can a statement involving inequalities such as 4x + 3 < x + 9 be simplified in the same way as the equation 4x + 3 = x + 9 can be solved to give x = 2?

Yes, but CARE IS NEEDED

  1. You can add or subtract a number on both sides of an inequality,

    2. e.g. 4 < 7

    so 4 + 2 < 7 = 2 (i.e. 6 < 9)

    or 4 – 1 < 7 – 1 (i.e. 3 < 6).

  2. You can multiply or divide both sides of an inequality by any POSITIVE number,

e.g. 3 < 5 so 2 ´ 3 < 2 ´ 5 or .

3. If you multiply or divide both sides of an inequality by a NEGATIVE number, you must reverse the direction of the inequality sign,

e.g. 2 < 4 but 2(-2) > 4(-2) since –4 > -8, i.e. –8 < -4

(see Figure 2.8).

 

-8 -6 -4 -2 0 2 4 6

 

Figure 2.8

4. If x > y then (provided x ¹ 0 and y ¹ 0) ,

e.g. 3 > 2 and

Example 2b. Using these properties of inequalities, simplify

4x + 3 < x + 9

Subtract 3 from both sides 4x + 3 –3 < x + 9 – 3,

so 4x < x + 6,

subtract x from both sides, 3x < 6,

divide both sides by 3 x < 2.

Example 2c. Simplify 3 5x 2x + 9.

Separate the inequality into two parts and find the values of x which satisfy both parts.

  1. 3 5x

    2. divide both sides by 5 .

  2. 5x
2x + 9

subtract 2x from both sides 3x 9

divide both sides by 3 x 3.

On the real number line, what sections represent both and x 3?

 

x

x 3

0 3

Figure 2.9

From Figure 2.9 the values of x which satisfy both inequalities are those in the interval [], i.e.

3 5x 2x + 9 can be simplified to 3.

 

Exercises 2.5 Solutions

Simplify the following inequalities:

  1. 3x – 2 < 7

  2. x + 2 < 7x – 1

  3. 3 – x ³ -4

  4. –7 2x + 5 7

  5. x + 1 3x + 2 < 5 + x

  6. Express the following statement mathematically using inequalities: ‘A mechanic expects to spend x hours on one job y hours on another but he does not want to spend more than eight hours total time on both jobs’.

 

 

DISTANCE ON THE REAL NUMBER LINE

The distance between two numbers (that is, the length of the interval between them) is obtained by subtracting the smaller one from the larger. 

      ____|______|______|____|______
         -1      0      2   2.6

Figure 2.10

e.g. the distance between 2.6 and 2.0 is 2.6 – 2.0 = 0.6 and the distance between 2 and –1 is 2 – (-1) = 2 + 1 = 3 (see Figure 2.10).

The symbol used to denote the distance between numbers a and b is

½ a - b½.

A distance must be either positive or zero; it cannot be negative.

e.g distance between 2.6 and 2.0 = ½ 2.6 - 2.0½ = 0.6,

distance between 2.0 and 2.6 = ½ 2.0 - 2.6½ = 0.6,

distance between 2 and (-1) = ½ 2 – (-1)½ = 0.6

note ½ 2 – (-1)½ can be written as ½ 2 + 1½ .

In general, ½ a - b½ and ½ b - a½ both represent the same distance. Also ½ a + b ½ means ½ a – (-b)½.

A special case of the use of this symbol is ½ x½ which denotes the distance between the number x and zero, i.e. ½ x½ = ½ x- 0½

e.g. |5| = distance between 5 and 0 = |5 – 0| = 5

|-2.5| = distance between (-2.5) and 0 = 2.5

|0| = distance between 0 and 0 = |0 – 0| = 0

Notice that

e.g. |3| = 3, |-3| = |0 – (-3)| = 3.

|x| is called the absolute value of x or the modulus of x.

 

Exercises 2.6 Solutions

  1. Calculate the distances between the following numbers:

    2. (a) 5 and 3 (b) 2.19 and 31.21

    (c) -1 and 1 (d) -2 and –5

    (e) 0 and 6 (f) 0 and -6

  2. If x = 2 and y = -5, evaluate the following:

(a) |x| (b) |y| (c) |x + y|

(d) |x| + |y| (e) |xy| (f) |x - y|

(g) |x| - |y| (h) |3|x| - 4|y| |

 

EQUATION INVOLVING |x|

If |x| = 6 then x must be some number at a distance of 6 units from 0 that is, x = 6 or x = -6 (see Figure 2.11).

6 6

X

-6 0 6

Figure 2.11: |x| = 6

 

 

To solve the equation

|x – 2| = 3

separate it into two parts using the fact that x – 2 = 3 or -(x - 2) = 3.

If x – 2 = 3 then x = 5.

If -(x – 2) = 3 then x – 2 = -3 so x = -1.

So | x – 2| = 3 has the two solutions, x =5 and x =-1.

Check this by substituting x = 5 into the left-hand side of the equation,

L.H.S. = |5 – 2| = |3| = 3

Since the right-hand side of the equation is 3, x = 5 is, a solution. Check x = -1 by substitution,

L.H.S. = |-1-2| = |-3| = 3 (=R.H.S.)

 

Exercises 2.7 Solutions

Solve the following equations:

1. |x| = 4

2. |3x + 2| = 8

3. |4x - 3| = 5

4. | + 2| = 1

INEQUALITITES INVOLVING |x|

You are 10 km from Spider Rock. Specimens of the species Spidercus bitercus are found anywhere within a distance of 3 km of Spider Rock. How far from you are they to be found?

Take your present position as zero (the origin) and let x be the distance between yourself and the animals (see Figure 2.12). 

        __you______________animals___________ SpiderRock____
           0                  x                    10

Figure 2.12

 

Then the distance between the animals and Spider Rock is |x – 10| and this is known to be less than 3 km.

|x – 10| < 3

To simplify this inequality, separate it into two parts

x – 10 < 3 and -(x – 10) < 3.

If x – 10 < 3, then x < 3 + 10, i.e. x < 13.

If -(x – 10) < 3, then x – 10 > -3, so x > 7.

 

Combining the two simplified inequalities x < 13 and x > 7 you have the answer to the question:

7 < x < 13

that is, the animals are found between 7 and 13 km from your present position.

 

Exercises 2.8 Solutions

Simplify the following inequalities:

  1. |x| < 2

  2. |x + 2| ³ 5

  3. |2x - 5| < 3