Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Topic 2.1 Number types
Number type
Natural
Integers
Symbol
Explanation
N
Positive whole numbers
Z
Whole numbers, positive, negative and 0.
Rational
Q
Any number that can be written as a fraction. This
includes all the integers.
Irrational
-
Any number that can NOT be written as a fraction.
Real
R
All the numbers listed above.
Examples
1,2,3, ….
….-3,-2,1,0,1,2,…
1, -3, 0,
1
3
2, 3, 
Any of the above
Topic 2.2 What is an exact number?
Sometimes you are required to write answers as exact numbers. This is when the number is written as either a
fraction or in square root from, not a rounded number.
For example
3 in exact form is 3 , not 1.73.
1
1
should be written as not 0.333.
3
3
Topic 2.2 Significant figures and decimal places
These two are easily mixed up.
 All non-zeros numbers are always significant.
 Zeros in the front of a number are never significant.
 Zeros in the middle are always significant.
 Zeros at the end are only significant if they follow the decimal point (e.g. 0.003420 has 4 significant
figures).
If you are rounding to 3 significant figures remember that you have to round up the last significant number if the
fourth number is 5 or larger. Below are some different examples of numbers rounded to 3 significant figures.
Number
6.5879
0.002562
10354
34.6582
To 3 significant figures
6.59
0.00256
10400
34.7
Decimal places are easier to understand (and explain) simply write down the number of digits required after the
decimal point. Remember the last number required may need to be rounded up. Below are some examples of
numbers rounded to 3 decimal places.
Number
6.5879
0.002562
0.20316
34.6582
To 3 decimal places
6.588
0.003
0.203
34.658
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Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Exam answers and working
On your final examination your answers must be to 3 significant figures or as an exact number. Students often
loose marks on the examination on this criteria. Here are some simple techniques:
 Fractions and square roots are always considered exact numbers. But remember that you should never
leave a decimal inside of a fraction. (e.g. change
2.5
5
to ).
3
6

Do not round off numbers whilst working, try to develop techniques with your GDC to keep numbers in
your calculator, or take the time to write out the numbers in full.

Leave fractions as fractions, do not change
1
to 0.333.
3
Exponents
23  the number three is the index or power. Very simply 23 = 2  2  2 = 8
In the table below are some notes and facts about exponents.
Exponent
What does it do?
power of 1
The number remains the same.
power of 0
The number becomes 1.
Examples
51 = 5
451 = 45
x1 = x
50 = 1
450 = 1
x0 = 1
1
are equivalent to square roots.
2
1
Powers of are equivalent to cube roots
3
Powers of
rational exponents
1
16 2  16  4
1
27 3  3 27  3
51 
negative exponents
Multiple
Divide
Keep the base, add the exponents
Keep the base, subtract the exponents
24  25  29
y3  y 2  y5
25
 23
2
2
y3
 y10
7
y
3 4
Keep the base, multiply the powers
4 2 
Example
2 
Power to a power
1
1

2
4 16
1
1
1
1

25 2  1 

2
25 5
25
The negative power simple turns the number
into a fraction.
Operations with exponents
Operation
Explanation
1
5
y 
 212
2 6
 y 12 
1
y12
2
Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Topic 2.3 Standard form (a.k.a. scientific notation) a  10n where 1  a  10
Standard form is used to write large or small numbers in an easier format. The IB does not use the term
“standard form” but instead often phrases questions “in the form a  10n where 1  a  10.”
To convert numbers to standard form:
 Let’s choose a number 2,780,000
 First choose the a, this is always between 1 and 10, round this to 3 significant figures. So 2,780,000
becomes 2.78
 Secondly calculate your n value. You do this by counting how many times you have moved the decimal
place.
 Since 2780000 is a large number, the exponent will be positive.
 So in the number 2,780,000 we have moved the decimal point 5 places left so n becomes 5.
 The final answer is 2.78  105
Below are some more examples of numbers converted to standard form.
Number
Standard form
3780124
= 3.78  106
0.002419
= 2.42  10-3
0.000075278
= 7.53  10-5
873.62
= 8.74  102
Calculations in scientific notation
When performing calculations using scientific notation be sure to apply the properties shown above with
exponents.
Multiplication: Multiply the a values together. Add the n values together and write again in the form a  10n.
Finally adjust the first a so that it is between 1 and 10, adjusting the n value accordingly.
Division: Divide the first a value by the second. Subtract the n values and write again in the form a  10n.
Finally adjust the first a value so that it is between 1 and 10, adjusting the n number accordingly.
Guided example
Let x = 5.61  104 and y = 9.75  10-8.
Calculate (a) xy
(b) x ÷ y
Answer (a)




Answer (b)




5.61  9.75 = 54.7
104  10-8 = 10-4
So we have 54.7  10-4 (note this is not in standard form)
Convert to standard form 5.47  10-3
5.61 ÷ 9.75 = 0.575
104 ÷ 10-8 = 1012
So we have 0.575  1012 (note this is not in standard form)
Convert to standard form 5.75  1011
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Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Topics 2.5 and 2.6 Sequences and series
You need to know about two types of series: arithmetic sequence (or arithmetic progression) and geometric
sequence (or geometric progression). For each one you must be able to calculate the next term and find the sum
of terms.
What is the difference between the two?
An arithmetic sequence is one where the numbers progress by adding by the same number each time, known as
the common difference. Sometimes the common difference is negative.
A geometric sequence is one where the numbers progress by multiplying by the same number each time, known
of the common ratio. Sometimes the common ratio is a fraction less than one.
Sequence
6, 9, 12, 15, 18, ……
1, 3, 9, 27, 81, ……
25, 20, 15, 10, 5,……
Type
Arithmetic
Geometric
Arithmetic
1 1 1 1 1
, , , , ,....
2 4 8 16 32
Geometric
Sometimes the first challenge with these questions is to establish if it is an AP (Arithmetic Progression) or a GP
(Geometric Progression). Listing out the terms given should help with this.
The following formulae are given in the formula booklet.
The nth term of an arithmetic sequence :
un = u1 + d(n – 1)
The sum of n terms of an arithmetic sequence:
Sn 
The nth term of a geometric sequence:
un = u1rn – 1
The sum of n terms of a geometric sequence:
The sum of an infinite geometric sequence:
n
2u1  d (n  1)   n u1  u n 
2
2
u1 (r n  1) u1 (1  r n )
Sn 

,r 1
r 1
1 r
u
S  1 , r 1
1 r
where
un = the nth term of a sequence
Sn = the sum of the first n terms of a sequence
u1 = the first term of a sequence
d = the common difference for an arithmetic sequence
r = the common ratio of a geometric sequence
n = the number of terms, or the position of the term
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Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Guided example
Devon has a savings scheme. She starts off by adding $10, the next month she adds $12, the following
month $14 and so on.
(a) Calculate how much Devon adds on the 14th month.
(b) Calculate how much Devon has saved after 2 years.
Answer (a)
 Firstly, establish if it is an AP or a GP by listing out the sequence: 10, 12, 14, 16, 18, etc.
 As the same number is being added each time the sequence is an AP.
 Find the numbers to go into the formula: a = 10, d = 2, n = 14 (as you are finding the 14th month).
 Now put all these numbers into the formula un = 10 + (13•2).
 The answer is $36.
Answer (b)
 We have already established that it is an AP, and we know all the values.

Use the formula for the sum of a formula: S 24 
24
 2 10   24  1  2  = $792.
2
Example
A swimmer is training for a long distance race. She will start on the first day by swimming
50m, and each day increase the number of meters she swims by 10% for the first 30 days.
(a) Calculate how far the swimmer must swim on the 30th day.
(b) Calculate how many meters the swimmer swum in 30 days.
Answer
5
Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Topic 2.7 Solutions of pairs of linear equations in two variables by use of a GDC
If you have a pair of linear equations, you can solve it with your GDC.
 2x  3y  4
and you want to solve for x and y.
5 x  4 y  17
For example, consider the pair of equations: 
Remember that graphically this means where these two lines intersect. So solve each of these equations for y,
2
4

 y   3 x  3
plug them into the calculator and Calculate > Intersection. 
. The solution to this is (5, -2).
5
17
y   x 

4
4
Topic 2.7 Quadratic Equations
A quadratic equation is an equation is one in the form ax2 + bx + c = 0, where a 0.
Some examples of quadratic equations:
x2 – 5x + 6 = 0,
x2 = 25,
x2 + 16x = 0,
2x2 – 3x + 16 = 0
You need to know how to factor, solve, graph, and interpret graphs of quadratic functions (topic 4.3).
Topic 2.7 is only about solving quadratic equations. Topic 4.3 has more about quadratic functions.
Factoring
There are several different types of factoring, depending on the equations.
You could take out the GREATEST COMMON FACTOR (GCF).
You could factor a DIFFERENCE OF SQUARES.
You could factor a TRINOMIAL into the product of two binomials
Below are some examples of quadratic equations factored:
Quadratic equation
factored
2
4x – 12
4(x2 – 3)
6 – 3x2
3(2 – x2)
6x2 + 24x
10x2 – 15x
4x2 – 9
(2x – 3)(2x + 3)
2
x + 7x + 12
(x + 4) (x + 3)
2
6x + 13x – 5
(3x – 1)(2x + 5)
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Topic 2 Revision Notes
Number and Algebra
IB Math Studies
Solving quadratic equations
One way to solve a quadratic equation is to factor it and then set each factor equal to zero and solve.
Below are some examples of solved quadratic equations.
Quadratic Equation
Factored
Solved
3 and x =  3
4x2 – 12 = 0
4(x2 – 3) = 0
x=
6 – 3x2 = 0
3(2 – x2) = 0
x=
4x2 – 9 = 0
(2x – 3)(2x + 3) = 0
2 and x =  2
x = 32 and x =  32
x + 7x + 12 = 0
(x + 4) (x + 3) = 0
x = -4 and x = -3
6x2 + 13x – 5 = 0
(3x – 1)(2x + 5) = 0
x2 + 10x – 25 = 0
(x + 5) (x + 5) = 0
2
x=
1
3
and x =  52
x = -5 and x = -5
In the final example the solution occurred twice so it’s called a repeated solution.
Guided example
A classroom is to be built in the shape of a rectangle. The width is 3 meters less than the length and the area is
28m2.
(a) If the length is x meters, write down an expression for the width.
(b) Write a quadratic equation using the area of the rectangle.
(c) Find the length and width of the classroom.
Solution
Answer (a)
• As the width is 3 m less than the length we will have the answer x – 3
Answer (b)
• To get the area = length  width, so in this case x(x – 3) = 28
• Multiplying this out we have x2 – 3x = 28. This is now a quadratic equation. Setting it equal to zero we get
x2 – 3x – 28 = 0.
Answer (c)
• Now factor the quadratic equation to get (x – 7) (x + 4) = 0.
• We have the solutions x = 7 and x = -4. As we are looking for a length, it cannot be negative so we must
choose x = 7.
• The length is 7 meters and the width is 4 meters.
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