1 ESO - UNIT 02 - POWERS AND SQUARE ROOTS

1. Unit 02 October
1. POWERS.
1.1. INDEX OR EXPONENT NOTATION.
Instead of writing 2 𝑥𝑥 2 𝑥𝑥 2 𝑥𝑥 2 𝑥𝑥 2 we can write 25
:
In 25
, the 2 is called the base number and the 5 is the index, power or
exponent. The index is the number of times the base number appears in the product.
This notation enables us to quickly write long lists of identical numbers being
multiplied together.
34
is the short way of writing 3 𝑥𝑥 3 𝑥𝑥 3 𝑥𝑥 3
106
is the short way of writing 1,000,000 = 10𝑥𝑥10𝑥𝑥10𝑥𝑥10𝑥𝑥10𝑥𝑥10
MATH VOCABULARY: Base Number, Index, Exponent. Distribution.
1.2. NAMING POWERS.
65
may be read as:
• Six to the fifth power
• Six to the power of five
• Six powered to five.
NOTE: During this curse use the first way.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.1

2. Unit 02 October
1.3. SQUARE AND CUBE POWERS.
We call Square Power a number to the second power. It represent the result of
multiplying a number by itself. The verb "to square" is used to denote this operation.
Squaring is the same as raising to the power 2.
We call Cube Power a number to the third power. It represent the result of the
number multiplied by itself twice. The verb "to cube" is used to denote this operation.
Cubing is the same as raising to the power 3.
MATH VOCABULARY: Square, Cube, Raise, To Factorize.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.2

3. Unit 02 October
1.4. POWERS OF TEN.
MULTIPLYING BY POWERS OF 10
When we multiply by:
10 we make a number 10 times larger
100 we make a number 100 times larger
1000 we make a number 1000 times larger
When we multiply by 1 0 … 0���
𝑛𝑛 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧
we add n zeros onto the end of the whole
number.
35 ∙ 10,000 = 350,000
When we use 10 as a base, the index shows the place value or number of zeros
following the one.
We remember that in expanded notation we write the number as the sum of
its place values.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.3

4. Unit 02 October
For example:
5,042 = (5 ∙ 1000) + (4 ∙ 10) + (2 ∙ 1)
Power notation is expanded notation written with powers of 10.
5,042 = (5 ∙ 103
) + (4 ∙ 101
) + (2 ∙ 100
)
NOTE: 100
= 1.
DIVIDING BY POWERS OF 10
When we divide by:
10 we make a number 10 times smaller
100 we make a number 100 times smaller
1000 we make a number 1000 times smaller
When we divide by 1 0 … 0���
𝑛𝑛 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧
we remove n zeros onto the end of the whole
number.
35,000 ÷ 1,000 = 35
230,000 ÷ 102
= 230,000 ÷ 100 = 2,300
USING POWERS OF 10 FOR SHORTENING LARGE NUMBERS
We can use the powers of 10 for shortening large numbers to simplify them.
For example a light-year has 9,460,800,000,000 km. We can rounded to
9,500,000,000,000 km. That is equal to 95 ∙ 100,000,000,000 = 95 ∙ 1011
𝑘𝑘𝑘𝑘.
MATH VOCABULARY: Power of 10, Expanded Notation, Power Notation, Larger, To
Remove, To Shorten, Kilometres, Capacity, Even Number, Odd Number, Gram.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.4

5. Unit 02 October
2. PROPERTIES OF POWERS.
2.1. MULTIPLICATION.
When powers with the same base are multiplied, the base remains unchanged
and the exponents are added.
𝒂𝒂𝒏𝒏
∙ 𝒂𝒂 𝒎𝒎
= 𝒂𝒂𝒏𝒏+𝒎𝒎
75
∙ 73
= (7 ∙ 7 ∙ 7 ∙ 7 ∙ 7) ∙ (7 ∙ 7 ∙ 7) = 75+3
= 78
When we have powers with different base but the same exponent, we multiply
the bases and keep the same exponent. We may also do it in the reciprocal way.
( 𝒂𝒂 ∙ 𝒃𝒃)𝒏𝒏
= 𝒂𝒂𝒏𝒏
∙ 𝒃𝒃𝒏𝒏
53
∙ 73
= (5 ∙ 5 ∙ 5) ∙ (7 ∙ 7 ∙ 7) = (5 ∙ 7) ∙ (5 ∙ 7) ∙ (5 ∙ 7) = (5 ∙ 7)3
= 353
64
= (3 ∙ 2)4
= 34
∙ 24
2.2. DIVISION.
When powers with the same base are divided, the base remains unchanged
and the exponents are subtracted.
𝒂𝒂𝒏𝒏
÷ 𝒂𝒂 𝒎𝒎
= 𝒂𝒂𝒏𝒏−𝒎𝒎
75
÷ 73
= (7 ∙ 7 ∙ 7 ∙ 7 ∙ 7) ÷ (7 ∙ 7 ∙ 7) = 75−3
= 72
When we have powers with different base but the same exponent, we divide
the bases and keep the same exponent. We may also do it in the reciprocal way.
( 𝒂𝒂 ÷ 𝒃𝒃)𝒏𝒏
= 𝒂𝒂𝒏𝒏
÷ 𝒃𝒃𝒏𝒏
Axel Cotón Gutiérrez Mathematics 1º ESO 2.5

6. Unit 02 October
153
÷ 33
= (15 ∙ 15 ∙ 15) ÷ (3 ∙ 3 ∙ 3) = (15 ÷ 3) ∙ (15 ÷ 3) ∙ (15 ÷ 3) = (15 ÷ 3)3
= �
15
3
�
3
= 53
2.3. POWER OF A POWER.
The exponents must be multiplied:
( 𝒂𝒂𝒏𝒏) 𝒎𝒎
= 𝒂𝒂𝒏𝒏∙𝒎𝒎
(23)5
= 23
∙ 23
∙ 23
∙ 23
∙ 23
= 23+3+3+3+3
= 23∙5
= 215
2.4. SPECIAL POWERS.
We have two special powers:
𝒂𝒂𝟎𝟎
= 𝟏𝟏 (𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒒𝒒𝒒𝒒𝒒𝒒 𝒂𝒂 ≠ 𝟎𝟎)
1,6780
= 1
𝒂𝒂𝟏𝟏
= 𝒂𝒂
1,6781
= 1,678
Solved exercise: Simplify:
(𝑎𝑎2
∙ 𝑎𝑎1)4
÷ (𝑎𝑎6
÷ 𝑎𝑎3)3
= (𝑎𝑎2+1)4
÷ (𝑎𝑎6−3)3
= (𝑎𝑎3)4
÷ (𝑎𝑎3)3
= 𝑎𝑎3∙4
÷ 𝑎𝑎3∙3
= 𝑎𝑎12
÷ 𝑎𝑎9
= 𝑎𝑎12−9
= 𝑎𝑎3
Axel Cotón Gutiérrez Mathematics 1º ESO 2.6

7. Unit 02 October
3. SQUARE ROOTS.
3.1. SQUARE ROOTS.
The square root of the square number 9 is written as √9. It is the positive
number which when squared gives 9. This symbol is called Radical Sign and the
number 9 is called Radicand.
Since 32
= 9 ⇒ √9 = 3
Examples: “Find out the following square”:
√36 = 6; √100 = 10; √64 = 8
A perfect square is a number that is the square of another natural number.
Examples: “Construct the list of the first eight perfect square numbers”.
They are: 1, 4, 9, 16, 25, 36, 49 and 64, because:
12
= 1; 22
= 4; 32
= 9; 42
= 16; 52
= 25; 62
= 36; 72
= 49;82
= 64
3.2. ESTIMATING SQUARE ROOTS.
Sometimes the square root of a number is not a natural number. In those cases
we can estimate the value of a non-exact square root giving the natural number which
is smaller but nearer that number.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.7

8. Unit 02 October
ESTIMATING BY TRIAL
Imagine you want to estimate by trial √3,900, we start for example with the
number 60:
602
= 3,600 < 3,900
612
= 3,721 < 3,900
622
= 3,844 < 3,900
632
= 3,969 > 3,900
So 3,900 is greater than 622
and smaller than 632
. That is 62 < √3,900 < 63.
We can write √3,900 ≈ 62.
THE SQUARE ROOT ALGORITHM
Let´s do it with an example. We will calculate �105,674
STEPS:
1. We separate in pairs, starting by the right side the radicand´s figures, and we
calculate the square root of the first pair (√10):
√10 . 56 . 74 3 ↤ 𝐴𝐴
3 ∙ 3 ↦ −9 6 ↤ 𝐵𝐵
1
𝐴𝐴 = √10 = 3 𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑅𝑅 = 1
𝐵𝐵 = 3 ∙ 2 = 6 (𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡ℎ𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑)
2. We down the following pair (56) and we look for the figure “C”, so that 6𝑪𝑪 × 𝑪𝑪 is
the closer number to 156
Axel Cotón Gutiérrez Mathematics 1º ESO 2.8

9. Unit 02 October
√10 . 56 . 74 3
3 ∙ 3 ↦ −9 6𝑪𝑪 × 𝑪𝑪
1 56
We can check that 6𝟐𝟐 ∙ 𝟐𝟐 = 124 is the closer number to 156, because
6𝟑𝟑 ∙ 𝟑𝟑 = 189 is greater than our target. So 𝑪𝑪 = 𝟐𝟐.
√10 . 56 . 74 3
3 ∙ 3 ↦ −9 6𝟐𝟐 × 𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏
156
62 × 2 = 124 → 124
032
3. We up the value of “C”, to the solution field, and we down the following pair and
we repeat the whole process.
√10 . 56 . 74 32
3 ∙ 3 ↦ −9 6𝟐𝟐 × 𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏
156 645 × 5 = 3,225
62 × 2 = 124 → 124
03274
3225
0049
We can check that 64𝟓𝟓 ∙ 𝟓𝟓 = 3,225 is the closer number to 3,274. 64 = 32 ∙ 2
4. We up the new value of “C”, to the solution field.
Answer: �105,674 = 325, and we have a remainder of 49. Proof: 3252
+
49 = 105,674
MATH VOCABULARY: Square Root, Root, Radical Sign, Radicand, Perfect Square, Trial,
Closer, Target, Proof.
Axel Cotón Gutiérrez Mathematics 1º ESO 2.9

10. Unit 02 October
Axel Cotón Gutiérrez Mathematics 1º ESO 2.10