Lesson Objective

Dear learner,

At the end of this lesson, you will be able to:

• Define real numbers.
• Compare two real numbers.
• Calculate the midpoint of a given interval by using the formula $ \frac{a+b}{2}$.
• Express real number intervals in inequalities, and number line diagrams.
• Use exponents and radicals rules to simplify the numbers.
• Round the numbers to the nearest numbers.

Key Terms

• Intervals
• Rational numbers
• Upper bounds
• lower bounds
• Rationalization

Brainstorming Questions

1 .Imagine you are comparing the prices of two types of local produce in a market in Addis Ababa: tomatoes and onions. If the cost per kilogram of tomatoes is $\frac{3\sqrt{2}}{5} ETB$ and the cost per kilogram of onions is $2.5 ETB$, determine which produce is cheaper. Use the properties of real numbers to make the comparison.

2. You are considering selling a piece of land between two locations: one at 8 kilometers and another at 15 kilometers from the city center of Addis Ababa. Determine the midpoint between these two locations to set an average price for the land.

Solution

1. To compare $ \frac{3\sqrt{2}}{5} ETB$ with 2.5 ETB, first convert $( \frac{3\sqrt{2}}{5} )$ to a decimal. The value of:

$ \sqrt{2} \approx 1.414 $, so: $\frac{3\sqrt{2}}{5} \approx \frac{3 \times 1.414}{5} \approx \frac{4.242}{5} \approx 0.8484 \text{ ETB}$.

Since 0.8484 ETB (for tomatoes) is less than 2.5 ETB (for onions), tomatoes are cheaper than onions. This comparison uses the trichotomy property, which states that for any two real numbers, one is less than, equal to, or greater than the other.
2. To find the midpoint between 8 km and 15 km, calculate the average:

$\text{Midpoint} = \frac{8 + 15}{2} = \frac{23}{2} = 11.5 \text{ kilometers}$.

Thus, the average distance is 11.5 kilometers. This method uses the concept of intervals to find a number between two given numbers.

4.1. Defining Real numbers

Definition:

A number is called a real number, if and only if it is either a rational number or an irrational number. The set of real numbers, denoted by R, can be described as the union of the sets of rational and irrational numbers. It can be expressed as
R = { x : x is a rational number or an irrational number} or diagrammatically,

i. Comparing real numbers

• You have seen that there is a one to one correspondence between a point on the number line and a real number.
• Suppose two real numbers 𝑎 and 𝑏 are provided. In this scenario, one of the following statements holds true:𝑎 < 𝑏, 𝑎 = 𝑏, or 𝑎 > 𝑏. This property is known as the trichotomy property in mathematics.
• For any three real numbers 𝑎, 𝑏, and 𝑐, if 𝑎 < 𝑏 and 𝑏 < 𝑐, then it follows that 𝑎 < 𝑐. This property is known as the transitive property of order in mathematics.
• Applying the above properties, we have Given two non-negative real numbers 𝑎 and 𝑏, if $a^{2} < b^{2}$, then 𝑎 < 𝑏.

Example 1:

Compare each pair

a) $\frac{\sqrt{3}}{5}$, 0.41

b) $\frac{2}{\sqrt{3}}$, 1.12

Solution:

a. When squaring the numbers $\frac{\sqrt{3}}{5}$ and 0.41, we obtain $\frac{3}{25}$ and 0.1681 respectively. Dividing 3 by 25 results in 0.12. Comparing 0.12 and 0.1681 reveals that 0.12 is less than 0.1681. Therefore, it can be concluded that $\frac{\sqrt{3}}{5}$ is less than 0.41.

Solution:

b. When squaring the numbers $\frac{2}{\sqrt{3}}$ and 1.12, we obtain $\frac{4}{3}$ and 1.2544 respectively. Dividing 4 by 3 results in 1.333….. Comparing 1.33…. and 1.2544 reveals that 1.333… is greater than 1.2544. Therefore, it can be concluded that $\frac{2}{\sqrt{3}}$ is greater than 1.12.

ii. Determining real numbers between two numbers

The number between a and b can be calculated as the average of a and b, which is given by the formula $\frac{a + b}{2}$. This formula provides the midpoint between the two numbers a and b.

Example 2:

Find a real number between 2 and 6.8.

Solution:

To find a real number between 2 and 6.8, we can simply take the average of the two numbers.

Average of 2 and 6.8 is given by $\frac{2 + 6.8} {2} = \frac{8.8}{2} = 4.4$

Therefore, a real number between 2 and 6.8 is 4.4.

Example 3:

Find at least two real numbers between −0.54 and −0.765.

Solution:

To find at least two real numbers between -0.54 and -0.765, we can follow these steps:

Step 1:

Find the average of the two given numbers:

$-0.54 + (-0.765) = -1.305$
$-\frac{1.305}{ 2} = -0.6525$

Step 2:

Choose a number between the average and the lower bound: Between -0.6525 and -0.54, we can choose -0.6 as one real number.

Step 3:

Choose a number between the average and the upper bound: Between -0.6525 and -0.765, we can choose -0.7 as another real number.

Therefore, two real numbers between -0.54 and -0.765 are -0.6 and -0.7.

4.2. Intervals

Dear learner,

A real interval is a set that includes all real numbers lying between two specified numbers. For instance, let us consider a real number 𝑥 that falls between two given real numbers 𝑎 and 𝑏 where.

The symbol ‘∞’ represents infinity, signifying endlessness or the absence of an end to the right, while ‘−∞’ denotes negative infinity, indicating endlessness or the absence of an end to the left. Real numbers within intervals can be represented using a point ‘a’ or ‘b’, along with ‘∞’ and ‘−∞’, as illustrated in the table below.

i) Real numbers between 3 and 5 including both end points.

ii) Real numbers between – 2 and $\frac{3}{2}$ including – 2 and excluding $\frac{3}{2}$.

iii) Real numbers on the right of 0 including 0.

Solution:

i) Real numbers between 3 and 5 including both endpoints:
a) Interval notation: $[3, 5]$
b) Inequalities: $3 ≤ x ≤ 5$
c) Number line representation:

ii) Real numbers between -2 and $\frac{3}{2}$ including -2 and excluding $\frac{3}{2}$:
a) Interval notation: [-2, $\frac{3}{2}$)
b) Inequalities: $ -2 ≤ x < \frac{3}{2}$
c) Number line representation:

iii) Real numbers on the right of 0 including 0:
a) Interval notation: [0, ∞)
b) Inequalities: $x \ge 0$
c) Number line representation:

4.2.1. Absolute values

Dear learner,

The absolute value of a real number, represented as |𝑥|, is defined as |𝑥| = 𝑥 if 𝑥 is greater than or equal to 0, and as -𝑥 if 𝑥 is less than.

Example 5:

Find the absolute value of each of the following.

a) – 5

b) $\sqrt{5} – 8$

Solution:

a) $\left| -5 \right| = 5$

b) $\left| \sqrt{5} – 8 \right| = – (\sqrt{5} – 8)$, because $(\sqrt{5} – 8) < 0$. Therefore: $\left| \sqrt{5} – 8 \right| = 8 – \sqrt{5}$

Example 6:

Find the distance between – 3 and 7?

Solution:

$7 – (- 3) = 7 + 3 = 10$

Example 7:

Determine the unknown ‘𝑥’ for a) $\left| x \right| – 5 = – 3$ b) $\left| x – 5 \right| = – 3$ c) $\left| x – 5 \right| – 5 = – 3$

Solution:

a) $\left| x \right| – 5 = – 3, \left| x \right| – 5 + 5 = – 3 + 5$ (add 5 to both sides), $\left| x \right| = 2$, therefore $x = 2$ or $x = – 2$.

b) $\left| x – 5 \right| = – 3$, absolute value shows the distance. thus, it can not be negative. Therefore, $\left| x – 5 \right| = – 3$ is undefined.

c) $\left| x – 5 \right| – 5 = – 3, \left| x – 5 \right| – 5 + 5 = – 3 + 5$ (add 5 to both sides),

$\left| x – 5 \right| = 2, x – 5 = 2$ or $x – 5 = – 2$.

$x = 2 + 5$ or $x = – 2 + 5$.

Therefore: $x = 7$ or $x = 3$.

4.3 Exponents and Radicals

If $a$ is a real number and $n$ is a positive integer, then,

$a \times a \times a \times a \times a \times … \times a = a^{n}$

$a^n$ is an exponential expression,

where $a$ is the base and $n$ is the exponent or power.

Example 8:

$3^{4} = 3 \times 3 \times 3 \times 3 = 81$,

$2^{6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

Note:

1. Zero exponent: If a ≠ 0, then $a^{0}$ = 1.
2. Negative exponent: If a ≠ 0 and 𝑛 is positive integer, then: $a^{-n} = \frac{1}{a^{n}}$

Definition

• If $a^{2} = b$, then $a$ is a square root of $b$.
• If$ a^{3} = b$, then $a$ is a cube root of $b$.
• If n is a positive integer and $a^{n} = b$, then a is called the $n^{th}$ root of $b$.
• If 𝑏 is any real number and 𝑛 is a positive integer greater than 1, then the principal 𝑛^𝑡ℎ root of 𝑏 is denoted by $\sqrt[n]{b}$ and is defined as the unique real number 𝑥 such that $x^{n}= b$.
• The expression $\sqrt[n]{b}$ is called a radical expression, where $\sqrt{}$ represents the radical sign, $n$ denotes the index, and $b$ is the radicand. When the index is not specified, the radical sign signifies the principal square root.
• If 𝑏 ∈ ℝ and n is an odd positive integer greater than 1, then

$b^{\frac{1}{n}} = \sqrt[n]{b}$.

• If 𝑏 ≥ 0 and n is an even positive integer greater than 1, then

$b^{\frac{1}{n}} = \sqrt[n]{b}$.

i. Law of exponents

Dear learner,

For any 𝑎, 𝑏 ∈ ℝ and 𝑛, 𝑚 ∈ ℕ, the following statement is true:

• $a^{n} \times a^{m} = a^{n+m}$
• $\frac{a^{n}}{a^{m}} = a^{(n-m)}$, where $a \neq 0$
• $(a^n)^m = (a^m)^n = a^{nm}$
• $(a \times b)^n = a^n \times b^n$
• $\frac{a^{n}}{b^{n}} = \left( \frac{a}{b}\right)^{n}$, for $(b \neq 0)$

Please note that these rules also apply to the $(1/n)^{th}$ power and rational power of the form

$a^{\frac{m}{n}} = (a^{\frac{1}{n}})^{m}$.

Example 9:

Use the above rules and simplify each of the following.

a. $5^{\frac{1}{3}} \times 5^{\frac{2}{3}} $

b. $\frac{7^{4}}{7^{2}}$

c. $11^{3}\times 3^{3} $

d. $64^{\frac{1}{3}}$.

Solution:

a. $5^{\frac{1}{3}} \times 5^{\frac{2}{3}} =5^{\frac{1}{3}+{\frac{2}{3}}} = 5^{\frac{3}{3}}=5$.

b. $\frac{7^{4}}{7^{2}}=7^{4-2}=7^2=49 $.

c. $11^{3}\times 3^{3}=(11\times 3)^3 =33^3=35,937$.

d. $64^{\frac{1}{3}}=(4^3)^{\frac{1}{3}}=4^1=4$.

ii. Addition and Subtraction of Radicals

Radicals that have the same index and the same radicand are said to be like radicals.

Example 10:

$-5\sqrt{7}, 3\sqrt{7}, \frac{-4}{11}\sqrt{7}, 0.75\sqrt{7}$ are like radicals and $\sqrt{5}, 3\sqrt{7}, \sqrt[3]{7}, \sqrt[4]{7}, …$ are all unlike radicals

Example 11:

Simplify $7\sqrt{27} – 12\sqrt{48} + 6\sqrt{75}$.

Solution:

$7\sqrt{27} – 12\sqrt{48} + 6\sqrt{75}$

$7\sqrt{9\ast 3} – 12\sqrt{16\ast 3} + 6\sqrt{25\ast 3}$

$7(3)\sqrt{3} – 12(4)\sqrt{3} + 6 (5)\sqrt{3}$

$21\sqrt{3} – 48\sqrt{3} + 30\sqrt{3}$

$51\sqrt{3} – 48\sqrt{3}$

$3\sqrt{3}$.

4.4 Limit of accuracy

Dear learner,

• Measurements are crucial in daily life for tasks like taking a child’s temperature, estimating time, measuring medicine, and determining weights, areas, and volumes. Sometimes, exact values may not be possible, leading to the need for approximate values.
• In this subtopic, you will learn mathematical concepts related to approximation, including rounding numbers, significant figures (s.f.), decimal place (d.p.), and accuracy.

i. Rounding

• Rounding off is a form of estimation used in daily life, Mathematics, and Physics. Physical quantities such as money amounts, distances, and lengths are estimated by rounding the actual number to the nearest whole number or specific decimal places.
• When rounding whole numbers, identify the “rounding digit.” For rounding to the nearest 10, the rounding digit is the second number from the right (tens place). For rounding to the nearest hundred, the third number from the right (hundreds place) is the rounding digit. To round off:

First, identify your rounding digit, then locate the digit to its right.

• If the digit is 0, 1, 2, 3, or 4, keep the rounding digit the same. Any digits to the right of the rounding digit become zero.
• If the digit is 5, 6, 7, 8, or 9, the rounding digit increases by one. All digits to the right of the rounding digit become zero.

Example 12:

67,612 people live in a town. Round this number to various levels of accuracy. 38,721
a. Nearest 100: The number rounds up to 67,600, as the rounding digit is 6. The digit to the right is 1, so the rounding digit remains the same and all digits to the right become zero.
b. Nearest 1,000: The number rounds up to 68,000. The rounding digit is 7, and the digit to the right is 6, causing the rounding digit to increase by one to 8, with all following digits becoming zero.
c. Nearest 10,000: The number rounds up to 70,000. The rounding digit is 6, and the digit to the right is 7, leading the rounding digit to increase by one to 7, with all following digits becoming zero.

In this scenario, it is unlikely that the exact number will be provided. Instead, the result is less precise but more user-friendly.

ii. Decimal place

A number can be approximated to a specific number of decimal places (d.p.), which indicates the figures written after a decimal point.

Note that: Approximating a number to 1 decimal place involves rounding to the nearest tenth, while approximating to 2 decimal places means rounding to the nearest hundredth.

Example 13:

Write 32.746 to a) 1 d.p b) 2 d.p c) 3 d.p d) 4 d.p

Solution:

a) To round 32.746 to 1 decimal place, we look at the digit in the second decimal place, which is 4. Since 4 is less than 5, we do not need to round up. Therefore, 32.746 rounded to 1 decimal place is 32.7.

b) To round 32.746 to two decimal places, we look at the third decimal place, which is 6. Since 6 is equal to or greater than 5, we round up the second decimal place. Therefore, 32.746 rounded to two decimal places is 32.75.

c) The number 32.746 round to 3 decimal places, we look at the digit in the fourth decimal place, which is 6. Since 6 is equal to or greater than 5, we round up the last digit in the third decimal place. Therefore, 32.746 rounded to 3 decimal places is 32.746.

d) To write 32.746 to 4 decimal places, we look at the digit in the 4th decimal place, which is not stated. Since the digit to the right of 6 is 0, we do not need to round up. Therefore, 32.746 to 4 decimal places is 32.7460.

Significant figures

Numbers can be approximated to a specific number of significant figures (s.f.). In the number 73.298, the 7 is the most significant figure with a value of 70, while the 8 is the least significant with a value of 8 thousandths. To show the accuracy of approximation using significant figures, count the digits from the first non-zero digit from left to right. This count represents the number of significant figures.

Example 14:

a. Write 80.573 to 3 s.f.

b. Write 0.0084 to 1 s.f.

c. Write 7038 to 1 s.f.

Solution:

a) First, 80.547 has five significant figures. To write the number in 3 s.f., we observe the fourth significant figure, which is 7. Since this number is not below 5, we round the third significant figure, 5, to 6. Therefore, the 3 s.f. form of 80.573 is 80.6.

b) n this case, the significant digits are 8 and 4. The number 8 is the most significant digit, and we will decide whether to round up 4. Since 4 is less than 5, we write 0.0084 as 0.008 to 1 significant figure.

c) When rounding 7038 to 1 significant figure, the result would be 7000.

Accuracy

Dear learners,

In this lesson, you will learn how to approximate upper and lower bounds for data with a specified accuracy, such as rounding numbers or expressing them to a specific number of significant figures.

Simply to find upper bound add zero at the last and add 5 from the last added zero, and to find the lower bound add zero at the last and subtract 5 at the last add zeros.

Example 15:

Find the upper and lower bound for a) 3.2 b) 3.20 c) 3.200

Solution:

a. the upper bound for 3.2 is add 0.05, 3.2 + 0.05 = 3.25 and the lower bound for 3.2 is subtract 0.05, 3.2 – 0.05 = 3.15.

b) the upper bound for 3.20 is add 0.005, 3.2 + 0.005 = 3.205 and the lower bound for 3.20 is subtract 0.005, 3.20 – 0.005 = 3.105.

c) the upper bound for 3.200 is add 0.0005, 3.200 + 0.0005 = 3.2005 and the lower bound for 3.200 is subtract 0.0005, 3.20 – 0.005 = 3.1005.

Effect of operation on accuracy

For addition and multiplication two numbers

i. To find upper bounds: – if it’s addition: add the two numbers upper bounds (if it’s multiplication: multiply the two numbers upper bounds).

Example 16:

Find the upper bound for a) $3.2 + 4.5$ b) $3.2 x 4.5$.

Solution:

a). Upper bound for $(3.2 + 4.5) = 3.25 + 4.55 = 7.80$.

b). Upper bound for $(3.2 \times 4.5) = 3.25 \times 4.55 = 14.7875$

ii. To find lower bounds: – if it’s addition: add the two numbers lower bounds (if it’s multiplication: multiply the two numbers lower bounds).

Example 17:

Find the lower bound for a) $3.2 + 4.5$ b) $3.2 \times 4.5$.

Solution:

a)The lower bound of $3.2 + 4.5 = 3.15 + 4.45 = 7.60$.

b) The lower bound of $3.2 \times 4.5 = 3.15 \times 4.45 = 14.0175$.

For subtraction and division two numbers

i. To find upper bounds: – if it’s subtraction: subtract the lower bound of the second number from the upper bound of the first number (or if it’s division: divide the upper bound of the numerator to the lower bound of the denominator).

Example 18:

Find the upper bound for a) 9.2 – 4.5 is 9.25 – 4.45 = 4.80 b) $\frac{9.2}{4.5}$.

Solution:

a) upper bound for $9.2 – 4.5 = 9.25 – 4.45 = 4.80$.

b) upper bound for $\frac{9.2}{4.5} = \frac{9.25}{4.45} = 2.08$.

ii. To find lower bounds: – if it’s subtraction: Subtract the upper bound second number from the lower bound of the first numbers (or if it’s division: divide the lower bound numerator to upper bound of denominator).

Example 19:

Find the lower bound for a) $9.2 – 4.5$ b) $\frac{9.2}{4.5}$.

Solution:

a) $9.2 – 4.5$ is $9.15 – 4.55 = 4.60$.

b) $\frac{9.2}{4.5}$ is $\frac{9.15}{4.55} = 2.01$.

2.4.5 Standard notation (Scientific notation)

Definition:

A number is said to be in scientific notation (or standard notation) if it is written as a product of the form $a \ast 10^𝑛$ where $1 ≤ a < 10$ and 𝑛 is an integer.

Example 20:

Express each of the following in standard notation. a) $0.000000234$ b) $567,200,000,000$

Solution:

a) $0.000000234 = 2.34 \times 10^{-7}$ (the decimal points are moved 7 times to right to write as $d\times 10^{n}$ where $1\le d < 10$ and $n\in Z$.

b) $567,200,000,000 = 5.672 \times 10^{11}$ (the decimal points are moved 11 times to left to write as $d\times 10^{n}$ where $1\le d < 10$ and $n\in Z$.

Example 21:

Write each of the following in ordinary decimal notation. a) $2.34 \times 10^{-7}$ b) $5.672 \times 10^{11}$.

Solution:

a) $2.34 \times 10^{-7} = 00000000002.34 \times 10^{-7}$, then move the decimal points to left 7 times, it will give as $0.000000234$.

b) $5.672 \times 10^{11} = 5.67200000000 \times 10^{11}$, then move the decimal points to right 11 times, it will give as 567,200,000,000.

4.6 Rationalization

Dear learner,

When dealing with ratios involving irrational denominators, calculating the quotient can be challenging. To simplify the process, it is often beneficial to convert the irrational denominator into a rational one. This conversion helps in making the calculations more manageable and facilitates a clearer understanding of the ratio.

The number used to multiply and rationalize the denominator is called the rationalizing factor, which is equivalent to 1.

Example 22:

Rationalize the denominator for each of the following.

a) $\frac{2}{\sqrt{5}}$

b) $\frac{4}{\sqrt[4]{2^3}}$

c) $\frac{3}{3-\sqrt{11}}$

Solution:

a) The rationalizing factor for $\frac{2}{\sqrt{5}}$ is $\frac{\sqrt{5}}{\sqrt{5}}$. Thus, to rationalize the denominator of $\frac{2}{\sqrt{5}}$, we multiply it by the rationalizing factor $\frac{\sqrt{5}}{\sqrt{5}}$.

Therefore,

$\frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}$

$= \frac{2\sqrt{5}}{\sqrt{5}\times \sqrt{5}} = \frac{2\sqrt{5}}{5}$.

b) First the rationalizing factor of $\frac{4}{\sqrt[4]{2^3}}$ is

$\frac{\sqrt[4]{2^{3}}}{\sqrt[4]{2^{3}}}$.

Thus,

$\frac{4}{\sqrt[4]{2}} = \frac{4}{\sqrt[4]{2}}\times \frac{\sqrt[4]{2^{3}}}{\sqrt[4]{2^{3}}}$

$= \frac{4\sqrt[4]{2^{3}}}{\sqrt[4]{2}\times \sqrt[4]{2^{3}}} = \frac{4\times \sqrt[4]{8}}{\sqrt[4]{2\times2^{3}}}$

$= \frac{4\sqrt[4]{8}}{\sqrt[4]{2^{4}}} = 2\sqrt[4]{8}}$.

c) The rationalizing factor for $\frac{3}{3-\sqrt{11}}$ is $\frac{3+\sqrt{11}}{3+\sqrt{11}}$. Therefore, we follow the steps below

Step 1:

Multiply by the rationalizing factor. The rationalizing factor of

$(3-\sqrt{11})$ is $\frac{3+\sqrt{11}}{3+\sqrt{11}}$. Now we have $\frac{3}{3-\sqrt{11}} \times \frac{3+\sqrt{11}}{3+\sqrt{11}}$.

Step 2:

Simplify the expression by multiplying the numerators and denominators.

$= \frac{3(3+\sqrt{11})}{(3-\sqrt{11})(3+\sqrt{11})}$.

Step 3:

Use the difference of squares formula $((a-b)(a+b) = a^{2} – b^{2}$ to simplify the denominator.

$= \frac{3(3+\sqrt{11})}{3^{2} – (\sqrt{11})^{2}}=\frac{3(3+\sqrt{11})}{9 – 11}=-\frac{3(3+\sqrt{11})}{2}$.

4.7. Applications on Real numbers

Applications: Real numbers are used in everyday life for measurements, finance, engineering, and science. They help model distances, temperatures, weights, and more, providing a complete numerical system for continuous quantities.

Example 23:

a. A store had 150 customers on Monday, and the number of customers increased by 25% on Tuesday. How many customers visited the store on Tuesday?

Solution:

The number of customers on $Tuesday = 150 + (0.25 \times 150) = 150 + 37.5 = 187.5$.
Since the number of customers must be a whole number, the store had 188 customers on Tuesday.

b. Question: Solve the equation $3x – 5 = 7$ for $x$. Is the solution a natural number, an integer, a rational number, an irrational number, or a real number?

Solution:

$3x – 5 = 7$
$3x = 7 + 5$
$3x = 12$
$x = \frac{12}{ 3}$
$x = 4$.

The solution $x = 4$ is a natural number, an integer, a rational number, and a real number.

c. Question: A recipe calls for $\frac{3}{4}$ cup of sugar. If you want to make 2 batches of the recipe, how many cups of sugar will you need in total?

Solution:

You will need 1.5 cups of sugar in total for 2 batches of the recipe $\frac{3}{4} \times 2 = 1.5 cups$.

d. Identify whether the following numbers are rational or irrational: √2, 0.75, -3, π.

Solution:

– √2 is an irrational number.
– 0.75 is a rational number.
– -3 is an integer, which is also a rational number.
– π is an irrational number.

e. Order the following numbers from least to greatest: $-2, 0.5, √3, -\frac{1}{2}, 3.14$.

Solution:

$-2, -\frac{1}{2}, 0.5, √3, 3.14$.

f. A small town has 530 flower pots. The gardener wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.

Solution:

Total number of flower pots is 530. Number of flower pots in each row is 21. Now, we have to find how many groups of 21 in 530. So, we use Euclid’s Division Lemma, that is dividing 530 by 21 as shown in the right. The result indicates, the gardener can arrange the flower pots in 25 rows with each row consisting of 21 pots. Also the remaining number of flower pots is 5.