## Solving SBT Lesson 2: Real numbers. Absolute Value of a Real Number (C2 Math 7 Horizon) – Math Book

Solving SBT Lesson 2: Real numbers. Absolute value of a real number (C2 Math 7 Horizon)
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### Solve problem 1 page 40 SBT Math 7 Creative horizon episode 1

Replace the sign with the symbol ∈ or ∉ for the correct statement.

3.9 ? Z

29% ? Q

$$\sqrt 7$$ ? Q

$$– \dfrac{4}{{99}}$$ ? Q

$$\sqrt 3$$ ? I.

$$\sqrt 5$$? CHEAP

$$\pi$$ ? I

Solution method

We use the definition of sets of numbers to fill in the symbols .

Detailed explanation

We have 3.9 which is a rational number that is not an integer, so 3.9 ∉ Z.

We have 29% = $$\dfrac{{29}}{{100}}$$ (where 29, 100 ℤ and 100 0) so 29% Q

We have $$\sqrt 7$$≈2,645751311 which is an infinite non-recurring decimal number so $$\sqrt 7$$ is an irrational number and an irrational number is not a rational number so $$\sqrt 7$$ Q

We have: $$– \dfrac{4}{{99}}$$ (where 4; 99 ∈ ℤ and 99 ≠ 0) so $$– \dfrac{4}{{99}}$$ ∈ Q

We have: $$\sqrt 3$$≈1,732050808… is an infinite non-cyclic decimal number so $$\sqrt 3$$ ∈ I

We have: $$\sqrt 5$$≈2,236067977… is an infinite non-recurring decimal number, so $$\sqrt 5$$ is an irrational number, where real numbers include rational and irrational numbers, so $$\sqrt 5$$ R

We have π ≈ 3.141592654… is an infinite non-recurring decimal number, so $$\pi$$ is irrational, so $$\pi$$ ∈ I

### Solve problem 2 page 40 SBT Math 7 Creative horizon episode 1

Arrange the following real numbers in order from smallest to largest: $$\dfrac{4}{5}$$;0,(8); $$\sqrt 3$$;$$– \pi$$;$$– 3,142$$; 2

Solution method

We write the numbers as decimals and then sort them in ascending order.

Detailed explanation

We have:

$$\begin{array}{l}\dfrac{4}{5} = 0.8\\0,(8) = 0.888…\\\sqrt 3 = 1.732…\\ – \pi \approx – 3, 14159….\end{array}$$

Because -3.142 < -3.14159…. < 0.8 < 0.888…< 1.732…< 1.74 < 2, so the numbers arranged in order from smallest to largest are: $$– 3,142;\, – \pi ;\,\dfrac{4}{5 };\,0,(8);\,\sqrt 3 ;\,1.74;\,2$$

### Solve problem 3 pages 40 SBT Math 7 Creative horizon episode 1

Indicate whether the following statements are true or false:

a) $$\sqrt 4$$;$$\sqrt 9$$;$$\sqrt {25}$$ are irrational numbers;

b) Irrational numbers are not real numbers;

c) $$– \dfrac{1}{2};\dfrac{2}{3}; – 0.45$$ are rational numbers;

d) The number 0 is an irrational number;

e) 0.1; 0; 9; 99% are rational numbers.

Solution method

An irrational number is a number that can be written as an infinite non-recurring decimal.

A rational number is a number that can be written as $$\dfrac{a}{b}$$ (with $$a,b \in Z; b \ne 0$$)

Detailed explanation

a) We have:

22 = 4 (2 > 0) so $$\sqrt 4$$ = 2 is a rational number, and a rational number is not an irrational number;

32 = 9 (3 > 0) so $$\sqrt 9$$ = 3 is a rational number, and a rational number is not an irrational number;

52 = 25 (5 > 0) so $$\sqrt {25}$$ = 5 is a rational number, and a rational number is not an irrational number.

It follows that $$\sqrt 4 ;\sqrt 9 ;\sqrt {25}$$ are rational numbers. Hence a) is false.

b) Real numbers include rational numbers and irrational numbers, so irrational numbers are real numbers. Hence b) is false.

c) We have:

$$– \dfrac{1}{2}$$ (where -1; 2 ∈ ℤ, 2 ≠ 0) is a rational number;

$$\dfrac{2}{3}$$ (where 3; 2 ∈ ℤ, 3 ≠ 0) is a rational number;

−0.45=$$– \dfrac{{45}}{{100}}$$ (where -45; 100 ∈ Z, 100 ≠ 0) is a rational number;

It follows that $$– \dfrac{1}{2};\dfrac{2}{3}; – 0.45$$ are rational numbers. Hence c) is true.

d) The number 0 is a rational number and zero is an irrational number. Hence d) is false.

e) We have: 0.1 = $$\dfrac{1}{{10}}$$ (where 1; 10 ∈ Z, 10 ≠ 0) is a rational number;

0 = $$\dfrac{0}{1}$$ (where 0; 1 ∈ ℤ, 10 ≠ 0) is a rational number;

9 = $$\dfrac{9}{1}$$ (where 9; 1 ∈ ℤ, 1 ≠ 0) is a rational number;

99% =$$\dfrac{{99}}{{100}}$$ (where 9; 100 ∈ Z, 100 ≠ 0) is a rational number.

Deduce 0.1; 0; 9; 99% are rational numbers. Hence e) is correct.

### Solution 4 page 41 SBT Math 7 Creative horizon episode 1

Please replace the sign ? with the appropriate numbers:

a) 9.289 > 9.2 ? 79;

b) -0.3489 > -0.34 ? 8.

Solution method

We apply the rule to compare two decimal numbers.

Compare the digits in the corresponding row of 2 numbers from left to right.

Detailed explanation

a) These two decimals have the same integer part, from left to right the first two decimal places are equal.

Because 9 > 7 so let 9.289 > 9.2?79, the number to enter can be: 0; first; 2; 3; 4; 5; 6; 7; 8.

So the appropriate numbers to replace the sign ? is 0; first; 2; 3; 4; 5; 6; 7; 8.

b) $$-0.3489 > -0.34 ? 8 \Leftrightarrow 0.3489 < 0.34?8$$

These two decimals have the same integer part, from left to right the first and second decimal places are equal.

Since 9 > 8, so 0.3489 < 0.34?8, then the number to enter can only be: 9.

So the appropriate numbers to replace the sign ? is 9.

### Solution 5 page 41 SBT Math 7 Creative horizon episode 1

Find the reciprocal of the following numbers: $$\pi$$; 25%; – 5;$$– \sqrt {11}$$; $$– \dfrac{3}{5}$$

Solution method

We use the definition of the reciprocal of a number.

Detailed explanation

The argument of $$\pi$$ is $$-\pi$$ ;

The opposite of 25% is – 25%;

The opposite of – 5 is – (– 5) = 5;

The argument of $$– \sqrt {11}$$ is $$– \left( { – \sqrt {11} } \right)$$= $$\sqrt {11}$$

The argument of $$– \dfrac{3}{5}$$ is $$– \left( { – \dfrac{3}{5}} \right)$$= $$\dfrac{3}{5}$$

### Solution 6 page 41 SBT Math 7 Creative horizon episode 1

Find the absolute value of the following numbers: $$\sqrt 9$$;– 23; – 90%;$$\dfrac{5}{4}$$;– $$\pi$$

Solution method

+ If $$a \ge 0$$ then $$|a| = a$$

+ If $$a < 0$$ then $$|a| = -a$$

Note: $$|a| \ge 0$$ for all real numbers $$a$$

Detailed explanation

We have:

Since $$\sqrt 9 = 3 >0$$ $$|\sqrt 9| =|3|=3$$;

Since – 23 < 0, so |– 23| = –(– 23) = 23;

Since – 90% < 0, so | – 90%| = – (– 90%) = 90%;

Since $$\dfrac{5}{4}>0$$ $$| {\dfrac{5}{4}}|$$=$$\dfrac{5}{4}$$

Since – $$\pi$$ < 0, so |– $$\pi$$ | = – (– $$\pi$$ ) = $$\pi$$ .

### Solve problem 7 page 41 SBT Math 7 Creative horizon episode 1

Arrange in order from smallest to largest the absolute value of the following numbers: – 1.99; 1.9; $$– \sqrt 3$$;$$1\dfrac{1}{9}$$

Solution method

We represent the numbers in decimal form and then find the absolute value and sort.

+ If $$a \ge 0$$ then $$|a| = a$$

+ If $$a < 0$$ then $$|a| = -a$$

Note: $$|a| \ge 0$$ for all real numbers $$a$$

Detailed explanation

+) We have:

Since – 1.99 < 0, so |– 1.99| = – ( – 1.99) = 1.99;

Since 1.9 > 0, so |1.9| = 1.9;

Since $$– \sqrt 3 < 0$$ $$\left( { – \sqrt 3 } \right)$$=$$– \left( { – \sqrt 3 } \right)$$=$$\ sqrt 3$$

Since $$1\dfrac{1}{9}$$> 0 then $$|1\dfrac{1}{9}|$$=$$1\dfrac{1}{9}$$

+) Compare absolute values:

Since 0 < 9, 1.90 < 1.99 or 1.9 < 1.99 (1)

We have: $$\sqrt 3 = 1.7320500808…$$ ; $$1\dfrac{1}{9}=1+\dfrac{1}{9}=1+0,(1)=1,(1)$$

Since 1 < 7 < 9, 1, (1) < 1.732050805… < 1.9 (2)

From (1) and (2) deduce 1,(1) < 1.732050805… < 1.9 < 1.99 or $$1\dfrac{1}{9}$$; $$\sqrt 3$$; 1.9; 1.99.

So, in order from smallest to largest, the absolute values ​​of the following numbers: – 1.99; 1.9; −$$\sqrt 3$$; $$1\dfrac{1}{9}$$ is: $$1\dfrac{1}{9}$$;$$\sqrt 3$$; 1.9; 1.99.

### Solve problem 8 page 41 SBT Math 7 Creative horizon episode 1

Find the value of x, knowing that: $$2\left| x \right| = \sqrt {12}$$

Solution method

Because of the nature of the absolute value, we will have to divide 2 cases of x, then use the sign conversion rule to find x.

Detailed explanation

$$\begin{array}{l}2\left| x \right| = \sqrt {12} \\ \Leftrightarrow \left| x \right| = \sqrt {12} :2\\ \Leftrightarrow \left| x \right| = \dfrac{{\sqrt {12} }}{2}\end{array}$$

$$\Rightarrow x = \dfrac{{\sqrt {12} }}{2}$$or $$x = – \dfrac{{\sqrt {12} }}{2}$$

So $$x = \dfrac{{\sqrt {12} }}{2}$$or $$x = – \dfrac{{\sqrt {12} }}{2}$$

### Solve lesson 9 page 41 SBT Math 7 Creative horizon episode 1

Find the value of y, knowing that: |2y – 5| = 0

Solution method

We use the definition of absolute value to find y

Note: |a| = 0 when a = 0

Detailed explanation

|2y – 5| = 0

2y – 5 = 0

2y = 5

y = 5 : 2

y = $$\dfrac{5}{2}$$

So y = $$\dfrac{5}{2}$$

### Solution 10 page 41 SBT Math 7 Creative horizon episode 1

Expression reduction: M =$$\sqrt {{a^2}}$$

Solution method

We use the definition of square root to reduce the expression M, we need to divide 2 cases by the number in the root greater than and less than 0.

Detailed explanation

TH1 : If a < 0 then –a > 0, we have : $${\left( { – a} \right)^2} = {a^2}$$so $$\sqrt {{a^2} } = – a$$

TH2 : If a $$\ge$$ 0, we have : $$\sqrt {{a^2}} = a$$

So M = $$\sqrt {{a^2}} = \left| a \right|$$when a < 0 and a > 0

### Solve lesson 11 page 41 SBT Math 7 Creative horizon episode 1

Given a square with an area of ​​5m2. Compare the length a of the side of the square with the length b = 2.361 m.

Solution method

Find the side length of the square and compare it with b = 2.361 m.

Note: The length of the side of a square with area a $$cm^2$$ is $$\sqrt{a}$$ $$cm$$

Detailed explanation

Since the area of ​​a square is equal to the square of the side length, the side length is equal to the arithmetic square root of the area.

The length a of the side of the square is:

a=$$\sqrt 5$$=2,236067977… (m)

We have: $$\sqrt 5$$=2,236067977…

Since 2 < 3, 2.236067977… < 2.361 or $$\sqrt 5$$ < 2.361.

So the length of side a of the square is $$\sqrt 5$$ and a < b.