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**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:

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

3^{2} = 9 (3 > 0) so \(\sqrt 9 \) = 3 is a rational number, and a rational number is not an irrational number;

5^{2} = 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 \(-\pi\) ;

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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 5m^{2}. 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.