Grade - 9

1,2,3,4,5,….. are called Natural Numbers.

It is denoted by N.

0,1,2,3,4,5,….. are called Whole Numbers.

It is denoted by W.

….-3,-2,-1,0,1,2,3,….. are called Integers.

It is denoted by Z.

Integers and be further classified as Positive (\(Z^+\)) and Negative Integers(\(Z^-\)).

Any number which can be expresses in \(\frac{p}{q}\) form where \(q\ne 0\) is a Rational Number.

It is denoted by Q.

Any number which cannot be expresses in \(\frac{p}{q}\) form where \(q\ne 0\) is an irrational Number.

It is denoted by I.

All rational numbers and all irrational numbers together make the collection of real numbers.

It is denoted by R.

I + R = I

I - R = I

R - I = I

R + R = R

R - R = R

\(I\times R = I\)

\(I\times I = I or R\)

\(\cfrac{R}{I} = I\)

\(\cfrac{I}{R} = I\)

For positive real numbers a and b,

\(\sqrt{a}\sqrt{b} = \sqrt{ab}\)

\(\sqrt{\cfrac{a}{b}} = \cfrac{\sqrt{a}}{\sqrt{b}}\)

\((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = (a-b)\)

\((a+\sqrt{b})(a-\sqrt{b}) = (a^2-b)\)

\((\sqrt{a}+\sqrt{b})^2 = (a + 2\sqrt{ab} + b)\)

If m and n are rational numbers and a,b are positive real numbers. Then,

\(a^ma^n = a^{m+n}\)

\((a^m)^n = a^{mn}\)

\(\cfrac{a^m}{a^n} = a^{m-n}\)

\(a^mb^m = (ab)^m\)

**Solution :**

0 can be expressed as \(\cfrac{0}{1}\) which is of the form \(\cfrac{p}{q}\).

Hence, 0 is a rational number.

**Solution :**

To find 6 rational numbers between 3 and 4, we multiply and divide the numbers by 6+1=7.

\(3\times \cfrac{7}{7} =\cfrac{21}{7}\)

\(4\times \cfrac{7}{7} =\cfrac{28}{7}\)

Hence, \(\cfrac{22}{7}, \cfrac{23}{7},\cfrac{24}{7},\cfrac{25}{7},\cfrac{26}{7}\) and \(\cfrac{27}{7}\) are the six rational numbers.

**Solution :**

To find 5 rational numbers between \(\cfrac{3}{5}\) and \(\cfrac{4}{5}\), we multiply and divide the numbers by 5+1=6.

\(\cfrac{3}{5}\times \cfrac{6}{6} =\cfrac{18}{30}\)

\(\cfrac{4}{5}\times \cfrac{6}{6} =\cfrac{24}{30}\)

Hence, \(\cfrac{19}{30}, \cfrac{20}{30},\cfrac{21}{30},\cfrac{22}{30}\) and \(\cfrac{23}{30}\) are the five rational numbers between \(\cfrac{3}{5}\) and \(\cfrac{4}{5}\).

**Solution :**

**i. TRUE**

Natural Number are 1,2,3,4,….

Whole Numbers are 0,1,2,3,4,…

We can say that whole numbers have all Natural numbers and Zero.

Hence, every natural number is a whole number but the revrse is not true.

**ii. FALSE**

Whole Numbers are 0,1,2,3,4,…

Integers are ….,-2,-1,0,1,2,….

We can clearly see that the engative integers are not whole numbers.

Hence, every integer is not a whole number but every whole number is an integer.

**iii. FALSE**

Whole Numbers are 0,1,2,3,4,…

Rational numbers are the one which can be expressed in the form of \(\cfrac{p}{q}\), where \(q\ne 0\).

We can clearly see that whole numbers does not include the fractions and the negative integers.

Hence, every rational number is not a whole number however, every whole number is a rational number.

**Solution :**

**i. TRUE**

Irrational Number : Numbers which cannot be represented in \(\cfrac{p}{q}\) form are called irrational numbers.

Real Numbers : Collection of rational and irrational numbers is called Real numbers.

Hence, Every irrational number is a real number however, every real number is not irrational.

**ii. FALSE**

The negative numbers cannot be expressed as square roots.

Hence, Every point on the number line is not of the form \(\sqrt{m}\), where m is a natural number.

**iii. FALSE**

Irrational Number : Numbers which cannot be represented in \(\cfrac{p}{q}\) form are called irrational numbers.

Real Numbers : Collection of rational and irrational numbers is called Real numbers.

Hence, every real number is not irrational.

**Solution :**

Square root of all positive integers are not irrational.

Examples:

\(\sqrt{4} = 2\), which is not a irrational.

\(\sqrt{9} = 3\), which is not a irrational.

We can see that, \(\sqrt{4}\) and \(\sqrt{9}\) is a rational number.

**Solution :**

We know that, \(\sqrt5 = \sqrt{2^2+1^2}\)

- Draw a line OA = 2 units

- Draw a line AB perpendicular to OA of 1 unit.

- Join point OB which is equal to \(\sqrt5\).

- Construct an ARC equal to OB by center at O and interescting the number line at D.

- OD here represents \(\sqrt5\) on number line.

**Solution:**

**Solution i. :**

\(\cfrac{36}{100} = 0.36\) which is a terminating decimal expansion.

**Solution ii. :**

\(\cfrac{1}{11} = 0.0909.... = 0.\overline{09}\) which is a non terminating decimal expansion.

**Solution iii. :**

\(4\frac{1}{8} = \cfrac{33}{8}= 4.125\) which is a terminating decimal expansion.

**Solution iv. :**

\(\frac{3}{13} = 0.\overline{230769}\) which is a terminating decimal expansion.

**Solution v. :**

\(\frac{2}{11} = 0.\overline{18}\) which is a terminating decimal expansion.

**Solution vi. :**

\(\frac{329}{400} = 0.8225\) which is a terminating decimal expansion.

**Solution :**

Given : \(\cfrac{1}{7} = 0.\overline{142857}\)

\(\cfrac{2}{7} = 2\times \cfrac{1}{7} = 2 \times0.\overline{142857} = 0.\overline{285714}\)

\(\cfrac{3}{7} = 3\times \cfrac{1}{7} = 3 \times0.\overline{142857} = 0.\overline{428571}\)

\(\cfrac{4}{7} = 4\times \cfrac{1}{7} = 4 \times0.\overline{142857} = 0.\overline{571428}\)

\(\cfrac{5}{7} = 5\times \cfrac{1}{7} = 5 \times0.\overline{142857} = 0.\overline{714285}\)

\(\cfrac{6}{7} = 6\times \cfrac{1}{7} = 6 \times0.\overline{142857} = 0.\overline{857142}\)

**Solution i. :**

\(0.\overline{6} = 0.666...\)

Assume, \(x = 0.666..\) ………………(1)

Multiply both sides of Eq(1) by 10, we get,

\(10x = 6.666...\)

\(10x = 6 + 0.666..\)

\(10x = 6 +x\)

\(9x = 6\)

\(x = \cfrac{2}{3}\)

\(0.\overline{6} =\cfrac{2}{3}\)

**Solution ii. :**

\(0.4\overline{7} = 0.4777...\)

As there is one repeating digit, multiplying (1) by 10 on both sides, we get,

Assume, \(x = 0.477..\) ………………(1)

Multiply both sides of Eq(1) by 10, we get,

\(10x = 4.77..\)

\(10x - x = 4.77... - 0.477...\)

\(9x = 4.3\)

\(x = \cfrac{43}{90}\)

\(0.4\overline{7}=\cfrac{43}{90}\)

**Solution iii. :**

\(0.\overline{001} = 0.001001...\)

As there is one repeating digit, multiplying (1) by 10 on both sides, we get,

Assume, \(x = 0.001..\) ………………(1)

Multiply both sides of Eq(1) by 1000, we get,

\(1000x = 1.001..\)

\(1000x - x = 1\)

\(999x = 1\)

\(x = \cfrac{1}{999}\)

\(0.\overline{001} = \cfrac{1}{999}\)

**Solution :**

Assume, \(x = 0.9999...\) ………………..(1)

As there is one repeating digit, multiplying (1) by 10 on both sides, we get,

Multiply both sides of Eq(1) by 1000, we get,

\(10x = 9.999....\) …………….(2)

Eq(2) - Eq(1), we get,

\(10x - x = 9.9999... - 0.9999....\)

\(9x = 9\)

\(x = 1\)

\(0.9999... = 1\)

Visualise \(4.\overline{26}\) on the number line, up to 4 decimal places.

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