# Number Systems - Class 9 - Mathematics

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## Exercise 1.1 - Number Systems | NCERT | Mathematics | Class 9

Is zero a rational number? Can you write it in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$ ?

Yes, zero is a rational number. It can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, by choosing $p=0$ and any non-zero $q$. For example, zero can be written as $\frac{0}{1}$, $\frac{0}{2}$, $\frac{0}{-3}$, and so on. In each case, $p = 0$ is an integer, and $q$ is any non-zero integer, fulfilling the conditions for a rational number.

Find six rational numbers between 3 and 4 .

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Sign up nowFind five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.

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Sign up nowState whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

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Sign up now## Exercise 1.2 - Number Systems | NCERT | Mathematics | Class 9

State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form $\sqrt{m}$, where $m$ is a natural number.

(iii) Every real number is an irrational number.

(i) **True**. An irrational number is a real number that cannot be expressed as a simple fraction. This means irrational numbers are considered to be part of the real number system but they cannot be exactly written as a ratio of two integers. The real number system contains both rational numbers (numbers that can be expressed as the quotient of two integers) and irrational numbers.

(ii) **False**. Not every point on the number line can be represented as $\sqrt{m}$, where $m$ is a natural number. For example, rational numbers and negative numbers cannot be represented in this form. While $\sqrt{m}$, for natural numbers $m$, does represent some points on the number line (specifically, the points corresponding to the square roots of natural numbers), there are many more points on the number line that do not fit this description, including all the rational numbers that are not perfect squares, all negative numbers, and other irrational numbers that are not square roots of natural numbers (like $\pi$ or $e$).

(iii) **False**. Not every real number is an irrational number. Real numbers include both rational numbers (which can be expressed as the quotient of two integers, including integers themselves as the special case where the denominator is 1) and irrational numbers (which cannot be expressed as the ratio of two integers). Rational numbers, such as 1/2, 2, or -5, are examples of real numbers that are not irrational.

Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

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Show how $\sqrt{5}$ can be represented on the number line.

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## Exercise 1.3 - Number Systems | NCERT | Mathematics | Class 9

Write the following in decimal form and say what kind of decimal expansion each has :

(i) $\frac{36}{100}$

(ii) $\frac{1}{11}$

(iii) $4 \frac{1}{8}$

(iv) $\frac{3}{13}$

(v) $\frac{2}{11}$

(vi) $\frac{329}{400}$

Here are the decimal forms of the given fractions along with the types of decimal expansion each has:

(i) $\frac{36}{100}$ in decimal form is $0.36$. This is a **terminating decimal** because it comes to a definite end.

(ii) $\frac{1}{11}$ in decimal form is approximately $0.090909...$, which can be represented as $0.\overline{09}$. This is a **repeating decimal** with a repeating period of $09$.

(iii) $4 \frac{1}{8}$ in decimal form is $4.125$. This is another example of a **terminating decimal**.

(iv) $\frac{3}{13}$ in decimal form is approximately $0.230769...$, which can be represented as $0.\overline{230769}$. This is a **repeating decimal** with a repeating period of $230769$.

(v) $\frac{2}{11}$ in decimal form is approximately $0.181818...$, which can be represented as $0.\overline{18}$. This is a **repeating decimal** with a repeating period of $18$.

(vi) $\frac{329}{400}$ in decimal form is $0.8225$. This too is a **terminating decimal**.

So, among the given fractions, (i), (iii), and (vi) have terminating decimal expansions, while (ii), (iv), and (v) have repeating decimal expansions.

You know that $\frac{1}{7}=0 . \overline{142857}$. Can you predict what the decimal expansions of $\frac{2}{7}, \frac{3}{7}$, $\frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ are, without actually doing the long division? If so, how?

[Hint : Study the remainders while finding the value of $\frac{1}{7}$ carefully.]

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Express the following in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

(i) $0 . \overline{6}$

(ii) $0.4 \overline{7}$

(iii) $0 . \overline{001}$

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Express $0.99999 \ldots$ in the form $\frac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

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What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

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Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0)$, where $p$ and $q$ are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property $q$ must satisfy?

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Write three numbers whose decimal expansions are non-terminating non-recurring.

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Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.

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Classify the following numbers as rational or irrational :

(i) $\sqrt{23}$

(ii) $\sqrt{225}$

(iii) 0.3796

(iv) $7.478478 \ldots$

(v) $1.101001000100001 \ldots$

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## Exercise 1.4 - Number Systems | NCERT | Mathematics | Class 9

Classify the following numbers as rational or irrational:

(i) $2-\sqrt{5}$

(ii) $(3+\sqrt{23})-\sqrt{23}$

(iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}$

(iv) $\frac{1}{\sqrt{2}}$

(v) $2 \pi$

To classify the given numbers as rational or irrational, we need to understand that a rational number can be expressed as the quotient of two integers, while an irrational number cannot be expressed in such a way.

(i) $2-\sqrt{5}$: This is irrational because $\sqrt{5}$ cannot be expressed as a ratio of two integers, and subtracting it from an integer doesn't change its irrational nature.

(ii) $(3+\sqrt{23})-\sqrt{23}$: This simplifies to $3$, which is rational because it can be expressed as $3/1$.

(iii) $\frac{2 \sqrt{7}}{7 \sqrt{7}}$: This simplifies to $\frac{2}{7}$, which is rational since it can be expressed as a ratio of the integers $2$ and $7$.

(iv) $\frac{1}{\sqrt{2}}$: This is irrational. Multiplying the numerator and denominator by $\sqrt{2}$ gives $\frac{\sqrt{2}}{2}$, which cannot be expressed as a ratio of two integers since $\sqrt{2}$ is irrational.

(v) $2 \pi$: This is irrational because $\pi$ is a well-known irrational number, and multiplying it by an integer ($2$ in this case) does not change its irrational nature.

So, summarizing:

(i) is irrational.

(ii) is rational.

(iii) is rational.

(iv) is irrational.

(v) is irrational.

Simplify each of the following expressions:

(i) $(3+\sqrt{3})(2+\sqrt{2})$

(ii) $(3+\sqrt{3})(3-\sqrt{3})$

(iii) $(\sqrt{5}+\sqrt{2})^{2}$

(iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

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Recall, $\pi$ is defined as the ratio of the circumference (say $c$ ) of a circle to its diameter (say $d$ ). That is, $\pi=\frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

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Represent $\sqrt{9.3}$ on the number line. (Hint use semi circle and do geometrically)

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Rationalise the denominators of the following:

(i) $\frac{1}{\sqrt{7}}$

(ii) $\frac{1}{\sqrt{7}-\sqrt{6}}$

(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$

(iv) $\frac{1}{\sqrt{7}-2}$

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## Exercise 1.5 - Number Systems | NCERT | Mathematics | Class 9

Find:

(i) $64^{\frac{1}{2}}$

(ii) $32^{\frac{1}{5}}$

(iii) $125^{\frac{1}{3}}$

The results for the given expressions are:

(i) $64^{\frac{1}{2}} = 8$

(ii) $32^{\frac{1}{5}} = 2$

(iii) $125^{\frac{1}{3}} = 5$

Find:

(i) $9^{\frac{3}{2}}$

(ii) $32^{\frac{2}{5}}$

(iii) $16^{\frac{3}{4}}$

(iv) $125^{\frac{-1}{3}}$

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Simplify:

(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$

(ii) $\left(\frac{1}{3^{3}}\right)^{7}$

(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

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## Extra Questions - Number Systems | NCERT | Mathematics | Class 9

Simplify $\frac{27^{\frac{2}{3}}}{9^{\frac{1}{2}} \times 3^{\frac{3}{2}}}$

Option A) $\sqrt{3}$

Option B) $\frac{1}{\sqrt{3}}$

Option C) $3\sqrt{3}$

Option D) $\frac{1}{3\sqrt{3}}$

To solve the given problem, we will first simplify the numerator

$$ 27^{\frac{2}{3}} = 3^{3 \times \frac{2}{3}} = 3^2 = 9 $$

Now, we will simplify the denominator:

$$ 9^{\frac{1}{2}} \times 3^{\frac{3}{2}} = 3^{2 \times \frac{1}{2}} times 3^{\frac{3}{2}} = 3 \times 3^{\frac{3}{2}} = 3^{\frac{5}{2}} $$

Putting them both together,

$ \frac{3^2}{3^{\frac{5}{2}}} = \frac{1}{3^{\frac{1}{2}}} = \frac{1}{\sqrt{3}} $

**Hence, the answer is option B: **$\frac{1}{\sqrt{3}}$

The roster form of the set $A = { \text{Whole numbers less than or equal to 8} }$ is __________.

Option A) $A = {0, 1, 2, 3, 4, 5, 6, 7, 8}$

Option B) $A = {1, 2, 3, 4, 5, 6, 7, 8}$

Option C) $A = {0, 1, 2, 3, 4, 5, 6, 7}$

Option D) $A = {2, 3, 4, 5, 6, 7, 8}$

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Can you explain how to make Roman numerals? How to form numbers of any choice.