NCERT Solutions for Class 9 Maths Chapter 1 Number Systems — Complete Guide with All Exercises

Here is a quick decision flowchart for classifying a number:

1. Is it a positive counting number? Then it is a natural number (and also whole, integer, rational, real).
2. Is it zero? Then it is a whole number (and also integer, rational, real), but NOT a natural number.
3. Is it a negative whole number? Then it is an integer (and also rational, real), but NOT a whole or natural number.
4. **Can it be written as $\frac{p}{q}$ with integers $p, q$ and $q \neq 0$?** Then it is rational (and real).
5. If none of the above — it is irrational (and real).

For the exam, always check: does the number have a terminating or recurring decimal? If yes, it is rational. If the decimal is non-terminating and non-recurring, it is irrational.

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

Solution:
Yes, zero is a rational number. We can express it as:

In each case, $p = 0$ (an integer) and $q$ is a non-zero integer, so the definition of a rational number is satisfied.

Key Insight: Any integer $n$ is rational because $n = \frac{n}{1}$. This is a common exam question — never say zero is "neither rational nor irrational."

Question: Find six rational numbers between 3 and 4.

Solution:
To find $n$ rational numbers between two numbers, express them with a denominator of $(n+1)$ or more.

Write $3 = \frac{21}{7}$ and $4 = \frac{28}{7}$.

Six rational numbers between them:

Alternative Method: We could also use $3 = \frac{30}{10}$ and $4 = \frac{40}{10}$, giving us $\frac{31}{10}, \frac{32}{10}, \ldots, \frac{39}{10}$ — nine rational numbers to choose any six from.

Key Insight: There are infinitely many rational numbers between any two rationals. The technique is to make the denominators large enough to create room.

Question: Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.

Solution:
Multiply numerator and denominator by 6 (one more than the number we need):

Five rational numbers between them:

Simplification check: $\frac{20}{30} = \frac{2}{3}$, $\frac{21}{30} = \frac{7}{10}$. In the exam, you do not need to simplify — the unsimplified form is perfectly acceptable.

Question: State whether every rational number can be located on the number line.

Solution:
Yes, every rational number can be located on the number line. Given a rational number $\frac{p}{q}$ (in lowest terms, $q > 0$):

1. Divide each unit interval on the number line into $q$ equal parts.
2. Starting from 0, count $p$ parts in the positive direction (if $p > 0$) or negative direction (if $p < 0$).

For example, to locate $\frac{3}{4}$: divide each unit into 4 equal parts, then count 3 parts from 0 in the positive direction.

Key Insight: This establishes that $\mathbb{Q}$ can be "embedded" into the number line. However, not every point on the number line is rational — the irrational points fill the "gaps."

Question: State whether the following statements are true or false. Give reasons.

(i) Every irrational number is a real number.

Solution: True. The set of real numbers $\mathbb{R}$ is defined as the union of rational numbers $\mathbb{Q}$ and irrational numbers $\mathbb{Q}'$. Since every irrational number belongs to $\mathbb{R}$ by definition, this statement is true.

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

Solution: False. Negative numbers like $-3$ are points on the number line but cannot be written as $\sqrt{m}$ for any natural number $m$ (since $\sqrt{m} \geq 0$ for all natural numbers $m$). Also, numbers like $\frac{1}{2}$ are on the number line but are not of the form $\sqrt{m}$ for any natural $m$.

(iii) Every real number is an irrational number.

Solution: False. For example, $2$ is a real number but it is rational (since $2 = \frac{2}{1}$). Real numbers include BOTH rational and irrational numbers.

Question: 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.

Solution:
No, the square roots of all positive integers are NOT irrational.

Counterexample: $\sqrt{4} = 2$, which is rational.

In general, $\sqrt{n}$ is rational if and only if $n$ is a perfect square (i.e., $n = 1, 4, 9, 16, 25, 36, \ldots$). For all other positive integers, $\sqrt{n}$ is irrational.

**Why is $\sqrt{2}$ irrational?** This can be proved by contradiction. Assume $\sqrt{2} = \frac{p}{q}$ in lowest terms. Squaring: $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$. This means $p^2$ is even, so $p$ is even. Let $p = 2k$. Then $4k^2 = 2q^2$, giving $q^2 = 2k^2$, so $q$ is also even. But this contradicts our assumption that $\frac{p}{q}$ was in lowest terms. Hence $\sqrt{2}$ is irrational.

Question: Locate $\sqrt{3}$ on the number line.

Solution (Spiral/Pythagoras Method):

Step 1: Mark $O$ at $0$ and $A$ at $1$ on the number line.

Step 2: At $A$, draw $AB \perp OA$ with $AB = 1$ unit.

Step 3: Join $OB$. By Pythagoras theorem:

Step 4: With centre $O$ and radius $OB = \sqrt{2}$, draw an arc to meet the number line at $C$. So $OC = \sqrt{2}$.

Step 5: At $C$, draw $CD \perp OC$ with $CD = 1$ unit.

Step 6: Join $OD$. By Pythagoras theorem:

Step 7: With centre $O$ and radius $OD = \sqrt{3}$, draw an arc to meet the number line. This point represents $\sqrt{3}$.

Key Insight: This "spiral of Theodorus" method can be extended to locate $\sqrt{4}$, $\sqrt{5}$, $\sqrt{6}$, etc. Each time, construct a perpendicular of length 1 at the previous point and use Pythagoras.

Question: Locate $\sqrt{5}$ on the number line using successive constructions.

Solution:
Continuing the spiral from the previous problem:

**For $\sqrt{4}$:** At the point $\sqrt{3}$ on the spiral, construct a perpendicular of length 1. The hypotenuse = $\sqrt{3+1} = \sqrt{4} = 2$. (This confirms the construction — we get exactly 2.)

**For $\sqrt{5}$:** At the point where $\sqrt{4} = 2$ was located in the spiral, construct a perpendicular of length 1. The hypotenuse:

Draw an arc with centre $O$ and radius $\sqrt{5}$ to locate the point on the number line.

Alternative (Direct) Method: Mark $O$ at $0$ and $A$ at $2$. At $A$, construct $AB \perp OA$ with $AB = 1$. Then $OB = \sqrt{4 + 1} = \sqrt{5}$. Transfer to the number line with a compass.

Question: 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) $\frac{329}{400}$

Solution:

(i) $\frac{36}{100} = 0.36$ — Terminating. (Denominator $100 = 2^2 \times 5^2$, only factors of 2 and 5.)

(ii) $\frac{1}{11} = 0.\overline{09}$ — Non-terminating recurring. (Denominator 11 is prime and not 2 or 5.)

(iii) $4\frac{1}{8} = \frac{33}{8} = 4.125$ — Terminating. (Denominator $8 = 2^3$, only factor of 2.)

(iv) $\frac{3}{13} = 0.\overline{230769}$ — Non-terminating recurring. (Denominator 13 is prime.)

(v) $\frac{2}{11} = 0.\overline{18}$ — Non-terminating recurring.

(vi) $\frac{329}{400} = 0.8225$ — Terminating. (Denominator $400 = 2^4 \times 5^2$.)

Key Rule: Check the denominator's prime factors after reducing the fraction. If only 2s and 5s appear, the decimal terminates. Otherwise, it recurs.

Question: Express $0.\overline{6}$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Solution:
Let $x = 0.6666\ldots$

Multiply both sides by 10:

Subtract the first equation from the second:

Verification: $\frac{2}{3} = 0.6666\ldots = 0.\overline{6}$ ✓

Key Technique: For a single repeating digit, multiply by 10. For two repeating digits, multiply by 100. For $n$ repeating digits, multiply by $10^n$.

Question: Express $0.\overline{001}$ in the form $\frac{p}{q}$.

Solution:
Let $x = 0.001001001\ldots$

The repeating block has 3 digits, so multiply by $10^3 = 1000$:

Subtract:

Answer: $0.\overline{001} = \frac{1}{999}$.

Question: Express $0.\overline{9}$ in the form $\frac{p}{q}$. Are you surprised by your answer?

Solution:
Let $x = 0.9999\ldots$

So $0.\overline{9} = 1$.

Yes, this is surprising but true! The number $0.9999\ldots$ (with infinitely many 9s) is exactly equal to 1, not "just less than 1." This is a rigorous mathematical fact, not an approximation.

Why? If $0.\overline{9} < 1$, then there would be a number between $0.\overline{9}$ and $1$. But there is no such number — you cannot find a decimal that is bigger than $0.9999\ldots$ and smaller than $1$.

Question: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$?

Solution:
When we divide by 17, the possible remainders at each step are $1, 2, 3, \ldots, 16$ (i.e., 16 possible non-zero remainders). Once a remainder repeats, the decimal cycle starts again.

Therefore, the maximum number of digits in the repeating block is $17 - 1 = 16$.

In fact, $\frac{1}{17} = 0.\overline{0588235294117647}$, which has exactly 16 repeating digits.

General Rule: For $\frac{1}{q}$, the repeating block has at most $q - 1$ digits. (This is related to Fermat's Little Theorem in number theory.)

Question: Visualise $3.765$ on the number line using successive magnification.

Solution:

Step 1: $3.765$ lies between $3$ and $4$. Divide the interval $[3, 4]$ into 10 equal parts. $3.765$ lies between $3.7$ and $3.8$.

Step 2: Magnify the interval $[3.7, 3.8]$ by dividing it into 10 equal parts. $3.765$ lies between $3.76$ and $3.77$.

Step 3: Magnify the interval $[3.76, 3.77]$ by dividing it into 10 equal parts. $3.765$ is the 5th mark, i.e., the midpoint.

Key Insight: With each magnification, we "zoom in" by a factor of 10. For a number with $n$ decimal places, we need $n$ levels of magnification.

Question: Visualise $4.\overline{26}$ on the number line, up to 4 decimal places.

Solution:
$4.\overline{26} = 4.262626\ldots$

Step 1: It lies between $4$ and $5$. Magnify $[4, 5]$: it lies between $4.2$ and $4.3$.

Step 2: Magnify $[4.2, 4.3]$: it lies between $4.26$ and $4.27$.

Step 3: Magnify $[4.26, 4.27]$: $4.2626$ lies between $4.262$ and $4.263$.

Step 4: Magnify $[4.262, 4.263]$: $4.2626$ lies between $4.2626$ and $4.2627$. We can mark $4.2626$ at the 6th division.

Key Insight: For recurring decimals, the process can continue indefinitely. We approximate to the desired number of decimal places.

Question: 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) $2\pi$

Solution:

(i) $2 - \sqrt{5}$: Since $\sqrt{5}$ is irrational and $2$ is rational, their difference is irrational. (A rational minus an irrational is always irrational.)

(ii) $(3 + \sqrt{23}) - \sqrt{23} = 3$: This simplifies to the integer $3$, which is rational.

(iii) $\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$: The $\sqrt{7}$ cancels, leaving a rational number.

(iv) $\frac{1}{\sqrt{2}}$: Since $\sqrt{2}$ is irrational and the reciprocal of a non-zero irrational is irrational, this is irrational.

(v) $2\pi$: Since $\pi$ is irrational and $2$ is a non-zero rational, their product is irrational.

Key Rules:
- Rational $\pm$ Irrational = Irrational
- Rational $\times$ Irrational (non-zero rational) = Irrational
- Irrational $\pm$ Irrational = Could be either (e.g., $\sqrt{2} + (-\sqrt{2}) = 0$, which is rational)
- Irrational $\times$ Irrational = Could be either (e.g., $\sqrt{2} \times \sqrt{2} = 2$, which is rational)

Question: 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) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$

Solution:

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

(ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
Using $(a+b)(a-b) = a^2 - b^2$:

(iii) $(\sqrt{5} + \sqrt{2})^2$
Using $(a+b)^2 = a^2 + 2ab + b^2$:

(iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
Using $(a-b)(a+b) = a^2 - b^2$:

Question: Rationalise the denominator of $\frac{1}{\sqrt{7}}$.

Solution:
Multiply numerator and denominator by $\sqrt{7}$:

Why do we rationalise? Having a surd in the denominator makes further calculations harder. Rationalisation moves the surd to the numerator, making the expression easier to work with.

The General Principle:
- For $\frac{1}{\sqrt{a}}$: multiply by $\frac{\sqrt{a}}{\sqrt{a}}$
- For $\frac{1}{a + \sqrt{b}}$: multiply by $\frac{a - \sqrt{b}}{a - \sqrt{b}}$ (the conjugate)
- For $\frac{1}{\sqrt{a} + \sqrt{b}}$: multiply by $\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}$

Question: Rationalise the denominator of $\frac{1}{\sqrt{5} + \sqrt{2}}$.

Solution:
The conjugate of $\sqrt{5} + \sqrt{2}$ is $\sqrt{5} - \sqrt{2}$.

Another Example: Rationalise $\frac{1}{\sqrt{7} - \sqrt{6}}$.

Another Example: Rationalise $\frac{1}{\sqrt{5} - 2}$.

Question: Rationalise the denominator of $\frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}}$.

Solution:
The conjugate of $7 + 4\sqrt{3}$ is $7 - 4\sqrt{3}$.

Numerator:

Denominator:

Answer: $\frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = 11 - 6\sqrt{3}$.

Key Insight: When the denominator simplifies to 1, the answer is just the numerator. This happens when the terms $a^2$ and $b^2$ are close in value.

Question: Simplify:
(i) $2^{2/3} \cdot 2^{1/5}$
(ii) $(3^{1/5})^4$
(iii) $\frac{7^{1/2}}{7^{1/3}}$
(iv) $13^{1/5} \cdot 17^{1/5}$

Solution:

(i) Using $a^m \cdot a^n = a^{m+n}$:

(ii) Using $(a^m)^n = a^{mn}$:

(iii) Using $\frac{a^m}{a^n} = a^{m-n}$:

(iv) Using $a^m \cdot b^m = (ab)^m$:

Question: Simplify:
(i) $\left(\frac{1}{3^3}\right)^7$
(ii) $\frac{11^{1/2}}{11^{1/4}}$
(iii) $(2^{1/3})^{1/4}$

Solution:

(i) $\left(\frac{1}{3^3}\right)^7 = \frac{1}{3^{21}} = 3^{-21}$

Using $(a^m)^n = a^{mn}$: $(3^{-3})^7 = 3^{-21}$.

(ii) $\frac{11^{1/2}}{11^{1/4}} = 11^{1/2 - 1/4} = 11^{1/4} = \sqrt[4]{11}$

(iii) $(2^{1/3})^{1/4} = 2^{1/3 \times 1/4} = 2^{1/12} = \sqrt[12]{2}$

Key Conversions to Remember:
- $\sqrt{a} = a^{1/2}$
- $\sqrt[3]{a} = a^{1/3}$
- $\sqrt[n]{a^m} = a^{m/n}$
- $\frac{1}{a^n} = a^{-n}$

Question: Simplify $\frac{3^{1/3} \times 3^{1/2}}{3^{1/6}}$.

Solution:

Answer: $\sqrt[3]{9}$.

Question: Simplify $(256)^{3/4}$.

Solution:

Key Strategy: Always try to express the base as a power of a small number first. This makes exponent arithmetic much cleaner.

Question: Prove that $\sqrt{2} + \sqrt{3}$ is irrational.

Solution:
Assume for contradiction that $\sqrt{2} + \sqrt{3}$ is rational. Let $\sqrt{2} + \sqrt{3} = r$, where $r$ is rational.

Squaring both sides:

Since $r$ is rational, $r^2$ is rational, $r^2 - 5$ is rational, and $\frac{r^2 - 5}{2}$ is rational. So $\sqrt{6}$ would be rational.

But $\sqrt{6}$ is irrational (since 6 is not a perfect square). This is a contradiction.

Therefore, $\sqrt{2} + \sqrt{3}$ is irrational. $\square$

Question: If $a = \frac{1}{\sqrt{3} + \sqrt{2}}$ and $b = \frac{1}{\sqrt{3} - \sqrt{2}}$, find $a^2 + b^2$.

Solution:
First, rationalise each:

Now:

Using $a^2 + b^2 = (a + b)^2 - 2ab$:

Answer: $a^2 + b^2 = 10$.

Question: Simplify: $\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}}$.

Solution:
Rationalise each term:

Adding all three:

Answer: The sum is $1$.

Key Insight: This is a telescoping sum — most terms cancel. Recognising this pattern saves enormous time in exams.

Question: If $2^x = 3^y = 6^z$, prove that $\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$.

Solution:
Let $2^x = 3^y = 6^z = k$.

Then:

Since $6 = 2 \times 3$:

Comparing exponents: