What are rational and irrational numbers?

A rational number can be written as p/q where p and q are integers and q ≠ 0 (e.g. 3/4, 0.5). An irrational number cannot be written as a simple fraction and has a non-terminating, non-repeating decimal (e.g. √2, π).

What are real numbers?

Real numbers are all rational and irrational numbers taken together — they include natural numbers, whole numbers, integers, fractions and irrational numbers.

What is the difference between a terminating and a non-terminating decimal?

A terminating decimal ends after a finite number of digits (e.g. 0.75); a non-terminating decimal continues without end — it may be recurring (e.g. 0.333…) or non-recurring (e.g. √2).

What is the law of exponents for multiplication?

When multiplying powers with the same base, add the exponents: aᵐ × aⁿ = aᵐ⁺ⁿ.

What is rationalising the denominator?

It is removing a surd from the denominator of a fraction by multiplying the numerator and denominator by a suitable factor — for example, multiplying 1/√2 by √2/√2 to get √2/2.

Samacheer Class 9 Maths - Real Numbers

Q: Which arrow best shows the position of (\frac{11}{3}) on the number line?

First convert the rational number into mixed form. $\frac{11}{3}=3\frac{2}{3}$

Q: Find any three rational numbers between

$-\frac{7}{11} \quad \text{and} \quad \frac{2}{11}$

Q: Express the following rational numbers into decimal form and state the kind of decimal expansion

# (i) (\frac{2}{7}) Solution $\frac{2}{7}=0.285714285714\ldots$

Q: Express (\frac{1}{13}) in decimal form. Find the length of the period of decimals.

Solution $\frac{1}{13}=0.076923076923\ldots$

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Your Progress — Unit 2: Real Numbers0% complete

Ex 2.1Rational Numbers3 questions

Q.1 Which arrow best shows the position of (\frac{11}{3}) on the number line? ▾

✓ Solution

First convert the rational number into mixed form.

\frac{11}{3}=3\frac{2}{3}

So, the number lies:

between (3) and (4)
closer to (4)

Therefore, the correct arrow is the one pointing at:

3\frac{2}{3}

on the number line.

Q.2 Find any three rational numbers between ▾

✓ Solution

-\frac{7}{11} \quad \text{and} \quad \frac{2}{11}

Solution

We need rational numbers greater than

-\frac{7}{11}

and less than

\frac{2}{11}

Possible rational numbers are:

-\frac{6}{11},; -\frac{5}{11},; -\frac{4}{11}

Answer

Any three rational numbers are:

-\frac{6}{11},; -\frac{5}{11},; -\frac{4}{11}

Q.3 Find any five rational numbers between ▾

✓ Solution

# (i) Between (\frac14) and (\frac15)

Step 1: Take LCM denominator

\frac14=\frac{25}{100}

\frac15=\frac{20}{100}

Now choose fractions between:

\frac{20}{100} \quad \text{and} \quad \frac{25}{100}

Answer

Five rational numbers are:

\frac{21}{100}, \frac{22}{100}, \frac{23}{100}, \frac{24}{100}, \frac{241}{1000}

(Any valid five rational numbers are acceptable.)

# (ii) Between (0.1) and (0.11)

Step 1: Write with same decimal places

$$0.1 = 0.100$$

$$0.11 = 0.110$$

Now choose numbers between them.

Answer

$$0.101,; 0.102,; 0.103,; 0.104,; 0.105$$

# (iii) Between (-1) and (-2)

We need numbers greater than (-2) and less than (-1).

Answer

-\frac32,; -\frac54,; -\frac76,; -\frac98,; -\frac{11}{10}

All these lie between (-2) and (-1).

Ex 2.2Irrational Numbers5 questions

Q.1 Express the following rational numbers into decimal form and state the kind of decimal expansion ▾

✓ Solution

# (i) (\frac{2}{7})

Solution

\frac{2}{7}=0.285714285714\ldots

The digits repeat continuously.

Answer

\frac{2}{7}=0.\overline{285714}

Type of decimal expansion:

Non-terminating recurring decimal

# (ii) (-5\frac{3}{11})

Step 1: Convert mixed fraction into improper fraction

-5\frac{3}{11} ============== -\frac{58}{11}

Step 2: Convert into decimal

\frac{58}{11}=5.272727\ldots

Therefore,

-\frac{58}{11}=-5.272727\ldots

Answer

-5\frac{3}{11}=-5.\overline{27}

Type of decimal expansion:

Non-terminating recurring decimal

# (iii) (\frac{22}{3})

Solution

\frac{22}{3}=7.333333\ldots

Answer

\frac{22}{3}=7.\overline{3}

Type of decimal expansion:

Non-terminating recurring decimal

# (iv) (\frac{327}{200})

Solution

\frac{327}{200}=1.635

Answer

\frac{327}{200}=1.635

Type of decimal expansion:

Terminating decimal

Q.2 Express (\frac{1}{13}) in decimal form. Find the length of the period of decimals. ▾

✓ Solution

Solution

\frac{1}{13}=0.076923076923\ldots

The repeating block is:

$$076923$$

This block contains 6 digits.

Answer

\frac{1}{13}=0.\overline{076923}

Length of the period:

$$6$$

Q.3 Express the rational number (\frac{1}{33}) in recurring decimal form using the recurring decimal expansion of (\frac{1}{11}). Hence write (\frac{71}{33}) in recurring decimal form. ▾

✓ Solution

Step 1: Use (\frac{1}{11})

We know:

\frac{1}{11}=0.\overline{09}

Step 2: Find (\frac{1}{33})

\frac{1}{33}=\frac{1}{3}\times\frac{1}{11}

=\frac13\times0.\overline{09}

=0.030303\ldots

Therefore,

\frac{1}{33}=0.\overline{03}

Step 3: Find (\frac{71}{33})

\frac{71}{33} ============= 2+\frac{5}{33}

Now,

\frac{5}{33}=5\times\frac{1}{33}

=5\times0.\overline{03}

=0.\overline{15}

Therefore,

\frac{71}{33}=2.\overline{15}

Answer

\frac{1}{33}=0.\overline{03}

\frac{71}{33}=2.\overline{15}

Q.4 Express the following decimal expressions into rational numbers ▾

✓ Solution

The decimal expressions are not visible in the question provided.
Please share the decimal numbers clearly to convert them into rational numbers.

Q.5 Without actual division, find which of the following rational numbers have terminating decimal expansion ▾

✓ Solution

A rational number has a terminating decimal expansion if the denominator in simplest form contains only the prime factors:

2 \text{ and/or } 5

# (i) (\frac{7}{128})

$$128=2^7$$

Only factor (2) is present.

Answer

\frac{7}{128}

has a terminating decimal expansion.

# (ii) (\frac{21}{15})

Simplify:

\frac{21}{15}=\frac75

Denominator:

$$5$$

Only factor (5).

Answer

\frac{21}{15}

has a terminating decimal expansion.

# (iii) (4\frac{9}{35})

Convert fractional part:

\frac{9}{35}

Denominator:

35=5\times7

Factor (7) is present.

Answer

4\frac{9}{35}

has a non-terminating recurring decimal expansion.

# (iv) (\frac{219}{2200})

Simplify:

\frac{219}{2200}

Factorize denominator:

2200=2^3\times5^2\times11

Factor (11) is present.

Answer

\frac{219}{2200}

has a non-terminating recurring decimal expansion.

Ex 2.3Decimal Representation to Identify Irrational Numbers3 questions

Q.1 Represent the following irrational numbers on the number line ▾

✓ Solution

# (i) (\sqrt3)

\sqrt3 \approx 1.732

So, on the number line:

Mark (0), (1), and (2)
Locate the point approximately at (1.732)

Answer

\sqrt3

lies between (1) and (2).

# (ii) (\sqrt{4.7})

\sqrt{4.7}\approx2.168

So, on the number line:

Mark (2) and (3)
Locate the point approximately at (2.168)

Answer

\sqrt{4.7}

lies between (2) and (3).

# (iii) (\sqrt{6.5})

\sqrt{6.5}\approx2.549

So, on the number line:

Mark (2) and (3)
Locate the point approximately at (2.549)

Answer

\sqrt{6.5}

lies between (2) and (3).

Q.2 Find any two irrational numbers between ▾

✓ Solution

# (i)

0.3010011000111\ldots

and

0.3020020002\ldots

Solution

Any non-terminating non-recurring decimals between them are irrational.

Possible answers:

0.301101110111\ldots

0.3015000500005\ldots

Both lie between the given numbers.

Answer

Two irrational numbers are:

0.301101110111\ldots

and

0.3015000500005\ldots

# (ii) Between (\frac67) and (\frac{12}{13})

Step 1: Convert into decimals

\frac67\approx0.857142\ldots

\frac{12}{13}\approx0.923076\ldots

Now choose irrational numbers between them.

Answer

Two irrational numbers are:

\sqrt{0.8}

and

\frac{\sqrt3}{2}

Approximate values:

\sqrt{0.8}\approx0.894

\frac{\sqrt3}{2}\approx0.866

Both lie between (0.857) and (0.923).

# (iii) Between (\sqrt2) and (\sqrt3)

Approximate values:

\sqrt2\approx1.414

\sqrt3\approx1.732

Choose irrational numbers between them.

Answer

Two irrational numbers are:

\sqrt{\frac52}

and

\sqrt{\frac83}

Approximate values:

\sqrt{\frac52}\approx1.581

\sqrt{\frac83}\approx1.633

Both lie between (\sqrt2) and (\sqrt3).

Q.3 Find any two rational numbers between ▾

✓ Solution

2.2360679\ldots

and

2.236505500\ldots

Solution

Any terminating decimals between the given numbers are rational numbers.

Possible answers:

$$2.2361$$

$$2.2364$$

Both lie between the given numbers.

Answer

Two rational numbers are:

$$2.2361$$

and

$$2.2364$$

Ex 2.4Real Numbers1 questions

Q.1 Represent the following numbers on the number line ▾

✓ Solution

# (i) (5.348)

Solution

Step 1

First note that:

$$5.348$$

lies between (5) and (6).

Step 2

Divide the portion between (5) and (6) into 10 equal parts using a magnifying glass.

This gives:

5.1,\ 5.2,\ 5.3,\ 5.4,\ldots

Now (5.348) lies between:

5.3 \text{ and } 5.4

Step 3

Divide the portion between (5.3) and (5.4) into 10 equal parts.

This gives:

5.31,\ 5.32,\ 5.33,\ldots,5.39

Now (5.348) lies between:

5.34 \text{ and } 5.35

Step 4

Divide the portion between (5.34) and (5.35) into 10 equal parts.

This gives:

5.341,\ 5.342,\ldots,5.349

Mark the eighth small division after (5.34), so the point is:

$$5.348$$

Step 5

Mark the point corresponding to:

$$5.348$$

on the number line.

Thus, (5.348) is marked exactly at the third decimal place on the number line.

# (ii) (6.\overline{4}) up to 3 decimal places

Step 1

6.\overline{4}=6.4444\ldots

Up to 3 decimal places:

$$6.444$$

The number lies between (6) and (7).

Step 2

Divide the portion between (6) and (7) into 10 equal parts.

Now (6.444) lies between:

6.4 \text{ and } 6.5

Step 3

Divide the portion between (6.4) and (6.5) into 10 equal parts.

Now (6.444) lies between:

6.44 \text{ and } 6.45

Step 4

Divide the portion between (6.44) and (6.45) into 10 equal parts.

Mark the fourth small division after (6.44), so the required approximation is:

$$6.444$$

Mark the point corresponding to:

$$6.444$$

on the number line.

Thus, (6.\overline{4}) is represented up to three decimal places as (6.444).

# (iii) (4.\overline{73}) up to 4 decimal places

Step 1

4.\overline{73}=4.73737373\ldots

Rounded to 4 decimal places:

$$4.7374$$

The number lies between (4) and (5).

Step 2

Divide the portion between (4) and (5) into 10 equal parts.

Now (4.7374) lies between:

4.7 \text{ and } 4.8

Step 3

Divide the portion between (4.7) and (4.8) into 10 equal parts.

Now (4.7374) lies between:

4.73 \text{ and } 4.74

Step 4

Divide the portion between (4.73) and (4.74) into 10 equal parts.

Now (4.7374) lies between:

4.737 \text{ and } 4.738

Step 5

Divide the portion between (4.737) and (4.738) into 10 equal parts.

Now (4.7374) is the fourth-decimal mark after rounding.

Mark the point corresponding to:

$$4.7374$$

on the number line.

Thus, (4.\overline{73}) is represented up to four decimal places as (4.7374).

Ex 2.5Radical Notation5 questions

# Important Formulae

\sqrt[n]{a}=a^{\frac{1}{n}}

a^{m}\times a^{n}=a^{m+n}

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

Q.1 Write the following in the form of (5^n) ▾

✓ Solution

# (i) (625)

Solution

625=5\times5\times5\times5

$$625=5^4$$

Answer

$$625=5^4$$

# (ii) (\frac15)

Solution

\frac15=5^{-1}

Answer

\frac15=5^{-1}

# (iii) (\sqrt5)

Using:

\sqrt a=a^{\frac12}

Solution

\sqrt5=5^{\frac12}

Answer

\sqrt5=5^{\frac12}

# (iv) (\sqrt[3]{125})

Since:

$$125=5^3$$

Solution

\sqrt[3]{125}=(5^3)^{\frac13}

$$=5^1$$

$$=5$$

Answer

$$5=5^1$$

Q.2 Write the following in the form of (4^n) ▾

✓ Solution

# (i) (16)

Solution

$$16=4^2$$

Answer

$$16=4^2$$

# (ii) (8)

Solution

$$8=2^3$$

Also,

$$4=2^2$$

Therefore,

8=(2^2)^{\frac32}

=4^{\frac32}

Answer

8=4^{\frac32}

# (iii) (32)

Solution

$$32=2^5$$

Since,

$$4=2^2$$

32=(2^2)^{\frac52}

=4^{\frac52}

Answer

32=4^{\frac52}

Q.3 Find the value of ▾

✓ Solution

The expressions are not visible in the uploaded question.
Please share the full expressions clearly (or upload the image again) so I can provide the exact validated solutions.

Q.4 Use a fractional index to write ▾

✓ Solution

# (i) (\sqrt5)

Solution

\sqrt5=5^{\frac12}

Answer

5^{\frac12}

# (ii) (\sqrt[2]{7})

Solution

\sqrt7=7^{\frac12}

Answer

7^{\frac12}

# (iii) ((\sqrt[3]{49})^5)

Solution

(\sqrt[3]{49})^5 ================ \left(49^{\frac13}\right)^5

Using:

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

=49^{\frac53}

Answer

49^{\frac53}

# (iv)

The expression is missing in the question provided.
Please share the complete expression.

Q.5 Find the 5th root of ▾

✓ Solution

# (i) (32)

Solution

$$32=2^5$$

Therefore,

\sqrt[5]{32}=2

Answer

$$2$$

# (ii) (243)

Solution

$$243=3^5$$

Therefore,

\sqrt[5]{243}=3

Answer

$$3$$

# (iii) (100000)

Solution

$$100000=10^5$$

Therefore,

\sqrt[5]{100000}=10

Answer

$$10$$

# (iv) (\frac{1024}{3125})

Solution

$$1024=4^5$$

$$3125=5^5$$

Therefore,

\sqrt[5]{\frac{1024}{3125}} =========================== \frac45

Answer

\frac45

Ex 2.6Surds6 questions

Q.1 Simplify using addition and subtraction properties of surds ▾

✓ Solution

# (i) (5\sqrt3+18\sqrt3-2\sqrt3)

Since all are like surds, combine coefficients.

Solution

= (5+18-2)\sqrt3

=21\sqrt3

Answer

21\sqrt3

# (ii) (4\sqrt[3]{5}+2\sqrt[3]{5}-3\sqrt[3]{5})

Solution

=(4+2-3)\sqrt[3]{5}

=3\sqrt[3]{5}

Answer

3\sqrt[3]{5}

# (iii) (3\sqrt{75}+5\sqrt{48}-\sqrt{243})

Step 1: Simplify each surd

\sqrt{75}=\sqrt{25\times3}=5\sqrt3

\sqrt{48}=\sqrt{16\times3}=4\sqrt3

\sqrt{243}=\sqrt{81\times3}=9\sqrt3

Step 2: Substitute

3(5\sqrt3)+5(4\sqrt3)-9\sqrt3

=15\sqrt3+20\sqrt3-9\sqrt3

=26\sqrt3

Answer

26\sqrt3

# (iv) (5\sqrt[3]{40}+2\sqrt[3]{625}-3\sqrt[3]{320})

Step 1: Simplify cube roots

\sqrt[3]{40}=\sqrt[3]{8\times5}=2\sqrt[3]{5}

\sqrt[3]{625}=\sqrt[3]{125\times5}=5\sqrt[3]{5}

\sqrt[3]{320}=\sqrt[3]{64\times5}=4\sqrt[3]{5}

Step 2: Substitute

5(2\sqrt[3]{5})+2(5\sqrt[3]{5})-3(4\sqrt[3]{5})

=10\sqrt[3]{5}+10\sqrt[3]{5}-12\sqrt[3]{5}

=8\sqrt[3]{5}

Answer

8\sqrt[3]{5}

Q.2 Simplify using multiplication and division properties of surds ▾

✓ Solution

# (i) (\sqrt3\times\sqrt5\times\sqrt2)

Solution

=\sqrt{3\times5\times2}

=\sqrt{30}

Answer

\sqrt{30}

# (ii) (\sqrt{35}\div\sqrt7)

Solution

=\sqrt{\frac{35}{7}}

=\sqrt5

Answer

\sqrt5

# (iii) (\sqrt[3]{27}\times\sqrt[3]{8}\times\sqrt[3]{125})

Solution

=3\times2\times5

$$=30$$

Answer

$$30$$

# (iv) ((7\sqrt a-5\sqrt b)(7\sqrt a+5\sqrt b))

Use:

genui{"math_block_widget_always_prefetch_v2":{"content":"(x-y)(x+y)=x^2-y^2"}}

Solution

=(7\sqrt a)^2-(5\sqrt b)^2

$$=49a-25b$$

Answer

$$49a-25b$$

# (v)

\sqrt{\frac{225}{729}} ---------------------- \sqrt{\frac{25}{144}} \div \sqrt{\frac{16}{81}}

Step 1: Simplify each surd

\sqrt{\frac{225}{729}}=\frac{15}{27}=\frac59

\sqrt{\frac{25}{144}}=\frac5{12}

\sqrt{\frac{16}{81}}=\frac49

Step 2: Perform division

\frac5{12}\div\frac49

=\frac5{12}\times\frac94

=\frac{45}{48}

=\frac{15}{16}

Step 3: Subtract

\frac59-\frac{15}{16}

LCM:

$$144$$

=\frac{80}{144}-\frac{135}{144}

=-\frac{55}{144}

Answer

-\frac{55}{144}

Q.3 Find the values correct to 3 decimal places ▾

✓ Solution

Given:

\sqrt2=1.414

\sqrt3=1.732

\sqrt5=2.236

\sqrt{10}=3.162

# (i) (\sqrt{40}-\sqrt{20})

Step 1: Simplify

\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}

\sqrt{20}=\sqrt{4\times5}=2\sqrt5

Step 2: Substitute values

$$2(3.162)-2(2.236)$$

$$=6.324-4.472$$

$$=1.852$$

Answer

$$1.852$$

# (ii) (\sqrt{300}+\sqrt{90}-\sqrt8)

Step 1: Simplify

\sqrt{300}=10\sqrt3

\sqrt{90}=3\sqrt{10}

\sqrt8=2\sqrt2

Step 2: Substitute values

$$10(1.732)+3(3.162)-2(1.414)$$

$$=17.320+9.486-2.828$$

$$=23.978$$

Answer

$$23.978$$

Q.4 Arrange surds in descending order ▾

✓ Solution

# (i) (3\sqrt5,\ 9\sqrt4,\ 6\sqrt3)

Step 1: Find approximate values

3\sqrt5=3(2.236)=6.708

9\sqrt4=9(2)=18

6\sqrt3=6(1.732)=10.392

Descending order

9\sqrt4>6\sqrt3>3\sqrt5

# (ii) (2\sqrt3\sqrt5,\ 3\sqrt4\sqrt7,\ \sqrt{\sqrt3})

Step 1: Simplify

2\sqrt3\sqrt5=2\sqrt{15}\approx7.746

3\sqrt4\sqrt7=3(2)\sqrt7=6\sqrt7\approx15.874

\sqrt{\sqrt3}=\sqrt[4]{3}\approx1.316

Descending order

3\sqrt4\sqrt7 > 2\sqrt3\sqrt5 > \sqrt{\sqrt3}

Q.5 Can you get a pure surd when you find ▾

✓ Solution

# (i) Sum of two surds

Yes.

Example:

\sqrt2+\sqrt3

is a pure surd.

# (ii) Difference of two surds

Yes.

Example:

\sqrt5-\sqrt2

is a pure surd.

# (iii) Product of two surds

Yes.

Example:

\sqrt2\times\sqrt3=\sqrt6

which is a pure surd.

# (iv) Quotient of two surds

Yes.

Example:

\frac{\sqrt6}{\sqrt2}=\sqrt3

which is a pure surd.

Q.6 Can you get a rational number when you compute ▾

✓ Solution

# (i) Sum of two surds

Yes.

Example:

\sqrt2+(-\sqrt2)=0

which is rational.

# (ii) Difference of two surds

Yes.

Example:

\sqrt5-\sqrt5=0

which is rational.

# (iii) Product of two surds

Yes.

Example:

\sqrt2\times\sqrt2=2

which is rational.

# (iv) Quotient of two surds

Yes.

Example:

\frac{\sqrt7}{\sqrt7}=1

which is rational.

Ex 2.7Rationalisation of Surds5 questions

Q.1 Rationalise the denominator ▾

✓ Solution

# (i) (\frac1{\sqrt{50}})

Solution

\frac1{\sqrt{50}} \times \frac{\sqrt{50}}{\sqrt{50}}

# [

\frac{\sqrt{50}}{50}
]

# [

\frac{\sqrt{25\times2}}{50}
]

# [

\frac{5\sqrt2}{50}
]

# [

\frac{\sqrt2}{10}
]

Answer

\frac{\sqrt2}{10}

# (ii) (\frac5{3\sqrt5})

Solution

\frac5{3\sqrt5} \times \frac{\sqrt5}{\sqrt5}

# [

\frac{5\sqrt5}{15}
]

# [

\frac{\sqrt5}{3}
]

Answer

\frac{\sqrt5}{3}

# (iii) (\frac{\sqrt{75}}{\sqrt{18}})

Solution

# [

\frac{\sqrt{3\times5\times5}}{\sqrt{3\times2\times3}}
]

# [

\frac{5\sqrt3}{3\sqrt2}
]

Rationalising:

\frac{5\sqrt3}{3\sqrt2} \times \frac{\sqrt2}{\sqrt2}

# [

\frac{5\sqrt6}{6}
]

Answer

\frac{5\sqrt6}{6}

# (iv) (\frac{3\sqrt5}{\sqrt6})

Solution

\frac{3\sqrt5}{\sqrt6} \times \frac{\sqrt6}{\sqrt6}

# [

\frac{3\sqrt{30}}{6}
]

# [

\frac{\sqrt{30}}{2}
]

Answer

\frac{\sqrt{30}}{2}

Q.2 Rationalise the denominator and simplify: (i) (√48 + √32) / (√27 − √18) (ii) (5√3 + √2) / (√3 + √2) (iii) (2√6 − √5) / (3√5 − 2√6) (iv) (√5 / (√6 + 2)) − (√5 / (√6 − 2)). ▾

Final simplified answers

(i) (√48 + √32) / (√27 − √18) = (20 + 8√6) / 3.
(ii) (5√3 + √2) / (√3 + √2) = 13 − 4√6.
(iii) (2√6 − √5) / (3√5 − 2√6) = (9 + 4√30) / 21.
(iv) √5/(√6+2) − √5/(√6−2) = −2√5.

Q.3 Find the value of (a) and (b) if ▾

✓ Solution

\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b

Solution

Rationalise the denominator.

\frac{\sqrt7-2}{\sqrt7+2} \times \frac{\sqrt7-2}{\sqrt7-2}

Using:

genui{"math_block_widget_always_prefetch_v2":{"content":"(a+b)(a-b)=a^2-b^2"}}

Numerator

(\sqrt7-2)^2

=7+4-4\sqrt7

=11-4\sqrt7

Denominator

(\sqrt7+2)(\sqrt7-2)

$$=7-4$$

$$=3$$

Therefore

\frac{\sqrt7-2}{\sqrt7+2} ========================= \frac{11-4\sqrt7}{3}

# [

-\frac43\sqrt7+\frac{11}{3}
]

Comparing with:

a\sqrt7+b

we get:

a=-\frac43

b=\frac{11}{3}

Answer

a=-\frac43,\qquad b=\frac{11}{3}

Q.4 If x = √5 + 2, find the value of x^2 + 1/x^2. ▾

✓ Solution

x=\sqrt5+2

find the value of

x^2+\frac1{x^2}

Step 1: Find (x^2)

x^2=(\sqrt5+2)^2

=5+4+4\sqrt5

=9+4\sqrt5

Step 2: Find (\frac1x)

\frac1x=\frac1{\sqrt5+2}

Rationalising:

\frac1{\sqrt5+2} \times \frac{\sqrt5-2}{\sqrt5-2}

=\frac{\sqrt5-2}{5-4}

=\sqrt5-2

Step 3: Find (\frac1{x^2})

\frac1{x^2}=(\sqrt5-2)^2

=5+4-4\sqrt5

=9-4\sqrt5

Step 4: Add

x^2+\frac1{x^2}

=(9+4\sqrt5)+(9-4\sqrt5)

$$=18$$

Answer

$$18$$

Q.5 Given √2 = 1.414, find the value of (8 − 5√2) / (3 − 2√2) correct to 3 decimal places. ▾

✓ Solution

\sqrt2=1.414

find the value of

\frac{8-5\sqrt2}{3-2\sqrt2}

correct to 3 decimal places.

Step 1: Rationalise the denominator

\frac{8-5\sqrt2}{3-2\sqrt2} \times \frac{3+2\sqrt2}{3+2\sqrt2}

Denominator

(3-2\sqrt2)(3+2\sqrt2)

$$=9-8$$

$$=1$$

Numerator

(8-5\sqrt2)(3+2\sqrt2)

=24+16\sqrt2-15\sqrt2-20

=4+\sqrt2

Step 2: Substitute value

4+\sqrt2

$$=4+1.414$$

$$=5.414$$

Answer

$$5.414$$

Ex 2.8Scientific Notation5 questions

# Important Form

A number is written in scientific notation as:

a\times10^n\quad\text{where }1\le a<10

Q.1 Represent the following numbers in scientific notation ▾

✓ Solution

# (i) (569430000000)

Move decimal point after first digit.

569430000000=5.6943\times10^{11}

Answer

5.6943\times10^{11}

# (ii) (2000.57)

2000.57=2.00057\times10^3

Answer

2.00057\times10^3

# (iii) (0.0000006000)

Move decimal 7 places right.

0.0000006000=6.000\times10^{-7}

Answer

6.000\times10^{-7}

# (iv) (0.0009000002)

0.0009000002=9.000002\times10^{-4}

Answer

9.000002\times10^{-4}

Q.2 Write the following numbers in decimal form ▾

✓ Solution

# (i) (3.459\times10^6)

Move decimal 6 places right.

$$=3459000$$

Answer

$$3459000$$

# (ii) (5.678\times10^4)

$$=56780$$

Answer

$$56780$$

# (iii) (1.00005\times10^{-5})

Move decimal 5 places left.

$$=0.0000100005$$

Answer

$$0.0000100005$$

# (iv) (2.530009\times10^{-7})

$$=0.0000002530009$$

Answer

$$0.0000002530009$$

Q.3 Represent the following numbers in scientific notation ▾

✓ Solution

# (i)

(300000)^2\times(20000)^4

Step 1: Write in scientific notation

300000=3\times10^5

20000=2\times10^4

Step 2: Apply powers

(3\times10^5)^2(2\times10^4)^4

=3^2\times10^{10}\times2^4\times10^{16}

=9\times16\times10^{26}

=144\times10^{26}

=1.44\times10^{28}

Answer

1.44\times10^{28}

# (ii)

(0.000001)^{11}\div(0.005)^3

Step 1: Convert to scientific notation

0.000001=10^{-6}

0.005=5\times10^{-3}

Step 2: Simplify

(10^{-6})^{11}\div(5\times10^{-3})^3

=10^{-66}\div(125\times10^{-9})

=\frac{10^{-66}}{125\times10^{-9}}

=\frac1{125}\times10^{-57}

=0.008\times10^{-57}

=8\times10^{-60}

Answer

8\times10^{-60}

# (iii)

\frac{(0.00003)^6(0.00005)^4}{(0.009)^3(0.05)^2}

Step 1: Scientific notation

0.00003=3\times10^{-5}

0.00005=5\times10^{-5}

0.009=9\times10^{-3}

0.05=5\times10^{-2}

Step 2: Apply powers

Numerator:

(3^6\times10^{-30})(5^4\times10^{-20})

=729\times625\times10^{-50}

=455625\times10^{-50}

Denominator:

(9^3\times10^{-9})(5^2\times10^{-4})

=729\times25\times10^{-13}

=18225\times10^{-13}

Step 3: Divide

\frac{455625\times10^{-50}}{18225\times10^{-13}}

=25\times10^{-37}

=2.5\times10^{-36}

Answer

2.5\times10^{-36}

Q.4 Represent the following information in scientific notation ▾

✓ Solution

# (i) World population

$$7000,000,000$$

=7\times10^9

Answer

7\times10^9

# (ii) One light year

9460528400000000\text{ m}

=9.4605284\times10^{15}\text{ m}

Since (1\text{ km}=1000\text{ m}),

9.4605284\times10^{15}\text{ m}=9.4605284\times10^{12}\text{ km}

Answer

9.4605284\times10^{12}\text{ km}

# (iii) Mass of an electron

$$0.00000000000000000000000000000091093822$$

=9.1093822\times10^{-31}

Answer

9.1093822\times10^{-31}\text{ kg}

Q.5 Simplify: (i) (2.75 × 10^7) + (1.23 × 10^8), (ii) (1.598 × 10^17) − (4.58 × 10^15), (iii) (1.02 × 10^10)(1.20 × 10^−3), (iv) (8.41 × 10^4) ÷ (4.3 × 10^5). ▾

✓ Solution

# (i)

(2.75\times10^7)+(1.23\times10^8)

Step 1: Equalise powers

2.75\times10^7 ============== 0.275\times10^8

Step 2: Add

(0.275+1.23)\times10^8

=1.505\times10^8

Answer

1.505\times10^8

# (ii)

(1.598\times10^{17})-(4.58\times10^{15})

Step 1: Equalise powers

4.58\times10^{15} ================= 0.0458\times10^{17}

Step 2: Subtract

(1.598-0.0458)\times10^{17}

=1.5522\times10^{17}

Answer

1.5522\times10^{17}

# (iii)

(1.02\times10^{10})(1.20\times10^{-3})

Multiply coefficients

1.02\times1.20=1.224

Add powers

10^{10}\times10^{-3}=10^7

Answer

1.224\times10^7

# (iv)

(8.41\times10^4)\div(4.3\times10^5)

Step 1: Divide coefficients

\frac{8.41}{4.3}\approx1.956

Step 2: Subtract powers

10^{4-5}=10^{-1}

Answer

1.956\times10^{-1}

# Activity – 3

Complete the table and arrange the planets in order of magnitude

# Completed Table

| Planet | Decimal Form (in km) | Scientific Notation (in km) |
| ------- | -------------------- | --------------------------- |
| Jupiter | 778000000 | (7.78\times10^8) |
| Mercury | 58000000 | (5.8\times10^7) |
| Mars | 228000000 | (2.28\times10^8) |
| Uranus | 2870000000 | (2.87\times10^9) |
| Venus | 108000000 | (1.08\times10^8) |
| Neptune | 4500000000 | (4.5\times10^9) |
| Earth | 150000000 | (1.5\times10^8) |
| Saturn | 1430000000 | (1.43\times10^9) |

# Arrangement in order of magnitude

(Closest to the Sun to farthest)

1. Mercury

5.8\times10^7

2. Venus

1.08\times10^8

3. Earth

1.5\times10^8

4. Mars

2.28\times10^8

5. Jupiter

7.78\times10^8

6. Saturn

1.43\times10^9

7. Uranus

2.87\times10^9

8. Neptune

4.5\times10^9

# Final Order

\text{Mercury} < \text{Venus} < \text{Earth} < \text{Mars} < \text{Jupiter} < \text{Saturn} < \text{Uranus} < \text{Neptune}

Ex 2.9Multiple Choice Questions20 questions

Q.1 If n is a natural number then √n is: (1) always a natural number; (2) always an irrational number; (3) always a rational number; (4) may be rational or irrational. ▾

Answer: (4) may be rational or irrational.

Reason: If n is a perfect square (e.g. n=9) √n is rational (√9=3). If n is not a perfect square (e.g. n=2) √n is irrational (√2).

Q.2 Which of the following is not true? ▾

✓ Solution

1. Every rational number is a real number.
2. Every integer is a rational number.
3. Every real number is an irrational number.
4. Every natural number is a whole number.

Solution

Real numbers include both rational and irrational numbers.

Hence statement (3) is false.

Answer

\boxed{(3)\ \text{Every real number is an irrational number}}

Q.3 Which one of the following regarding the sum of two irrational numbers is true? ▾

✓ Solution

1. always an irrational number
2. may be a rational or irrational number
3. always a rational number
4. always an integer

Solution

Example:

\sqrt2+\sqrt3

is irrational.

But,

\sqrt2+(-\sqrt2)=0

is rational.

Answer

\boxed{(2)\ \text{may be a rational or irrational number}}

Q.4 Which one of the following has a terminating decimal expansion? ▾

✓ Solution

1. (\frac5{64})
2. (\frac89)
3. (\frac{14}{15})
4. (\frac1{12})

Solution

A rational number has a terminating decimal if the denominator contains only factors (2) and/or (5).

$$64=2^6$$

Hence,

\frac5{64}

has a terminating decimal.

Answer

\boxed{(1)\ \frac5{64}}

Q.5 Which one of the following is an irrational number? ▾

✓ Solution

1. (\sqrt{25})
2. (\sqrt{\frac94})
3. (\frac7{11})
4. (\pi)

Solution

\sqrt{25}=5

\sqrt{\frac94}=\frac32

\frac7{11}

are rational numbers.

\pi

is irrational.

Answer

\boxed{(4)\ \pi}

Q.6 An irrational number between 2 and 2.5 is: (1) √11; (2) √5; (3) √2.5; (4) √8. ▾

Answer: (2) √5.

Reason: 2^2 = 4 and (2.5)^2 = 6.25; 5 lies between 4 and 6.25, so √5 lies between 2 and 2.5 and is irrational.

Q.7 The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates with one place of decimal is: (1) 1/10; (2) 3/10; (3) 3; (4) 30. ▾

Answer: (2) 3/10.

Reason: (1/3) × (3/10) = 1/10 = 0.1, which is a terminating decimal with one decimal place.

Q.8 If 1/7 = 0.(142857), find the value of 5/7. ▾

Multiply the repeating decimal by 5: 5 \times 0.(142857) = 0.(714285).

Answer: 0.(714285)

Q.9 Find the odd one out ▾

✓ Solution

1. (\sqrt{32}\times\sqrt2)
2. (\sqrt{27}\div\sqrt3)
3. (\sqrt{72}\times\sqrt8)
4. (\sqrt{54}\div\sqrt{18})

Solution

(1)

\sqrt{32}\times\sqrt2 ===================== # \sqrt{64} 8

(2)

\sqrt{27}\div\sqrt3 =================== # \sqrt9 3

(3)

\sqrt{72}\times\sqrt8 ===================== # \sqrt{576} 24

(4)

\sqrt{54}\div\sqrt{18} ====================== \sqrt3

which is irrational.

Hence it is different from others.

Answer

\boxed{(4)\ \sqrt{54}\div\sqrt{18}}

Q.10 Question 10 ▾

✓ Solution

The complete question is not visible.
Only the solution fragment:

0.343434\ldots+0.344444\ldots

is visible.

Please share the full question/options for exact validation.

Q.11 Which of the following statement is false? ▾

✓ Solution

1. The square root of (25) is (5) or (-5)
2. (-\sqrt{25}=-5)
3. (\sqrt{25}=5)
4. (\sqrt{25}=\pm5)

Solution

The symbol:

\sqrt{25}

represents only the principal positive square root.

Hence,

\sqrt{25}=5

not (\pm5).

Answer

\boxed{(4)\ \sqrt{25}=\pm5}

Q.12 Which one of the following is not a rational number? ▾

✓ Solution

1. (\sqrt{\frac8{18}})
2. (\frac73)
3. (\sqrt{0.01})
4. (\sqrt{13})

Solution

\sqrt{\frac8{18}} ================= # \sqrt{\frac49} \frac23

rational.

\frac73

rational.

\sqrt{0.01} =========== \frac1{10}

rational.

\sqrt{13}

irrational.

Answer

\boxed{(4)\ \sqrt{13}}

Q.13 (\sqrt{27}+\sqrt{12}=) ▾

✓ Solution

1. (\sqrt{39})
2. (5\sqrt6)
3. (5\sqrt3)
4. (3\sqrt5)

Solution

\sqrt{27}=3\sqrt3

\sqrt{12}=2\sqrt3

3\sqrt3+2\sqrt3 =============== 5\sqrt3

Answer

\boxed{(3)\ 5\sqrt3}

Q.14 If $\sqrt{80}=k\sqrt5$, then k = ? ▾

Answer: 4

Q.15 (4\sqrt7\times2\sqrt3=) ▾

✓ Solution

Solution

4\sqrt7\times2\sqrt3 ==================== 8\sqrt{21}

Answer

\boxed{(2)\ 8\sqrt{21}}

Q.16 Rationalise: $\dfrac{2\sqrt{3}}{3\sqrt{2}}$. ▾

Multiply numerator and denominator by $\sqrt2$:

$\dfrac{2\sqrt3}{3\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{2\sqrt6}{6}=\dfrac{\sqrt6}{3}.$

Q.17 Simplify: $(2\sqrt5-\sqrt2)^2$. ▾

Use (a-b)^2 = a^2 - 2ab + b^2 with a = 2√5, b = √2:

$(2\sqrt5)^2 - 2(2\sqrt5)(\sqrt2) + (\sqrt2)^2 = 20 - 4\sqrt{10} + 2 = 22 - 4\sqrt{10}.$

Q.18 Question 18 ▾

✓ Solution

(0.000729)^{-3/4}\times(0.09)^{-3/4}

Solution

0.000729=\left(\frac3{10}\right)^6

0.09=\left(\frac3{10}\right)^2

Thus,

\left(\frac3{10}\right)^{-9/2} \times \left(\frac3{10}\right)^{-3/2}

# [

\left(\frac3{10}\right)^{-6}
]

# [

\left(\frac{10}{3}\right)^6
]

Answer

\boxed{\frac{10^6}{3^6}}

Q.19 If $\sqrt{9x}=3\sqrt[3]{9^2}$, find $x$. ▾

Write powers of 3: $9=3^2$. Then RHS = $3\cdot9^{2/3}=3\cdot3^{4/3}=3^{7/3}.$

So $(9x)^{1/2}=3^{7/3}$. Square both sides: $9x=3^{14/3}$.

Since $9=3^2$, divide: $x=3^{14/3-2}=3^{8/3}.$

Hence $x=3^{8/3}$ (approximately $18.72$).

Q.20 The length and breadth of a rectangular plot are $5\times10^5$ m and $4\times10^4$ m respectively. Find the area. ▾

Area = $(5\times10^5)(4\times10^4)=20\times10^9=2\times10^{10}$ m².

Brain Grain Grade 9 Maths

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Unit 1: Set Language

Unit 3: Algebra

Samacheer Class 9 Maths - Real Numbers

Real Numbers — key concepts & quick answers

Solution

Answer

Step 1: Take LCM denominator

Answer

Step 1: Write with same decimal places

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Solution

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Step 1: Convert mixed fraction into improper fraction

Step 2: Convert into decimal

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Step 1: Use (\frac{1}{11})

Step 2: Find (\frac{1}{33})

Therefore,

Step 3: Find (\frac{71}{33})

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Step 1: Convert into decimals

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