Ch 3 The World of Numbers NCERT Solutions Ganita Manjari Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers - Ganita Manjari

Chapter 3 The World of Numbers NCERT Solutions for Class 9 Maths is prepared by studyrankers faculty which will be helpful in covering the entire syllabus and solving the difficult problems given in exercise. You can also Download PDF of Class 9 Maths Chapter Ch 3 The World of Numbers NCERT Solutions which will prove useful guide in making a student confident. The chapter is taken from the new NCERT Mathematics textbook, Ganita Manjari.

We have also provided Chapter 3 The World of Numbers Revision Notes which will help you in completing your homework on time. These NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily. Students can also check Extra Questions Answer for The World of Numbers Class 9 Maths to prepare for their examination completely.

NCERT Solutions for Chapter 3 The World of Numbers Class 9 Maths

Page No. 42

Exercise 3.1

1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Answer

Given:

2 bags of spices = 15 copper ingots

12 bags of spices = ? copper ingots

Since \(12 = 6 \times 2\), the number of bags has increased 6 times.

Therefore, the number of copper ingots will also increase 6 times.

\(15 \times 6 = 90\)

Therefore, the merchant will receive 90 copper ingots.

2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Answer

The numbers 11, 13, 17 and 19 are all prime numbers.

A prime number has exactly two factors:

• 1
• The number itself

The next three prime numbers after 19 are:

\(\boxed{23,\;29,\;31}\)

Therefore, the next three numbers are 23, 29 and 31.

3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Answer

No, Natural Numbers are not closed under subtraction.

Example 1:

\(9-4=5\)

Since 5 is a natural number, the result belongs to the set of natural numbers.

Example 2:

\(4-9=-5\)

Since \(-5\) is not a natural number, the result does not belong to the set of natural numbers.

Therefore, Natural Numbers are not closed under subtraction.

4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Answer

There are 4 fingers (excluding the thumb).

Each finger has 3 joints.

Total joints that can be counted:

\(4 \times 3 = 12\)

Therefore, one hand can be used to count up to 12.

This counting method is believed to have contributed to the development of the base-12 (duodecimal) number system. The other hand could then be used to count groups of twelve.

 

Page No. 46

Think and Reflect

Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative represents the removal of that debt.

(Hint: If someone takes away (–) four of your debts that are each worth ₹3 (that is, –3), you are effectively ₹12 richer! Therefore, (–3) × (– 4) = +12.)

Answer

A negative number can be thought of as a debt or loss.

Multiplying by another negative means removing or taking away that debt.

If a debt is removed, the person's financial position improves, resulting in a positive gain.

Example:

Suppose you have four debts of ₹3 each.

Total debt = -₹12

If someone removes these four debts, the operation can be written as:

\((-3)\times(-4)\)

This means removing four debts of ₹3 each.

As a result, your debt disappears and your financial position improves by ₹12.

Therefore,

\((-3)\times(-4)=+12\)

Hence, multiplying a negative number by another negative number always gives a positive number.

Page No. 46

Exercise Set 3.2

1. The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?

Answer

Given:

Temperature at noon = \(4^\circ\text{C}\)

Drop in temperature = \(15^\circ\text{C}\)

Midnight temperature

\(=4-15\)

\(=-11^\circ\text{C}\)

Therefore, the midnight temperature is \(-11^\circ\text{C}\).

2. A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.

Answer

Represent the transactions using integers.

Debt of ₹850 = \(-850\)

Profit of ₹1200 = \(+1200\)

Loss of ₹450 = \(-450\)

Required equation:

\((-850)+1200+(-450)\)

\(=350-450\)

\(=-100\)

Therefore, the trader's final financial standing is ₹100 in debt.

3. Calculate the following using Brahmagupta's laws:

(i) \((-12)\times5\)

(ii) \((-8)\times(-7)\)

(iii) \(0-(-14)\)

(iv) \((-20)\div4\)

Answer

(i) \((-12)\times5=-60\)

(ii) \((-8)\times(-7)=56\)

(iii) \(0-(-14)\)

\(=0+14\)

\(=14\)

(iv) \((-20)\div4=-5\)

4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., \(10-(-5)=15\)).

Answer

Suppose you have ₹10 with you and you also have a debt of ₹5.

If someone clears your debt of ₹5, your money effectively increases by ₹5.

This can be written as

\(10-(-5)=10+5=15\)

Removing a debt has the same effect as gaining money.

Therefore, subtracting a negative number is the same as adding a positive number.

 

Page No. 47

Think and Reflect

Can you explain why we need \(q \ne 0\) in the definition of a rational number?

Answer

A rational number is defined as a number of the form \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers and \(q \ne 0\).

This condition is necessary because division by zero is not defined in mathematics.

If the denominator is zero, the value of the fraction cannot be determined.

Example:

\(\dfrac{8}{2}=4\) ✅ (Defined)

\(\dfrac{8}{0}\) ❌ (Not Defined)

Since fractions with denominator zero do not represent any valid number, they cannot be considered rational numbers.

Therefore, the denominator must always be non-zero, i.e., \(q \ne 0\).

Concept Used: A rational number is any number that can be expressed as a fraction whose numerator and denominator are integers, provided the denominator is not zero.

Definition:

\(\text{Rational Number}=\dfrac{p}{q},\quad p,q\in\mathbb{Z},\;q\ne0\)

Why is division by zero not allowed?

If division by zero were allowed, there would be no unique answer because multiplying any number by 0 always gives 0.

For example, if

\(\dfrac{6}{0}=x\)

then

\(0\times x=6\)

But \(0\times x=0\) for every value of \(x\), which is impossible.

Hence, division by zero is undefined.

📝 Quick Remember

Fraction	Rational Number?
\(\dfrac{7}{3}\)	✅ Yes
\(\dfrac{-5}{8}\)	✅ Yes
\(\dfrac{9}{0}\)	❌ No (Division by zero is undefined)

 

Think and Reflect

1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Answer

When two rational numbers have different denominators, we first find the Least Common Multiple (LCM) of the denominators.

Then, we convert both fractions into equivalent fractions having the same denominator.

After making the denominators equal, we can easily add or subtract the numerators.

Example:

Add

\(\dfrac{1}{2}+\dfrac{1}{3}\)

Step 1: Find the LCM of 2 and 3.

LCM = 6

Step 2: Convert both fractions into equivalent fractions.

\(\dfrac{1}{2}=\dfrac{3}{6}\)

\(\dfrac{1}{3}=\dfrac{2}{6}\)

Step 3: Add the numerators.

\(\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)

Therefore, the required answer is \(\dfrac{5}{6}\).

 

2. Verify the distributive law for rational numbers.

Answer

The distributive law states that

\(a\times(b+c)=a\times b+a\times c\)

Let

\(a=\dfrac12,\quad b=\dfrac13,\quad c=\dfrac16\)

Left-Hand Side (LHS)

\(\dfrac12\times\left(\dfrac13+\dfrac16\right)\)

\(=\dfrac12\times\dfrac36\)

\(=\dfrac12\times\dfrac12\)

\(=\dfrac14\)

Right-Hand Side (RHS)

\(\left(\dfrac12\times\dfrac13\right)+\left(\dfrac12\times\dfrac16\right)\)

\(=\dfrac16+\dfrac1{12}\)

\(=\dfrac2{12}+\dfrac1{12}\)

\(=\dfrac3{12}\)

\(=\dfrac14\)

Since

LHS = RHS = \(\dfrac14\)

Hence, the distributive law is verified.

 

Page No. 49

Exercise Set 3.3

1. Prove that the following rational numbers are equal:

(i) \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\)

(ii) \(\dfrac54\) and \(\dfrac{10}{8}\)

(iii) \(-\dfrac35\) and \(-\dfrac6{10}\)

(iv) \(\dfrac93\) and \(3\)

Answer

(i) \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\)

Using cross multiplication,

\(2\times6=12\)

\(3\times4=12\)

Since both products are equal,

\(\dfrac23=\dfrac46\)

Hence, the given rational numbers are equal.

(ii) \(\dfrac54\) and \(\dfrac{10}{8}\)

\(5\times8=40\)

\(4\times10=40\)

Since both products are equal,

\(\dfrac54=\dfrac{10}{8}\)

Hence, the given rational numbers are equal.

(iii) \(-\dfrac35\) and \(-\dfrac6{10}\)

\((-3)\times10=-30\)

\(5\times(-6)=-30\)

Since both products are equal,

\(-\dfrac35=-\dfrac6{10}\)

Hence, the given rational numbers are equal.

(iv) \(\dfrac93\) and \(3\)

Write \(3\) as \(\dfrac31\).

\(9\times1=9\)

\(3\times3=9\)

Since both products are equal,

\(\dfrac93=3\)

Hence, the given rational numbers are equal.

2. Find the sum:

(i) \(\dfrac25+\dfrac3{10}\)

(ii) \(\dfrac7{12}+\dfrac58\)

(iii) \(-\dfrac47+\dfrac3{14}\)

Answer

(i) \(\dfrac25+\dfrac3{10}\)

LCM of 5 and 10 = 10

\(\dfrac25=\dfrac4{10}\)

\(\dfrac4{10}+\dfrac3{10}=\dfrac7{10}\)

(ii) \(\dfrac7{12}+\dfrac58\)

LCM of 12 and 8 = 24

\(\dfrac7{12}=\dfrac{14}{24}\)

\(\dfrac58=\dfrac{15}{24}\)

\(\dfrac{14}{24}+\dfrac{15}{24}=\dfrac{29}{24}\)

(iii) \(-\dfrac47+\dfrac3{14}\)

LCM of 7 and 14 = 14

\(-\dfrac47=-\dfrac8{14}\)

\(-\dfrac8{14}+\dfrac3{14}=-\dfrac5{14}\)

 

3. Find the difference:

(i) \(\dfrac56-\dfrac14\)

(ii) \(\dfrac{11}{8}-\dfrac34\)

(iii) \(-\dfrac79-\left(-\dfrac23\right)\)

Answer

(i) \(\dfrac56-\dfrac14\)

LCM of 6 and 4 = 12

\(\dfrac56=\dfrac{10}{12}\)

\(\dfrac14=\dfrac3{12}\)

\(\dfrac{10}{12}-\dfrac3{12}=\dfrac7{12}\)

(ii) \(\dfrac{11}{8}-\dfrac34\)

\(\dfrac34=\dfrac68\)

\(\dfrac{11}{8}-\dfrac68=\dfrac58\)

(iii) \(-\dfrac79-\left(-\dfrac23\right)\)

\(-\dfrac79+\dfrac23\)

\(\dfrac23=\dfrac69\)

\(-\dfrac79+\dfrac69=-\dfrac19\)

 

4. Find the product:

(i) \(\dfrac23\times\dfrac3{10}\)

(ii) \(\dfrac7{11}\times\dfrac58\)

(iii) \(-\dfrac47\times\dfrac5{14}\)

Answer

(i) \(\dfrac23\times\dfrac3{10}\)

\(\dfrac23\times\dfrac3{10}=\dfrac6{30}=\dfrac15\)

(ii) \(\dfrac7{11}\times\dfrac58\)

\(\dfrac7{11}\times\dfrac58=\dfrac{35}{88}\)

(iii) \(-\dfrac47\times\dfrac5{14}\)

\(-\dfrac47\times\dfrac5{14}=-\dfrac{20}{98}\)

\(=-\dfrac{10}{49}\)

5. Find the quotient:

(i) \(\dfrac23\div\dfrac3{10}\)

(ii) \(\dfrac7{11}\div\dfrac58\)

(iii) \(-\dfrac47\div\dfrac5{14}\)

Answer

(i) \(\dfrac23\div\dfrac3{10}\)

\(\dfrac23\div\dfrac3{10}\)

\(=\dfrac23\times\dfrac{10}3\)

\(=\dfrac{20}9\)

(ii) \(\dfrac7{11}\div\dfrac58\)

\(\dfrac7{11}\div\dfrac58\)

\(=\dfrac7{11}\times\dfrac85\)

\(=\dfrac{56}{55}\)

(iii) \(-\dfrac47\div\dfrac5{14}\)

Divide by multiplying with the reciprocal of the second rational number.

\(-\dfrac47\div\dfrac5{14}\)

\(=-\dfrac47\times\dfrac{14}{5}\)

Cancel the common factor 7.

\(=-\dfrac45\times2\)

\(=-\dfrac85\)

Therefore, the required quotient is \(-\dfrac85\).

Concept Used: Division of rational numbers is performed by multiplying the first rational number by the reciprocal of the second rational number.

Formula Used:

\(\dfrac ab\div\dfrac cd=\dfrac ab\times\dfrac dc\)

 

6. Show that:

\(\left(\dfrac12+\dfrac34\right)\times\dfrac83=\dfrac12\times\dfrac83+\dfrac34\times\dfrac83\)

Answer

Left-Hand Side (LHS)

\(\left(\dfrac12+\dfrac34\right)\times\dfrac83\)

\(=\left(\dfrac24+\dfrac34\right)\times\dfrac83\)

\(=\dfrac54\times\dfrac83\)

\(=\dfrac{40}{12}\)

\(=\dfrac{10}{3}\)

Right-Hand Side (RHS)

\(\dfrac12\times\dfrac83+\dfrac34\times\dfrac83\)

\(=\dfrac43+2\)

\(=\dfrac43+\dfrac63\)

\(=\dfrac{10}{3}\)

Since,

LHS = RHS = \(\dfrac{10}{3}\)

Hence, the distributive law is verified.

Concept Used: Multiplication distributes over addition.

Formula Used:

\(a\times(b+c)=ab+ac\).

 

7. Simplify the following using the distributive property:

\(\dfrac79\left(\dfrac67-\dfrac34\right)\)

Answer

Using the distributive property,

\(\dfrac79\left(\dfrac67-\dfrac34\right)\)

\(=\dfrac79\times\dfrac67-\dfrac79\times\dfrac34\)

\(=\dfrac69-\dfrac{21}{36}\)

\(=\dfrac23-\dfrac7{12}\)

\(=\dfrac8{12}-\dfrac7{12}\)

\(=\dfrac1{12}\)

Therefore, the simplified value is \(\dfrac1{12}\).

8. Find the rational number \(x\) such that:

\(\dfrac56\left(x+\dfrac35\right)=\dfrac56x+\dfrac12\)

Answer

Left-Hand Side

\(\dfrac56\left(x+\dfrac35\right)\)

Apply the distributive property.

\(=\dfrac56x+\dfrac56\times\dfrac35\)

\(=\dfrac56x+\dfrac{15}{30}\)

\(=\dfrac56x+\dfrac12\)

This is exactly equal to the Right-Hand Side.

Therefore, the given equation is true for every rational number \(x\).

Hence, \(x\) can be any rational number.

 

Page No. 51

Think and Reflect

Try and represent \(\dfrac{8}{5}\) and \(-\dfrac{7}{4}\) on a number line.

Answer

(i) Representation of \(\dfrac85\)

Step 1: Convert the improper fraction into a mixed fraction.

\(\dfrac85=1\dfrac35\)

Step 2: Locate the integers 1 and 2 on the number line.

Step 3: Divide the interval between 1 and 2 into 5 equal parts.

Step 4: Move 3 parts to the right of 1.

This point represents \(\dfrac85\)

 

(ii) Representation of \(-\dfrac74\)

Step 1: Convert the improper fraction into a mixed fraction.

\(-\dfrac74=-1\dfrac34\)

Step 2: Locate the integers \(-2\) and \(-1\) on the number line.

Step 3: Divide the interval into 4 equal parts.

Step 4: Move 3 parts to the left of \(-1\) (or equivalently, 1 part to the right of \(-2\)).

This point represents \(-\dfrac74\)

Page No. 52

Exercise Set 3.4

1. Represent the rational numbers \(\dfrac23\), \(-\dfrac54\) and \(1\dfrac12\) on a single number line.

Answer

(i) Representation of \(\dfrac23\)

Since \(\dfrac23\) lies between 0 and 1, divide the interval from 0 to 1 into 3 equal parts and mark the second part.

(ii) Representation of \(-\dfrac54\)

Convert the improper fraction into a mixed fraction.

\(-\dfrac54=-1\dfrac14\)

It lies between \(-2\) and \(-1\).

Divide the interval into 4 equal parts and mark the point one-fourth to the left of \(-1\) (or three-fourths to the right of \(-2\)).

(iii) Representation of \(1\dfrac12\)

Convert the mixed fraction into an improper fraction.

\(1\dfrac12=\dfrac32\)

It lies between 1 and 2.

Divide the interval into 2 equal parts and mark the midpoint.

Therefore, all three rational numbers can be represented accurately on a single number line.

2. Find three distinct rational numbers that lie strictly between \(-\dfrac12\) and \(\dfrac14\).

Answer

First, make the denominators equal.

LCM of 2 and 4 = 4

\(-\dfrac12=-\dfrac24\)

Now we have

\(-\dfrac24\) and \(\dfrac14\)

To obtain more rational numbers between them, multiply both fractions by 2.

\(-\dfrac24=-\dfrac48\)

\(\dfrac14=\dfrac28\)

The rational numbers lying strictly between them are:

\(-\dfrac38,\;-\dfrac28,\;-\dfrac18,\;0,\;\dfrac18\)

Any three of these may be chosen.

One possible answer is:

\(\boxed{-\dfrac38,\;-\dfrac18,\;\dfrac18}\)

Therefore, the required three rational numbers are \(-\dfrac38,\;-\dfrac18,\;\dfrac18\).

 

3. Simplify the expression: \(\left(-\dfrac14\right)+\left(\dfrac5{12}\right)\).

Answer

Given,

\(\left(-\dfrac14\right)+\left(\dfrac5{12}\right)\)

LCM of 4 and 12 = 12

\(-\dfrac14=-\dfrac3{12}\)

Now add the fractions.

\(-\dfrac3{12}+\dfrac5{12}=\dfrac2{12}\)

Simplify the fraction.

\(\dfrac2{12}=\dfrac16\)

Therefore, the simplified value is \(\dfrac16\).

 

4. A tailor has \(15\dfrac34\) metres of fine silk. If making one kurta requires \(2\dfrac14\) metres of silk, exactly how many kurtas can he make?

Answer

Convert the mixed fractions into improper fractions.

\(15\dfrac34=\dfrac{63}{4}\)

\(2\dfrac14=\dfrac94\)

Number of kurtas

\(=\dfrac{63}{4}\div\dfrac94\)

\(=\dfrac{63}{4}\times\dfrac49\)

Cancel the common factor 4.

\(=\dfrac{63}{9}\)

\(=7\)

Therefore, the tailor can make 7 kurtas.

 

5. Find three rational numbers between 3.1415 and 3.1416.

Answer

Write both numbers with one more decimal place.

\(3.1415=3.14150\)

\(3.1416=3.14160\)

Now choose any three numbers lying strictly between them.

One possible answer is:

\(3.14151,\;3.14152,\;3.14153\)

Therefore, three rational numbers between the given numbers are 3.14151, 3.14152 and 3.14153.

6. Can you think of other way(s) to find a rational number between any two rational numbers?

Answer

Yes. One simple method is to take the average (mean) of the two rational numbers.

If the two rational numbers are \(a\) and \(b\), then

\(\dfrac{a+b}{2}\)

is a rational number lying between them.

Example:

Find a rational number between \(\dfrac12\) and \(\dfrac34\).

\(\dfrac{\dfrac12+\dfrac34}{2}\)

\(=\dfrac{\dfrac24+\dfrac34}{2}\)

\(=\dfrac{\dfrac54}{2}\)

\(=\dfrac58\)

Since

\(\dfrac12<\dfrac58<\dfrac34\)

\(\dfrac58\) is a rational number between them.

Therefore, taking the average is an easy method to find a rational number between any two rational numbers.

 

Page No. 53

Think and Reflect

Can \(\sqrt{2}\) be written as a rational number \(\dfrac{p}{q}\)?

Answer

No. The number \(\sqrt{2}\) cannot be written in the form

\(\dfrac{p}{q}\)

where \(p\) and \(q\) are integers and \(q\ne0\).

The decimal expansion of \(\sqrt{2}\) is

\(\sqrt{2}=1.41421356\ldots\)

This decimal expansion is non-terminating and non-repeating.

Since rational numbers always have decimal expansions that are either terminating or non-terminating repeating, \(\sqrt{2}\) cannot be a rational number.

Hence, \(\sqrt{2}\) is an irrational number.

Therefore, \(\sqrt{2}\) cannot be written as \(\dfrac{p}{q}\).

Think and Reflect

Try to prove the irrationality of \(\sqrt3\) using the approach of proof by contradiction. Will the same approach work for \(\sqrt5\), \(\sqrt7\), or \(\sqrt{10}\)?

Answer

Yes. The same method of proof by contradiction can be used to prove that \(\sqrt3\), \(\sqrt5\), \(\sqrt7\), and \(\sqrt{10}\) are irrational numbers.

Proof for \(\sqrt3\):

Step 1: Assume that \(\sqrt3\) is a rational number.

Then it can be written in the form

\(\sqrt3=\dfrac pq\)

where \(p\) and \(q\) are integers having no common factor and \(q\ne0\).

Step 2: Squaring both sides,

\(3=\dfrac{p^2}{q^2}\)

or

\(p^2=3q^2\)

This shows that \(p^2\) is divisible by 3.

Hence, \(p\) is also divisible by 3.

Let

\(p=3k\)

where \(k\) is an integer.

Step 3: Substitute \(p=3k\) into the equation.

\((3k)^2=3q^2\)

\(9k^2=3q^2\)

\(3k^2=q^2\)

This shows that \(q^2\) is also divisible by 3.

Hence, \(q\) is divisible by 3.

Step 4: Therefore, both \(p\) and \(q\) are divisible by 3.

This contradicts our assumption that \(p\) and \(q\) have no common factor.

Hence, our original assumption is false.

Therefore, \(\sqrt3\) is an irrational number.

The same proof can be applied to \(\sqrt5\), \(\sqrt7\), and \(\sqrt{10}\). In each case, assuming the number is rational leads to a contradiction. Therefore, these numbers are also irrational numbers.

Think and Reflect

Try to extend this method for constructing line segments of lengths \(\sqrt3\) and \(\sqrt5\) using a ruler and a compass. Generalise this method to construct a line segment of any length of the form \(\sqrt n\), where \(n\) is a positive integer.

Answer

The construction of \(\sqrt2\) can be extended to construct line segments of lengths \(\sqrt3\), \(\sqrt5\), and, in general, \(\sqrt n\) by repeatedly using the Pythagoras Theorem.

Construction of \(\sqrt3\)

1. Construct a line segment OA of length \(\sqrt2\).
2. At one end of this line segment, draw a perpendicularAB of length 1 unit.
3. Join the free end of the perpendicular to the opposite end of the original line segment.
4. The new hypotenuse OB has length \(\sqrt3\).

 

Construction of \(\sqrt5\)

1. Construct a line segment OA of length 2 units.
2. Draw a perpendicular AB of length 1 unit at one end.
3. Join the free end of the perpendicular to the other end of the 2-unit line segment.
4. The hypotenuse OB obtained has length \(\sqrt5\).

Since \(2^2+1^2=4+1=5\)

the hypotenuse is \(\sqrt5\).

Generalisation for Constructing \(\sqrt n\)

To construct a line segment of length \(\sqrt n\), where \(n\) is a positive integer:

1. Construct a line segment of length \(\sqrt{\,n-1\,}\).
2. Draw a perpendicular of length 1 unit at one end.
3. Join the free end of the perpendicular to the opposite end.
4. The hypotenuse obtained has length \(\sqrt n\).

This process can be repeated successively to construct \(\sqrt2,\;\sqrt3,\;\sqrt4,\;\sqrt5,\ldots\).

Therefore, by repeatedly applying the Pythagoras Theorem, a line segment of length \(\sqrt n\) can be constructed for any positive integer \(n\).

Concept Used: The construction is based on the Pythagoras Theorem, which relates the lengths of the sides of a right-angled triangle.

Formula Used:

\(\text{Hypotenuse}^2=\text{Base}^2+\text{Perpendicular}^2\)

or,

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

Examples:

• \(\sqrt2=\sqrt{1^2+1^2}\)
• \(\sqrt3=\sqrt{(\sqrt2)^2+1^2}\)
• \(\sqrt5=\sqrt{2^2+1^2}\)

Think and Reflect

The decimal expansion of \(\dfrac{p}{q}\) will be terminating precisely when the prime factors of \(q\) are only 2, only 5, or both 2 and 5. Can you explain why?

Answer

Yes. A rational number \(\dfrac{p}{q}\) has a terminating decimal expansion only when the denominator (after simplifying the fraction) has no prime factors other than 2 and 5.

This is because our decimal number system is based on 10, and \(10=2\times5\)

If the denominator contains only the prime factors 2 and/or 5, it can always be converted into a denominator of the form \(10,\;100,\;1000,\;\ldots\)

Since these denominators are powers of 10, the decimal expansion comes to an end after a finite number of digits.

Example 1:

\(\dfrac38\)

Since

\(8=2^3\)

Multiply numerator and denominator by 125.

\(\dfrac38=\dfrac{375}{1000}=0.375\)

Hence, the decimal expansion is terminating.

Example 2:

\(\dfrac7{20}\)

Since

\(20=2^2\times5\)

\(\dfrac7{20}=0.35\)

This is also a terminating decimal.

Example 3:

\(\dfrac13\)

The denominator contains the prime factor 3.

Its decimal expansion is

\(\dfrac13=0.3333\ldots\)

This decimal never ends.

Therefore, it is a non-terminating repeating decimal.

Hence, a rational number has a terminating decimal expansion only when the denominator (in simplest form) contains no prime factors other than 2 and/or 5.

 

2. Perform the long division for \(\dfrac{1}{13}\). Identify the repeating block of digits. Does it show cyclic properties if you evaluate \(\dfrac{2}{13}\)? Now compute \(\dfrac{3}{13}\), \(\dfrac{4}{13}\), etc. What do you notice?

Answer

Performing the long division,

\(\dfrac{1}{13}=0.076923076923\ldots\)

The repeating block of digits is

\(076923\).

Now evaluate the following fractions:

Fraction	Decimal Expansion	Repeating Block
\(\dfrac1{13}\)	0.076923...	076923
\(\dfrac2{13}\)	0.153846...	153846
\(\dfrac3{13}\)	0.230769...	230769
\(\dfrac4{13}\)	0.307692...	307692

We observe that each repeating block is obtained by a cyclic rotation of the digits 076923.

Therefore, the decimal expansion of \(\dfrac1{13}\) exhibits a cyclic pattern.

 

3. Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.

(i) \(\sqrt{81}\)

Answer

\(\sqrt{81}=9=\dfrac91\)

Classification: Rational Number.

 

(ii) \(\sqrt{12}\)

Answer

\(\sqrt{12}=2\sqrt3\)

Since \(\sqrt3\) is irrational, \(\sqrt{12}\) is also irrational.

Classification: Irrational Number.

 

(iii) \(0.33333\ldots\)

Answer

This is a repeating decimal.

\(0.33333\ldots=\dfrac13\)

Classification: Rational Number.

 

(iv) \(0.123451234512345\ldots\)

Answer

The block 12345 repeats continuously.

Hence, it is a repeating decimal.

Classification: Rational Number.

(An explicit fraction exists because every repeating decimal is rational.)

 

(v) \(1.01001000100001\ldots\)

Answer

The decimal does not repeat a fixed block. The number of zeros keeps increasing.

Classification: Irrational Number.

 

(vi) \(23.560185612239874790120\)

Answer

This is a terminating decimal.

Every terminating decimal is rational.

Removing the decimal point gives

\(\dfrac{23560185612239874790120}{1000000000000000000000}\)

which can be simplified further.

Classification: Rational Number.

Quick Rules

• Perfect square root → Rational
• Non-perfect square root → Irrational
• Terminating decimal → Rational
• Repeating decimal → Rational
• Non-terminating, non-repeating decimal → Irrational

 

4. The number \(0.\overline9\) (which means \(0.9999\ldots\)) is a rational number. Using algebra (let \(x=0.\overline9\), multiply by 10, and subtract), explain why \(0.\overline9\) is exactly equal to 1.

Answer

Let

\(x=0.\overline9\)

Multiply both sides by 10.

\(10x=9.\overline9\)

Subtract the first equation from the second.

\(10x-x=9.\overline9-0.\overline9\)

\(9x=9\)

\(x=1\)

But \(x=0.\overline9\).

Therefore,

\(0.\overline9=1\)

Concept Used: Infinite repeating decimals can be converted into fractions using algebra.

Formula Used:

Let the repeating decimal be \(x\), multiply by a suitable power of 10, then subtract to eliminate the repeating part.

 

5. We have seen that the repeating block of \(\dfrac17\) is a cyclic number. Try to find more numbers (\(n\)) whose reciprocals \(\left(\dfrac1n\right)\) produce decimals with repeating blocks that are cyclic.

Answer

Some numbers whose reciprocals produce cyclic repeating blocks are:

• \(13\)
• \(17\)
• \(19\)
• \(23\)
• \(29\)

For example,

\(\dfrac1{17}=0.0588235294117647\ldots\)

The repeating digits form a cyclic pattern.

These numbers are examples of numbers whose reciprocals produce cyclic decimals.

Note: Not every prime number has this property.

Think and Reflect

Consider this puzzle: What is the square root of \(-1\)? We know that \(1\times1=1\). We also know that \((-1)\times(-1)=1\). There is no Real Number that, when multiplied by itself, results in a negative number. Thus, \(\sqrt{-1}\) cannot exist on the real number line.

Answer

There is no real number whose square is equal to \(-1\).

For any real number \(x\),

\(x^2 \ge 0\)

This means the square of every real number is always zero or positive.

For example,

• \(2^2=4\)
• \((-2)^2=4\)
• \(0^2=0\)

Since no real number gives \(-1\) when multiplied by itself, \(\sqrt{-1}\) cannot be represented on the real number line.

To solve this problem, mathematicians introduced a new type of number called an Imaginary Number.

The symbol \(i=\sqrt{-1}\) is called the imaginary unit.

Using this symbol, \(i^2=-1\)

Imaginary numbers, together with real numbers, form the set of Complex Numbers, which are studied in higher classes.

Therefore, \(\sqrt{-1}\) is not a real number; it is represented by the imaginary unit \(i\).

Page No. 64

End of Chapter Exercises

1. Convert the following rational numbers into the form of a terminating decimal or non-terminating repeating decimal, whichever the case may be, by the process of long division.

(i) \(\dfrac{3}{50}\)

Answer

Divide 3 by 50 using the long division method.

Step 1:

Since 50 is greater than 3, write a decimal point and add zeros.

\(3.0000 \div 50\)

Step 2:

\(300 \div 50 = 6\)

Write 6 in the tenths place.

Remainder = 0

Therefore,

\(\dfrac{3}{50}=0.06\)

Since the remainder becomes zero, the decimal expansion terminates.

Therefore, \(\dfrac{3}{50}=0.06\), which is a terminating decimal.

 

(ii) \(\dfrac{2}{9}\)

Answer

Divide 2 by 9 using the long division method.

Step 1: Since 9 is greater than 2, write a decimal point and add zeros.

\(2.0000\div9\)

Step 2:

\(20\div9=2\)

Remainder = 2

Again,

\(20\div9=2\)

Remainder = 2

The same remainder repeats continuously.

Therefore,

\(\dfrac29=0.\overline2\) or \(\dfrac29=0.222222\ldots\)

Therefore, \(\dfrac29\) is a non-terminating repeating decimal.

2. Prove that \(\sqrt5\) is an irrational number.

Answer

We shall use the method of proof by contradiction.

Step 1: Assume that \(\sqrt5\) is a rational number.

Then it can be written in the form

\(\sqrt5=\dfrac pq\)

where \(p\) and \(q\) are integers having no common factor and \(q\ne0\).

Step 2: Squaring both sides,

\(5=\dfrac{p^2}{q^2}\)

or

\(p^2=5q^2\)

This means that \(p^2\) is divisible by 5.

Hence, \(p\) is also divisible by 5.

Let \(p=5k\)

where \(k\) is an integer.

Step 3: Substitute \(p=5k\) into the equation.

\((5k)^2=5q^2\)

\(25k^2=5q^2\)

\(5k^2=q^2\)

This shows that \(q^2\) is also divisible by 5.

Hence, \(q\) is divisible by 5.

Step 4:

Thus, both \(p\) and \(q\) are divisible by 5.

This contradicts our assumption that \(p\) and \(q\) have no common factor.

Therefore, our assumption is false.

Hence, \(\sqrt5\) is an irrational number.

3. Convert the following decimal numbers into the form of \(\dfrac{p}{q}\).

(i) \(12.6\)

Answer

Let \(x=12.6\)

Since there is one digit after the decimal point, multiply both sides by 10.

\(10x=126\)

\(x=\dfrac{126}{10}\)

Simplify the fraction.

\(=\dfrac{63}{5}\)

Therefore, \(12.6=\dfrac{63}{5}\).

(ii) \(0.0120\)

Answer

Let\(x=0.0120\)

There are four digits after the decimal point.

\(x=\dfrac{120}{10000}\)

Simplify the fraction.

\(=\dfrac{3}{250}\)

Therefore, \(0.0120=\dfrac{3}{250}\).

(iii) \(3.\overline{052}\)

Answer

Let \(x=3.\overline{052}\)

The repeating block has 3 digits.

Multiply both sides by \(1000\).

\(1000x=3052.\overline{052}\)

Subtract the original equation.

\(1000x-x=3052.\overline{052}-3.\overline{052}\)

\(999x=3049\)

\(x=\dfrac{3049}{999}\)

Therefore, \(3.\overline{052}=\dfrac{3049}{999}\).

(iv) \(1.\overline{235}\)

Answer

Let

\(x=1.\overline{235}\)

Multiply both sides by \(1000\).

\(1000x=1235.\overline{235}\)

Subtract the original equation.

\(999x=1234\)

\(x=\dfrac{1234}{999}\)

Therefore, \(1.\overline{235}=\dfrac{1234}{999}\).

(v) \(0.\overline{23}\)

Answer

Let

\(x=0.\overline{23}\)

Multiply both sides by 100.

\(100x=23.\overline{23}\)

Subtract the original equation.

\(100x-x=23\)

\(99x=23\)

\(x=\dfrac{23}{99}\)

Therefore, \(0.\overline{23}=\dfrac{23}{99}\).

(vi) \(2.0\overline{5}\)

Answer

Let

\(x=2.0\overline5=2.05555\ldots\)

There is 1 non-repeating digit (0) and 1 repeating digit (5).

Multiply by 10 to move the decimal after the non-repeating part.

\(10x=20.\overline5\)

Now multiply by 10 again.

\(100x=205.\overline5\)

Subtract the two equations.

\(100x-10x=205.\overline5-20.\overline5\)

\(90x=185\)

\(x=\dfrac{185}{90}\)

Simplify the fraction.

\(=\dfrac{37}{18}\)

Therefore, \(2.0\overline5=\dfrac{37}{18}\).

(vii) \(2.12\overline5\)

Answer

Let

\(x=2.12555\ldots\)

There are 2 non-repeating digits (12) and 1 repeating digit (5).

Multiply by 100.

\(100x=212.5555\ldots\)

Multiply by 10 again.

\(1000x=2125.5555\ldots\)

Subtract.

\(1000x-100x=2125.5555\ldots-212.5555\ldots\)

\(900x=1913\)

\(x=\dfrac{1913}{900}\)

Therefore, \(2.12\overline5=\dfrac{1913}{900}\).

(viii) \(3.12\overline5\)

Answer

Let

\(x=3.12555\ldots\)

Multiply by 100.

\(100x=312.5555\ldots\)

Multiply by 10 again.

\(1000x=3125.5555\ldots\)

Subtract.

\(900x=2813\)

\(x=\dfrac{2813}{900}\)

Therefore, \(3.12\overline5=\dfrac{2813}{900}\).

(ix) \(2.\overline{1625}\)

Answer

Let

\(x=2.\overline{1625}\)

The repeating block contains 4 digits.

Multiply by \(10000\).

\(10000x=21625.\overline{1625}\)

Subtract the original equation.

\(10000x-x=21625.\overline{1625}-2.\overline{1625}\)

\(9999x=21623\)

\(x=\dfrac{21623}{9999}\)

Therefore, \(2.\overline{1625}=\dfrac{21623}{9999}\).

 

4. Locate the following rational numbers on the number line.

(i) \(0.532\)

Answer

The number \(0.532\) lies between 0 and 1.

Step 1: Mark the points 0 and 1 on the number line.

Step 2: Divide the interval from 0 to 1 into 10 equal parts to represent tenths.

Step 3: Locate 0.5.

Step 4: Divide the interval from 0.5 to 0.6 into 10 equal parts to represent hundredths.

Step 5: Locate 0.53.

Step 6: Divide the interval from 0.53 to 0.54 into 10 equal parts to represent thousandths.

Step 7: Mark the second division after 0.53.

This point represents \(0.532\)

Therefore, \(0.532\) is located on the number line at the second thousandth division after \(0.53\).

 

(ii) \(1.1\overline5\)

Answer

The decimal \(1.1\overline5=1.15555\ldots\) lies between 1.1 and 1.2.

Step 1: Mark the integers 1 and 2 on the number line.

Step 2: Divide the interval from 1 to 2 into 10 equal parts.

This gives the tenths:

\(1.1,\;1.2,\;1.3,\ldots\)

Step 3: Locate the interval from 1.1 to 1.2.

Step 4: Divide this interval into 10 equal parts to obtain hundredths.

The number \(1.15555\ldots\) lies between \(1.15\) and \(1.16\)

Step 5: Divide the interval from 1.15 to 1.16 into 10 equal parts.

The point lies slightly to the right of 1.155 because the digit 5 continues indefinitely.

This point represents \(1.1\overline5\)

Therefore, \(1.1\overline5\) is located between \(1.15\) and \(1.16\) on the number line.

 

5. Find 6 rational numbers between 3 and 4.

Answer

Write the given numbers with a common denominator.

\(3=\dfrac{30}{10}\)

\(4=\dfrac{40}{10}\)

The fractions lying between them are

\(\dfrac{31}{10},\;\dfrac{32}{10},\;\dfrac{33}{10},\;\dfrac{34}{10},\;\dfrac{35}{10},\;\dfrac{36}{10}\)

or, in decimal form,

\(3.1,\;3.2,\;3.3,\;3.4,\;3.5,\;3.6\)

Therefore, six rational numbers between 3 and 4 are

\(\boxed{\dfrac{31}{10},\;\dfrac{32}{10},\;\dfrac{33}{10},\;\dfrac{34}{10},\;\dfrac{35}{10},\;\dfrac{36}{10}}\)

 

6. Find 5 rational numbers between \(\dfrac25\) and \(\dfrac35\).

Answer

Multiply both fractions by 6 to obtain a larger common denominator.

\(\dfrac25=\dfrac{12}{30}\)

\(\dfrac35=\dfrac{18}{30}\)

The five rational numbers lying between them are

\(\dfrac{13}{30},\;\dfrac{14}{30},\;\dfrac{15}{30},\;\dfrac{16}{30},\;\dfrac{17}{30}\)

Therefore, the required rational numbers are

\(\boxed{\dfrac{13}{30},\;\dfrac{14}{30},\;\dfrac{15}{30},\;\dfrac{16}{30},\;\dfrac{17}{30}}\) 

7. Find 5 rational numbers between \(\dfrac16\) and \(\dfrac25\).

Answer

LCM of 6 and 5 is 30.

\(\dfrac16=\dfrac5{30}\)

\(\dfrac25=\dfrac{12}{30}\)

There are not enough numbers between them.

Multiply both fractions by 2.

\(\dfrac16=\dfrac{10}{60}\)

\(\dfrac25=\dfrac{24}{60}\)

Five rational numbers between them are

\(\dfrac{11}{60},\;\dfrac{12}{60},\;\dfrac{13}{60},\;\dfrac{14}{60},\;\dfrac{15}{60}\)

Therefore, the required rational numbers are

\(\boxed{\dfrac{11}{60},\;\dfrac{12}{60},\;\dfrac{13}{60},\;\dfrac{14}{60},\;\dfrac{15}{60}}\)

 

8. If \(\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{16}{15}\), find the rational number \(x\).

Answer

Given,

\(\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{16}{15}\)

Take the LCM of 3 and 5.

\(\dfrac{5x+3x}{15}=\dfrac{16}{15}\)

\(\dfrac{8x}{15}=\dfrac{16}{15}\)

Multiply both sides by 15.

\(8x=16\)

\(x=2\)

Therefore, the required rational number is \(2\).

 

9. Let \(a\) and \(b\) be two non-zero rational numbers such that \(a+\dfrac1b=0\). Without assigning any numerical values, determine whether \(ab\) is positive or negative. Justify your answer.

Answer

Given,

\(a+\dfrac1b=0\)

Therefore,

\(a=-\dfrac1b\)

Multiply both sides by \(b\).

\(ab=-1\)

Since

\(-1\)

is a negative number,

\(ab\) is also negative.

Therefore, \(ab\) is negative.

 

10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form \(\dfrac{p}{10^4}\), where \(p\) is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by \(2^4\) or \(5^4\)? Give reasons.

Answer

If the last non-zero digit occurs in the fourth decimal place, then the number can be written as \(\dfrac{p}{10000}=\dfrac{p}{10^4}\) where \(p\) is an integer.

Since the fourth decimal place is the last non-zero digit, \(p\) cannot be divisible by 10. Otherwise, the fraction could be simplified by removing a factor of 10, and the last non-zero digit would no longer remain in the fourth decimal place.

Now write the fraction in its lowest form.

The denominator may lose common factors with the numerator during simplification.

Therefore, the denominator in the lowest form need not be divisible by \(2^4\) or \(5^4\).

Example:

\(\dfrac{2500}{10000}=\dfrac14\)

The original denominator is

\(10^4=10000\)

but after simplification the denominator becomes only

\(4=2^2\)

Hence, it is not divisible by \(2^4\) or \(5^4\).

Therefore, the denominator in the lowest form is not necessarily divisible by \(2^4\) or \(5^4\).

11. Without performing division, determine whether the decimal expansion of \(\dfrac{18}{125}\) is terminating or non-terminating. If it terminates, state the number of decimal places.

Answer

First, check whether the fraction is in its lowest form.

Since the HCF of 18 and 125 is 1, the fraction is already in its lowest form.

Hence, the fraction is already in its lowest form.

Prime factorise the denominator.

\(125=5^3\)

The denominator contains only the prime factor 5.

Therefore, the decimal expansion is terminating.

To determine the number of decimal places, write the denominator as a power of 10.

\(125\times8=1000=10^3\)

Hence,

\(\dfrac{18}{125}=\dfrac{18\times8}{125\times8}=\dfrac{144}{1000}=0.144\)

Therefore, \(\dfrac{18}{125}\) has a terminating decimal expansion with 3 decimal places.

Concept Used: A rational number has a terminating decimal expansion if the denominator (in lowest form) contains only the prime factors 2 and/or 5.

Rule:

If the denominator is \(2^m\times5^n\), then the number of decimal places is

\(\max(m,n)\)

 

12. A rational number in its lowest form has denominator \(2^3\times5\). How many decimal places will its decimal expansion have? Explain your answer.

Answer

The denominator is

\(2^3\times5\)

To convert it into a power of 10, both prime factors must have the same exponent.

Multiply the denominator by \(5^2\).

\(2^3\times5\times5^2=2^3\times5^3=10^3\)

Hence, the denominator becomes

\(1000\)

Therefore, the decimal expansion has 3 decimal places.

Hence, the required answer is 3 decimal places.

Formula Used:

If the denominator is \(2^m\times5^n\), then

Number of Decimal Places \(=\max(m,n)\)

Here,

\(m=3,\quad n=1\)

Therefore,

\(\max(3,1)=3\)

 

13. Let \(a=\dfrac{7}{12}\) and \(b=\dfrac56\). Express both \(a\) and \(b\) in the form \(\dfrac{k_1}{m}\) and \(\dfrac{k_2}{m}\), where \(k_1\), \(k_2\), and \(m\) are integers and \(k_2-k_1>6\). Using the same denominator \(m\), write exactly five distinct rational numbers lying between \(a\) and \(b\) keeping an integer numerator. Explain why the condition \(k_2-k_1>n+1\) is necessary to find \(n\) such rational numbers between two rational numbers using this method.

Answer

Given,

\(a=\dfrac7{12}\)

\(b=\dfrac56\)

The LCM of 12 and 6 is 12.

\(\dfrac56=\dfrac{10}{12}\)

The difference between the numerators is

\(10-7=3\)

This is less than 6.

Multiply both fractions by 3.

\(a=\dfrac{21}{36}\)

\(b=\dfrac{30}{36}\)

Now,

\(30-21=9>6\)

Hence, the required form is

\(k_1=21,\quad k_2=30,\quad m=36\)

The five rational numbers lying between them are

\(\boxed{\dfrac{22}{36},\;\dfrac{23}{36},\;\dfrac{24}{36},\;\dfrac{25}{36},\;\dfrac{26}{36}}\)

Why is the condition \(k_2-k_1>n+1\) necessary?

Between \(k_1\) and \(k_2\), there are

\(k_2-k_1-1\)

integers.

To obtain exactly \(n\) rational numbers, we must have

\(k_2-k_1-1\ge n\)

or

\(k_2-k_1\ge n+1\)

Thus, the given condition ensures that enough integer numerators are available to form the required rational numbers.

 

14. Three rational numbers \(x\), \(y\), and \(z\) satisfy \(x+y+z=0\) and \(xy+yz+zx=0\). Show that all the rational numbers \(x\), \(y\), and \(z\) must be simultaneously zero.

Answer

Given,

\(x+y+z=0\)

and

\(xy+yz+zx=0\)

Now,

\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\)

Substitute the given values.

\(0=x^2+y^2+z^2+2(0)\)

\(x^2+y^2+z^2=0\)

The square of every rational number is non-negative.

The sum of three non-negative numbers can be zero only if each of them is zero.

Therefore,

\(x^2=0,\quad y^2=0,\quad z^2=0\)

Hence,

\(x=0,\quad y=0,\quad z=0\)

Therefore, all three rational numbers must be simultaneously zero.

Identity Used:

\((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\)

 

15. Show that the rational number \(\dfrac{a+b}{2}\) lies between the rational numbers \(a\) and \(b\).

Answer

We have to show that \(a<\dfrac{a+b}{2}\)

Step 1: Compare \(\dfrac{a+b}{2}\) with \(a\).

Since, a<b

a+a < a+b

2a < a+b

Dividing both sides by 2,

\(a<\dfrac{a+b}{2}\)

Step 2: Compare \(\dfrac{a+b}{2}\) with \(b\).

Again Since, 

2a < a+b

Dividing both sides by 2,

\(\dfrac{a+b}{2}\)

Combining the two results,

\(a<\dfrac{a+b}{2}\)

Hence, the rational number \(\dfrac{a+b}{2}\) always lies between \(a\) and \(b\).

 

16. Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14, which is referred to as the square root spiral.

Answer

The square root spiral is constructed using the Pythagoras Theorem.

Each new right triangle has:

• One side equal to 1 unit.
• The other side equal to the hypotenuse of the previous triangle.

Applying the Pythagoras Theorem successively:

Triangle	Calculation	Hypotenuse
1st	\(\sqrt{1^2+1^2}\)	\(\sqrt2\)
2nd	\(\sqrt{(\sqrt2)^2+1^2}\)	\(\sqrt3\)
3rd	\(\sqrt{(\sqrt3)^2+1^2}\)	2
4th	\(\sqrt{2^2+1^2}\)	\(\sqrt5\)
5th	\(\sqrt{(\sqrt5)^2+1^2}\)	\(\sqrt6\)
6th	\(\sqrt{(\sqrt6)^2+1^2}\)	\(\sqrt7\)
7th	\(\sqrt{(\sqrt7)^2+1^2}\)	\(2\sqrt2=\sqrt8\)
8th	\(\sqrt{(\sqrt8)^2+1^2}\)	3=\sqrt9
9th	\(\sqrt{3^2+1^2}\)	\(\sqrt{10}\)

Hence, the hypotenuses obtained successively are

\(\boxed{\sqrt2,\;\sqrt3,\;2,\;\sqrt5,\;\sqrt6,\;\sqrt7,\;\sqrt8,\;3,\;\sqrt{10}}\)