NCERT Class 9 Maths Chapter 3 The World of Numbers

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NCERT Class 9 Maths Chapter 3 The World of Numbers

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The World of Numbers

Chapter – 3

Exercise Set 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?

Ans: Given:

2 bags of spices → 15 ingots

12 bags of spices → ? ingots

First, find how many times 2 bags fit into 12 bags:

12 ÷ 2 = 6

So, the trader makes 6 such exchanges.

Total ingots received:

6 × 15 = 90

The trader 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.

Ans: The numbers 11,13,17,19 are all prime numbers (numbers that have only two factors: 1 and itself).

Next three prime numbers after 19 are 23, 29, 31.

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

Ans: Natural numbers are 1, 2, 3, 4,…

Check subtraction:

5 – 3 = 2 (natural number)

3 – 5 = -2 (not a natural number)

4 – 5 = -1 (not a natural number)

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?

Ans: Each finger (except thumb) has 3 joints.

Number of fingers used =4 (excluding thumb)

Total joints = 4 × 3 = 12

So, we can count up to 12 using one hand.

Relation to base-12 system:

Since counting reaches 12 on one hand, it naturally leads to a base-12 (duodecimal) counting system used in ancient times.

Therefore:

Total count = 12

This explains the origin of base-12 counting system.

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?

Ans: The temperature at noon is 4 °C. By midnight, it drops by 15° C, which means we subtract 15 from 4:

4 – 15 = -11

So, the midnight temperature is -11 °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.

Ans: Debt = -₹850

Profit = +₹1200

Loss = -₹450

Equation: -₹850 + ₹1200 – ₹450

Step-by-step calculation:

= ₹350 – ₹450

= -₹100

Therefore, his final financial standing is -₹100 (a loss of ₹100).

3. Calculate the following using Brahmagupta’s laws:

(i) (-12) × 5

Ans: (-12) × 5

The product of a debt and a fortune is a debt(A negative multiplied by a positive is negative):

-12 × 5 = -60

(ii) (-8) × (-7)

Ans: (-8) × (-7)

The product of two debts is a fortune (A negative multiplied by a negative is positive):

-8 × -7 = 56

(iii) 0 – (-14)

Ans: 0 – (-14)

Zero minus debt is a fortune (Subtracting a negative becomes addition):

0 + 14 = 14

(iv) (-20) ÷ 4

Ans: (-20) + 4

A debt divided by fortune is debt (A negative divided by a positive is negative):

-20 + 4 = -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).

Ans: Consider you have 10.

A negative number represents debt.

So, -5 means you owe 5.

Now, 10 – (-5) means removing a debt of 5.

If your debt is removed, your money increases by 5.

So, 10 – (-5) = 10 + 5 = 15

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

Exercise Set 3.3

1. Prove that the following rational numbers are equal:

(i) 2/3 and 4/6

Ans: Two rational numbers a/b and c/d are said to be equal if ad = bc.

2/3 and 4/6

a = 2, b = 3 and c = 4, d = 6

Here, 2 × 6 = 3 × 4

⇒ 12 = 12 [a × d = b × c]

So, both are equal.

(ii) 5/4 and 10/8

Ans: First Fraction: 5/4 = 5/4

Second Fraction: 10/8 = 5/4 (dividing numerator and denominator by 2)

Therefore, 5/4 and 10/8 are equal.

(iii) -3/5 and -6/10

Ans: -35  and -610;

a = -3, b = 5 and c = -6, d = 10

⇒ -30 = -30 [a × d = b × c]

So, both are equal.

(iv) 9/3 and 3

Ans: First Fraction: 9/3 = 3

Hence, 9/3 and 3 are equal.

2. Find the sum:

(i) 2/5 + 3/10

Ans: To find the sum of rational numbers, first make the denominators the same (LCM of denominators), then add the numerators.

2/5 + 3/10

LCM of 5 and 10 = 10

2/5 = 4/10

4/10 +3/10=7/10

(ii) 7/12 + 5/8

Ans: LCM of 12 and 8 = 24

Simplifying the first number: 7/12 = 14/24 [Making the same denominator]

Simplifying the second number: 5/8 = 15/24 [Making the same denominator]

Now, the sum

= 7/12 + 5/8

= 14/24 + 15/24

= 29/24

Therefore, the sum of 7/12 + 5/8 is 29/24.

(iii) -4/7 + 3/14

Ans: LCM of 7 and 14 = 14

-47 = -814

-814 + 314 = -514

3. Find the difference:

(i) 5/6 – 1/4

Ans: LCM of 6 and 4 = 12

Simplifying the first number: 5/6 = 10/12 [Making the same denominator]

Simplifying the Second number: 1/4 = 3/12 [Making the same denominator]

So, the difference

= 5/6 – 1/4

= 10/12 – 3/12

= 7/12

Therefore, the difference is 7/12.

(ii) 11/8 – 3/4

Ans: LCM of 8 and 4 = 8

34 = 68

118 – 68 = 58

(iii) -7/9 – (-2/3)

Ans: -7/9 – (-2/3) = -7/9 + 2/3

LCM of 9 and 3 = 9

Simplifying the Second number: 2/3 = 6/9 [Making the same denominator]

So, the difference

= -7/9 – (-2/3)

= -7/9 + 6/9

= -1/9

Therefore, the difference is -1/9.

4. Find the product:

(i) 2/3 × 3/10

Ans: To find the product of rational numbers, multiply the numerators and multiply the denominators, then simplify if possible.

23 × 310

2 × 33 × 10 = 630 = 15

(ii) 7/11 × 5/8

Ans: 7/11 × 5/8 

7 × 5/11 × 8 = 35/88

(iii) -4/7 × 5/14

Ans: -4/7 × 5/14 

-4 × 5/7 × 14 = −20/98 = -10/49

5. Find the quotient:

(i) 2/3 ÷ 3/10

Ans: To divide fractions, multiply by the reciprocal.

2/3 ÷ 3/10

= 2/3 × 10/3

= (2 × 10)/(3 × 3)

= 20/9

Therefore, the quotient is 20/9.

(ii) 7/11 ÷ 5/8 

Ans: 7/11 ÷ 5/8 = 7/11 × 8/5 × 8 = 56/55

(iii) -4/7 ÷ 5/14

Ans: -4/7 ÷ 5/14 = -4/7 × 14/5 = -85

6. 

Ans: To show:

7. Simplify the following using the distributive property:

Ans: Using distributive property:

7/9(6/7 – 3/4)

= 7/9 × 6/7 – 7/9 × 3/4

= 6/9 – 21/36

= 2/3 – 7/12

= 8/12 – 7/12 [LCM of 3 and 12 = 12 and 2/3 = 8/12]

= 1/12

Therefore, the simplified value of 7/9(6/7 – 3/4) is 1/12.

8. Find the rational number x such that:

Ans: Apply the distributive property on the LHS:

5/6(x + 3/5) = 5/6x + 5/6 ×3/5

Simplify the product:

5/6 × 3/5 = 3/6 = 1/2

So, LHS becomes:

5/6(x + 3/5) = 5/6x + 1/2

So, the equation becomes:

5/6x + 1/2 = 5/6x + 1/2

Subtract 5/6 x from both sides: 1/2 = 1/2

This is always true, so the equation holds true for all rational numbers.

So, x can be any rational number.

Exercise Set 3.4

1. Represent the rational numbers 2/3, -5/4 and 1 1/2 on a single number line.

Ans: To represent the rational numbers 2/3, -5/4 and 1 1/2 on a single number line, first identify their positions.

Identify the values

2/3 lies between 0 and 1.

-5/4 = -1 1/4, which lies between -2 and -1.

1 1/2 = 3/2, which lies between 1 and 2.

Use suitable divisions

Mark integers -2, -1,0, 1 and 2 on the number line.

For -5/4: Divide the interval from -2 to -1 into 4 equal parts and mark one part to the left of -1 (-114)

For 2/3: Divide the interval from 0 to 1 into 3 equal parts and mark the second part.

For 1 1/2: Divide the interval from 1 to 2 into 2 equal parts and mark the midpoint.

Thus, the number line:

2. Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Ans: First, take the average of -1/2 and 1/4:

(-1/2 + 1/4)2 = -2/4 + 1/4/2 = -1/4/2 = -1/8

So, -1/8 lies between -1/2 and 1/4.

Now find the average of -1/2 and –1/8:

(-1/2 – 1/8)/2 = -4/8 – 1/8/2 = -5/8/2 = -5/16

Next, find the average of -18 and 14:

(-1/8 + 1/4)2 = -1/8 + 2/8/2 = 1/8/2 = 1/16

Thus, three distinct rational numbers strictly between -1/2 and 1/4 are:

-5/16, -1/8, 1/16

(Answer may vary)

3. Simplify the expression:

Ans: We take LCM of 4 and 12, which is 12.

-1/4 = -3/12

Now,

-3/12 + 5/12 = 2/12 = 1/6

4. A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can 4 he make?

Ans: Convert mixed fractions to improper fractions:

15 3/4 = 63/4, 2 1/4 = 9/4

Now divide:

63/4 ÷ 9/4 = 63/4 × 4/9 = 63/9 = 7

Therefore, he can make 7 kurtas.

5. Find three rational numbers between 3.1415 and 3.1416.

Ans: We need three rational numbers strictly between 3.1415 and 3.1416.

One possible correct set is:

3.14151, 3.14152, 3.14153

All three are terminating decimals, so they are rational numbers, and each lies between 3.1415 and 3.1416.

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

Ans: One simple way is to make the denominators the same. Suppose the two rational numbers are a/b and c/d with a/b < c/d.

Convert them to equivalent fractions with a common denominator, then choose any fraction whose numerator lies between the two numerators. Another way is to multiply both numbers by the same positive number to create more space, then choose a number in between and divide back.

Example: To find a rational number between 1/3 and 1/2, write them with denominator 6:

1/3 = 2/6, 1/2 = 3/6

Now a number between them is:

5/12

since

2/6 < 5/12 < 3/6

We can also express the rational numbers in decimal form and then find the rational numbers between them.

So, besides the average method, we can use common denominators, scaling, by expressing in decimal form or simply pick a fraction with a suitable numerator after rewriting.

Exercise Set 3.5

1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Ans: 7/20

20 = 2² × 5

Only 2 and 5 ⇒ terminating decimal

Long division:

7/20 = 0.35

4/15

15 = 3 × 5

Contains 3 ⇒ repeating decimal

Long division:

415 = 0.2666…

13/250

250 = 2 × 5³

Only 2 and 5 ⇒ terminating decimal

Long division:

13/250 = 0.052

7/20 → terminating (0.35)

4/15 → repeating (0.2) overline (6)

13/250 → terminating (0.052)

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

Ans: Long division for 113:

So, the repeating block is: 076923

Now,

3. Classify the following numbers as rational or irrational:

(i) √81

(ii) √12

(iii) 0.33333…

(iv) 0.123451234512345…

(v) 1.01001000100001… (Notice the pattern: Is it repeating a single block?)

(vi) 23.560185612239874790120

Find the explicit fractions in case they are rational.

Ans: (i) √81

√81 = 9

So it is rational.

Explicit fraction: 9 = 9/1

(ii) √12

√12 = √4  ×3 = 2√3

Since √3 is irrational, 2√3 is also irrational.

So √12 is irrational.

(iii) 0.33333…

This is a repeating decimal.

Let x= 0.33333…

Then 10x = 3.33333…

Subtract:

10x – x = 3.33333… -0.33333…

9x = 3

x = 3/9 = 1/3

So it is rational. Explicit fraction:

0.33333…= ⅓

(iv) 0.123451234512345…

The block 12345 repeats again and again. So, this is a repeating decimal, hence rational.

Let x = 0.1234512345…

Then 100000 x = 12345.1234512345…

Subtract:

100000x – x = 12345

99999 x = 12345

x = 12345/99999

This can be simplified as 3:

12345/99999 = 4115/33333

So, the number is rational.

Explicit fraction:

0.123451234512345… = 4115/33333

(v) 1.01001000100001… → Irrational No fixed repeating block (pattern keeps increasing zeros), so non-repeating ⟹ irrational.

(vi) 23.560185612239874790120

This is a terminating decimal because it ends after a finite number of digits. So, it is rational.

As a fraction:

= 23560185612239874790120/10²¹

This may also be simplified to:

589004640305996869753/25000000000000000000

4. The number (which means 0.99999…) is a rational number. Using algebra (let x = , multiply by 10, and subtract), explain why is exactly equal to 1.

Ans: Let

x =

= 0.99999…

Multiply by 10:

10x = 9.99999…

Now subtract the first equation from the second:

10x – x = 9.99999… -0.99999…

9x = 9

So,

x = 1

Final conclusion:

= 1

5. We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.

Ans: A cyclic number is one where its repeating digits rotate when multiplied by numbers.

We already know:

The block 142857 is cyclic because multiplying it by 2, 3, 4, 5, or 6 gives rotations of the same digits.

Now, to find more such numbers, we look for values of n where:

The decimal expansion of 1/n is repeating.

The repeating block has special rotation properties.

These usually occur when:

n is a prime number.

The length of the repeating cycle is n -1 (called a full pretend prime).

Examples:

1. 1/17 = 0.0588235294117647 17

The repeating block has 16 digits and shows cyclic-like properties.

2. 1/19 = 0.052631578947368421 19

A number n (more precisely, a prime p) has this cyclic property when the decimal expansion of 1/p repeats with the maximum possible length, which is P – 1. Such primes are called full repent primes. These primes generate repeating decimals where the digit block cycles under multiplication, forming cyclic numbers.

END-OF-CHAPTER EXERCISES

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

(i) 3/50

Ans: 3/50 = 0.06 (Terminating decimal)

(ii) 2/9

Ans: 29 = 0.2222…

(Non-terminating repeating decimal)

2. Prove that √5 is an irrational number.

Ans: We will use proof by contradiction.

Assume that √5 is rational.

Then it can be written in the form:

√5 = p/q

where p and q are integers having no common factor (i.e., in lowest form), and q 0.

Squaring both sides:

5 = p²/q²

p² = 5q²

This means p² is divisible by 5, so p must also be divisible by 5.

Let p = 5k for some integer k.

Substitute back:

(5k)² = 5q²

25k² = 5q²

Divide both sides by 5:

5k² = q²

This shows q² is divisible by 5, so q is also divisible by 5. So, both p and q are divisible by 5, which contradicts our assumption that p/q is in lowest form.

Thus, our assumption is wrong.

Hence, √5 is irrational.

3. Convert the following decimal numbers into the form p/q.

(i) 12.6

Ans: Let x = 12.6

x = 12.6 = 126/10

Simplify:

126/10 = 63/5

12.6 = 63/5

(ii) 0.0120

Ans: 0.0120 = 120/10000 = 3/250

Ans: Let

x =

= 3.0525252…

There is 1 non-repeating digit (0) and 2 repeating digits (52).

Multiply by 10:

10x = 30.525252…

Multiply by 1000:

1000x = 3052.525252…

Now subtract:

1000x – 10x = 3052.525252… -30.525252…

990x = 3022

x = 3022/990

Simplify:

x = 1511/495

= 1511/495

Ans: Let x = 1.2353535…

10x = 12.3535…

1000x = 1235.3535…

1000x – 10x = 1223

⇒ 990x = 1223

⇒ x = 1223/990

Ans: Let

x =

= 0.232323…

Since 2 digits repeat, multiply by 100:

100x = 23.232323…

Now subtract:

100x – x = 23.232323… -0.232323…

99x = 23

x = 23/99

= 23/99

Ans: Let x = 20.5555…

100x = 2.0555…

100x – x = 20.5555…. -2.0555….

9x = 18.5

⇒ x = 18.5/9 =185/90=37/18

Ans: Let x = 2.125555…

100.x = 212.5555…

1000x = 2125.5555…

1000x – 100x = 2125.5555… -212.5555…

900x = 1913

⇒ x = 1913900

Ans: Let x = 3.125555…

100x = 312.5555…

1000x = 3125.5555…

900x = 2813

x = 2813/900

Ans: Let x = 2.162516251625…

10000x = 21625.1625…

1000x – x = 21625.1625… -2.1625…

9999x = 21623

x = 21623/9999

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

(i) 0.532

(ii) 1.15

Ans: (i) 0.532 lies between 0.53 and 0.54. Divide the segment from 0.53 to 0.54 into 10 equal parts; the 2nd mark after 0.53 gives 0.532

(ii) Convert the repeating decimal to a fraction first: Let x – 1.15555…

Then 10x = 11.5555…

100x = 115.555…

Subtract:

100x – 10x= 115.5555… – 11.555…

= 10.4

90x = 104

x = 104/90=52/45

So,

1.15555….. =52/45

52/45 = 1 + 7/45

It lies between 1 and 2. Divide the segment from 1 to 2 into 45 equal parts. Count 7 parts after 1.

5. Find 6 rational numbers between 3 and 4.

Ans: We can write 3 and 4 with a common denominator:

3 = 30/10, 4 = 40/10

Now choose any 6 fractions between them:

31/10, 32/10, 33/10, 34/10, 35/10, 36/10

3.1, 3.2, 3.3, 3.4, 3.5, 3.6

6. Find 5 rational numbers between 2/5 and 3/5.

Ans: Make denominators larger to create more:

2/5 = 20/50, 3/5 = 30/50

Now take any 5 rational numbers between them:

21/50, 22/50, 23/50, 24/50, 25/50

So the required rational numbers are:

21/50, 22/50, 23/50, 24/50, 25/50

(Answer may vary)

7. Find 5 rational numbers between 1/6 and 2/5.

Ans: We first make the denominators the same:

1/6 = 5/30, 2/5 = 12/30

Now choose 5 rational numbers between them:

6/30, 7/30, 8/30, 9/30, 10/30

(These can also be simplified if needed.)

8. If x/3 + x/5 = 16/15, find the rational number x.

Ans: x/3 + x/5 = 16/15

Take LCM of 3 and 5 = 15

5x + 3x/15 = 16/15

8x/15 = 16/15

Multiply both sides by 15:

8x = 16

x = 2

9. Let a and b be two non-zero rational numbers such that a + 1/b = 0.

Without assigning numerical values, determine whether ab is positive or negative. Justify your answer.

Ans: Given a + 1/b = 0

Rearrange: a = -1/b

Now multiply both sides by b:

ab = -1

Since -1 is a negative number, so, 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, p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

Ans: Let the number be x.

Since its decimal expansion terminates and the last non-zero digit is in the 4th decimal place, we can write it as

x = p/10⁴

where p is an integer.

Also, since the last digit is non-zero, p is not divisible by 10 (otherwise there would be trailing zeros).

Second part:

Write 10⁴ = 2⁴ × 5⁴.

When the fraction p/10⁴ is reduced to lowest form, some common factors (2 or 5) between p and 10^4/4 may cancel.

So, the denominator in lowest form will be of the form:

2^a × 5^b where a ≤ 4, b ≤ 4

Therefore, it is NOT necessary that the denominator is divisible by 2⁴ or 5⁴.

Example:

0.1250 = 1250/10000 = 1/8

Here denominator = 8 = 2³,

not divisible by 2⁴ or 5⁴.

Yes, it can be written as p10², with p not divisible by 10.

No, the denominator in lowest form need not be divisible by 2⁴ or 5⁴.

11. Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

Ans: A rational number p/q (in lowest form) has a terminating decimal expansion if the prime factorisation of the denominator q contains only powers of 2 and/or 5.

First, check if the fraction is in lowest form:

HCF (18, 125) = 1

So it is already in lowest term.

Now factorise the denominator:

125 = 5³

Since the denominator contains only the prime factor 5, the decimal expansion is terminating.

To find the number of decimal places, we need to convert the denominator into a power of 10.

10³ = 2³ × 5³

Multiply numerator and denominator by 2³ = 8:

18/125 = 18 × 8/125 × 8 = 144/1000

Thus, the decimal has 3 decimal places.

12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.

Ans: Denominator = 2³ × 5¹.

For a rational number in lowest form, the number of decimal places in its terminating decimal is

max (power of 2, power of 5 )

Here:

max(3, 1) = 3

So, we multiply numerator and denominator by 5² to make denominator 10³.

13. Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ – k₁ > 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 k₂ – k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Ans: Given

a = 7/12, b = 5/6

Step 1: Make same denominator m

LCM of 12 and 6 is 12:

a = 7/12, b = 10/12

Here k₂ – k₁ = 10 – 7 = 3 (not > 6), so multiply both by 3:

a = 21/36, b = 30/36

Now:

k₁ = 21, k₂ = 30, m = 36, k₂ – k₁ = 9 > 6

Step 2: Five rational numbers between them

Numbers between 21 and 30:

22, 23, 24, 25, 26

So required numbers:

22/36, 23/36, 24/36, 25/36, 26/36

Step 3: Why k₂ – k₁ > n + 1 ?

Between k₁ and k₂, the number of integers available is:

k₂ – k₁ – 1

To get n numbers:

k₂ – k₁ – 1 ≥ n ⇒ k₂ – k₁ ≥ n + 1

Hence, condition k₂ – k₁ > n + 1 ensures enough integers exist to form n rational numbers.

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

Ans: Given:

x + y + z = 0 ⇒ z = -(x + y)

Substitute this in the second equation

xy + yz + zx = 0:

xy + y (-x – y) + x (-x – y) = 0

Simplify:

xy – xy – y² – x² – xy = 0

-(x² + y² + xy) = 0

x² + y² + xy = 0 …(1)

Now consider:

(x + y)² = x² + y² + 2xy

Using (1), substitute x² + y² = -xy:

(x + y)² = -xy + 2xy = xy

So, (x + y)² = xy …(2)

But from z = -(x + y), we also have:

z² = (x + y)²

So from (2):

z² = xy …(3)

Now similarly, by symmetry:

x² = yz, y² = zx

Add all three:

x² + y² + z² = xy + yz + zx

But given:

xy + yz + zx = 0

So,

x² + y² + z² = 0

Since squares of rational numbers are non-negative, the only way their sum is zero is:

x² = y² = z² = 0

⇒ x = y = z = 0

z = y = z = 0

15. Show that the rational number (a + b)/2 lies between the rational numbers a and b.

Ans: Assume a < b.

Multiply by 2:

2a < a + b < 2b

Divide throughout by 2:

a < a + b/2 < b

Hence, a + b/2 lies between a and b.

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

Ans: Each new triangle is formed by taking the previous hypotenuse as one side and 1 as the other side.

We can find the hypotenuse of each triangle using the Baudhayana- Pythagoras Theorem:

(hypotenuse)² = (base)² + (perpendicular)²

For triangle 1: Sides: 1, 1

Hypotenuse = √1² + 1² = √2

For triangle 2: Sides: √2, 1

Hypotenuse = √(2)² + 1² = √2 + 1 = √3

Similarly:

Triangle 3: Sides: √3 , 1; Hypotenuse: 2

Triangle 4: Sides: 2, 1; Hypotenuse: √5

Triangle 5: Sides: √5 , 1; Hypotenuse: √6

Triangle 6: Sides: √6, 1; Hypotenuse: √7

Triangle 7: Sides: √7, 1; Hypotenuse: √8

Triangle 8: Sides: √8,, 1; Hypotenuse: 3

Triangle 9: Sides: 3, 1; Hypotenuse: √10

Triangle 10: Sides: √10, 1; Hypotenuse: √11

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