Chapter 3: The World of Numbers

Page 43 – Exercise Set 3.1

Question 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:

For every 2 bags of spices, the merchant receives 15 copper ingots.

Number of groups of 2 bags in 12 bags:

$$12÷2=6$$

Copper ingots received:

$$6\times 15=90$$

Therefore, the merchant will leave with 90 copper ingots.

Question 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:

$$11,13,17,19$$

are all prime numbers between 10 and 20.

A prime number has only two factors: 1 and itself.

The next three prime numbers are:

$$23,29,31$$

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

Question 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.

Natural numbers are:

$$1,2,3,4,...$$

Example 1:

$$5-3=2$$

Here, 2 is a natural number.

Example 2:

$$3-5=-2$$

Here, -2 is not a natural number.

Example 3:

$$3-3=0$$

Here, 0 is also not a natural number if natural numbers are taken as starting from 1.

Therefore, natural numbers are not closed under subtraction.

Question 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:

On one hand, the thumb is used to count the joints of the other four fingers.

Each finger has 3 joints.

So, total joints counted are:

$$4\times 3=12$$

Therefore, one can count up to 12 on one hand.

This relates to the ancient base-12 counting system because 12 was a useful counting unit. Even today, we use 12 in many places, such as:

$$12\mathrm{\text{ months in a year}}12\mathrm{\text{ items in a dozen}}$$

Think and Reflect – Page 46

Question. 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 number 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, $(-3)\times (-4)=+12$.

Answer:

A negative number can represent a debt.

For example:

$$-3$$

means a debt of ₹3.

Now, multiplying by a negative number can be understood as removing something.

So,

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

means removing 4 debts of ₹3 each.

If one debt is ₹3, then 4 debts are:

$$4\times 3=12$$

When these debts are removed, the person becomes ₹12 richer.

Therefore,

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

Hence, a negative times a negative gives a positive because removing a debt increases the value.

Exercise Set 3.2 – Page 46

Question 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:

Temperature at noon:

$$4^{\circ }C$$

Drop in temperature:

$${15}^{\circ }C$$

Midnight temperature:

$$4-15=-11$$

Therefore, the midnight temperature is:

$$\boxed{-{11}^{\circ }C}$$

Question 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:

Loan or debt of ₹850 is represented as:

$$-850$$

Profit of ₹1,200 is represented as:

$$+1200$$

Loss of ₹450 is represented as:

$$-450$$

So, the equation is:

$$-850+1200-450$$

Now calculate:

$$-850+1200=350350-450=-100$$

Therefore, the trader’s final financial standing is:

$$\boxed{-100}$$

So, he has a debt of ₹100.

Question 3. Calculate the following using Brahmagupta’s laws:

Question 3(i). $\mathbf{(}\mathbf{-}\mathrm{𝟏𝟐}\mathbf{)}\mathbf{\times }\mathrm{𝟓}$

Answer:

A debt multiplied by a fortune gives a debt.

$$(-12)\times 5=-60$$

Therefore,

$$\boxed{-60}$$

Question 3(ii). $\left(\mathbf{-}\mathrm{𝟖}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{-}\mathrm{𝟕}\right)$

Answer:

A debt multiplied by a debt gives a fortune.

$$(-8)\times (-7)=56$$

Therefore,

$$\boxed{56}$$

Question 3(iii). $\mathrm{𝟎}\mathbf{-}\mathbf{(}\mathbf{-}\mathrm{𝟏𝟒}\mathbf{)}$

Answer:

Subtracting a negative number is the same as adding a positive number.

$$0-(-14)=0+14=14$$

Therefore,

$$\boxed{14}$$

Question 3(iv). $\mathbf{(}\mathbf{-}\mathrm{𝟐𝟎}\mathbf{)}\mathbf{÷}\mathrm{𝟒}$

Answer:

A negative number divided by a positive number gives a negative number.

$$(-20)÷4=-5$$

Therefore,

$$\boxed{-5}$$

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

Answer:

A negative number can represent a debt.

Suppose a person has ₹10.

A debt of ₹5 can be represented as:

$$-5$$

Now, if this debt of ₹5 is removed, it means the person is no longer required to pay ₹5. So, the person becomes ₹5 better off.

Mathematically:

$$10-(-5)$$

means ₹10 and removal of a debt of ₹5.

Removing a debt increases the amount by ₹5.

So,

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

Therefore,

$$\boxed{10-(-5)=15}$$

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

Page 47 – Think and Reflect

Question. Can you explain why we need $𝐪\mathbf{\ne }\mathrm{𝟎}$in the definition of a rational number?

Answer:

A rational number is written in the form:

$$\frac{p}{q}$$

where $p$and $q$are integers.

We need:

$$q\ne 0$$

because division by zero is not defined.

For example:

$$\frac{5}{0}$$

has no mathematical meaning because no number multiplied by 0 can give 5.

Therefore, in a rational number, the denominator cannot be zero.

Page 49 – Think and Reflect

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

Answer:

To make denominators equal, we find the LCM of the denominators.

Example:

$$\frac{1}{2}+\frac{1}{3}$$

The LCM of 2 and 3 is 6.

So,

$$\frac{1}{2}=\frac{3}{6}\frac{1}{3}=\frac{2}{6}$$

Now the denominators are equal.

$$\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$$

Question 2. Verify the distributive law for rational numbers.

Answer:

The distributive law is:

$$p(q+r)=pq+pr$$

Let:

$$p=\frac{1}{2},q=\frac{2}{3},r=\frac{1}{3}$$

Left-hand side:

$$p(q+r)=\frac{1}{2}(\frac{2}{3}+\frac{1}{3})=\frac{1}{2}\left(1\right)=\frac{1}{2}$$

Right-hand side:

$$pq+pr=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$$

Since LHS = RHS, the distributive law is verified.

Page 49–50 – Exercise Set 3.3

Question 1. Prove that the following rational numbers are equal:

Question 1(i). $\frac{\mathrm{𝟐}}{\mathrm{𝟑}}$and $\frac{\mathrm{𝟒}}{\mathrm{𝟔}}$

Answer:

Two rational numbers $\frac{a}{b}$and $\frac{c}{d}$are equal if:

$$ad=bc$$

Here,

$$\frac{2}{3}\mathrm{\text{ and }}\frac{4}{6}2\times 6=123\times 4=12$$

Since both products are equal,

$$\frac{2}{3}=\frac{4}{6}$$

Question 1(ii). $\frac{\mathrm{𝟓}}{\mathrm{𝟒}}$and $\frac{\mathrm{𝟏𝟎}}{\mathrm{𝟖}}$

Answer:

$$5\times 8=404\times 10=40$$

Since both products are equal,

$$\frac{5}{4}=\frac{10}{8}$$

Question 1(iii). $\mathbf{-}\frac{\mathrm{𝟑}}{\mathrm{𝟓}}$and $\mathbf{-}\frac{\mathrm{𝟔}}{\mathrm{𝟏𝟎}}$

Answer:

$$(-3)\times 10=-305\times (-6)=-30$$

Since both products are equal,

$$-\frac{3}{5}=-\frac{6}{10}$$

Question 1(iv). $\frac{\mathrm{𝟗}}{\mathrm{𝟑}}$and $\mathrm{𝟑}$

Answer:

Write 3 as:

$$3=\frac{3}{1}$$

Now compare:

$$\frac{9}{3}\mathrm{\text{ and }}\frac{3}{1}9\times 1=93\times 3=9$$

Since both products are equal,

$$\frac{9}{3}=3$$

Question 2. Find the sum:

Question 2(i). $\frac{\mathrm{𝟐}}{\mathrm{𝟓}}\mathbf{+}\frac{\mathrm{𝟑}}{\mathrm{𝟏𝟎}}$

Answer:

$$\frac{2}{5}+\frac{3}{10}$$

LCM of 5 and 10 is 10.

$$\frac{2}{5}=\frac{4}{10}\frac{4}{10}+\frac{3}{10}=\frac{7}{10}$$

Therefore,

$$\frac{2}{5}+\frac{3}{10}=\frac{7}{10}$$

Question 2(ii). $\frac{\mathrm{𝟕}}{\mathrm{𝟏𝟐}}\mathbf{+}\frac{\mathrm{𝟓}}{\mathrm{𝟖}}$

Answer:

LCM of 12 and 8 is 24.

$$\frac{7}{12}=\frac{14}{24}\frac{5}{8}=\frac{15}{24}\frac{14}{24}+\frac{15}{24}=\frac{29}{24}$$

Therefore,

$$\frac{7}{12}+\frac{5}{8}=\frac{29}{24}$$

Question 2(iii). $\mathbf{-}\frac{\mathrm{𝟒}}{\mathrm{𝟕}}\mathbf{+}\frac{\mathrm{𝟑}}{\mathrm{𝟏𝟒}}$

Answer:

LCM of 7 and 14 is 14.

$$-\frac{4}{7}=-\frac{8}{14}-\frac{8}{14}+\frac{3}{14}=-\frac{5}{14}$$

Therefore,

$$-\frac{4}{7}+\frac{3}{14}=-\frac{5}{14}$$

Question 3. Find the difference:

Question 3(i). $\frac{\mathrm{𝟓}}{\mathrm{𝟔}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟒}}$

Answer:

LCM of 6 and 4 is 12.

$$\frac{5}{6}=\frac{10}{12}\frac{1}{4}=\frac{3}{12}\frac{10}{12}-\frac{3}{12}=\frac{7}{12}$$

Therefore,

$$\frac{5}{6}-\frac{1}{4}=\frac{7}{12}$$

Question 3(ii). $\frac{\mathrm{𝟏𝟏}}{\mathrm{𝟖}}\mathbf{-}\frac{\mathrm{𝟑}}{\mathrm{𝟒}}$

Answer:

$$\frac{3}{4}=\frac{6}{8}\frac{11}{8}-\frac{6}{8}=\frac{5}{8}$$

Therefore,

$$\frac{11}{8}-\frac{3}{4}=\frac{5}{8}$$

Question 3(iii). $\mathbf{-}\frac{\mathrm{𝟕}}{\mathrm{𝟗}}\mathbf{-}(-\frac{\mathrm{𝟐}}{\mathrm{𝟑}})$

Answer:

$$-\frac{7}{9}-(-\frac{2}{3})=-\frac{7}{9}+\frac{2}{3}=-\frac{7}{9}+\frac{6}{9}=-\frac{1}{9}$$

Therefore,

$$-\frac{7}{9}-(-\frac{2}{3})=-\frac{1}{9}$$

Question 4. Find the product:

Question 4(i). $\frac{\mathrm{𝟐}}{\mathrm{𝟑}}\mathbf{\times }\frac{\mathrm{𝟑}}{\mathrm{𝟏𝟎}}$

Answer:

$$\frac{2}{3}\times \frac{3}{10}=\frac{2\times 3}{3\times 10}=\frac{6}{30}=\frac{1}{5}$$

Question 4(ii). $\frac{\mathrm{𝟕}}{\mathrm{𝟏𝟏}}\mathbf{\times }\frac{\mathrm{𝟓}}{\mathrm{𝟖}}$

Answer:

$$\frac{7}{11}\times \frac{5}{8}=\frac{35}{88}$$

Question 4(iii). $\mathbf{-}\frac{\mathrm{𝟒}}{\mathrm{𝟕}}\mathbf{\times }\frac{\mathrm{𝟓}}{\mathrm{𝟏𝟒}}$

Answer:

$$-\frac{4}{7}\times \frac{5}{14}=-\frac{20}{98}=-\frac{10}{49}$$

Question 5. Find the quotient:

Question 5(i). $\frac{\mathrm{𝟐}}{\mathrm{𝟑}}\mathbf{÷}\frac{\mathrm{𝟑}}{\mathrm{𝟏𝟎}}$

Answer:

$$\frac{2}{3}÷\frac{3}{10}=\frac{2}{3}\times \frac{10}{3}=\frac{20}{9}$$

Question 5(ii). $\frac{\mathrm{𝟕}}{\mathrm{𝟏𝟏}}\mathbf{÷}\frac{\mathrm{𝟓}}{\mathrm{𝟖}}$

Answer:

$$\frac{7}{11}÷\frac{5}{8}=\frac{7}{11}\times \frac{8}{5}=\frac{56}{55}$$

Question 5(iii). $\mathbf{-}\frac{\mathrm{𝟒}}{\mathrm{𝟕}}\mathbf{÷}\frac{\mathrm{𝟓}}{\mathrm{𝟏𝟒}}$

Answer:

$$-\frac{4}{7}÷\frac{5}{14}=-\frac{4}{7}\times \frac{14}{5}=-\frac{8}{5}$$

Question 6. Show that: $(\frac{\mathrm{𝟏}}{\mathrm{𝟐}}+\frac{\mathrm{𝟑}}{\mathrm{𝟒}})\mathbf{\times }\frac{\mathrm{𝟖}}{\mathrm{𝟑}}\mathbf{=}\frac{\mathrm{𝟏}}{\mathrm{𝟐}}\mathbf{\times }\frac{\mathrm{𝟖}}{\mathrm{𝟑}}\mathbf{+}\frac{\mathrm{𝟑}}{\mathrm{𝟒}}\mathbf{\times }\frac{\mathrm{𝟖}}{\mathrm{𝟑}}$.

Answer:

Left-hand side:

$$(\frac{1}{2}+\frac{3}{4})\times \frac{8}{3}=(\frac{2}{4}+\frac{3}{4})\times \frac{8}{3}=\frac{5}{4}\times \frac{8}{3}=\frac{40}{12}=\frac{10}{3}$$

Right-hand side:

$$\frac{1}{2}\times \frac{8}{3}+\frac{3}{4}\times \frac{8}{3}=\frac{8}{6}+\frac{24}{12}=\frac{4}{3}+2=\frac{4}{3}+\frac{6}{3}=\frac{10}{3}$$

Since LHS = RHS, the statement is proved.

Question 7. Simplify the following using the distributive property: $\frac{\mathrm{𝟕}}{\mathrm{𝟗}}(\frac{\mathrm{𝟔}}{\mathrm{𝟕}}-\frac{\mathrm{𝟑}}{\mathrm{𝟒}})$.

Answer:

$$\frac{7}{9}(\frac{6}{7}-\frac{3}{4})$$

Using distributive property:

$$=\frac{7}{9}\times \frac{6}{7}-\frac{7}{9}\times \frac{3}{4}=\frac{6}{9}-\frac{21}{36}=\frac{2}{3}-\frac{7}{12}=\frac{8}{12}-\frac{7}{12}=\frac{1}{12}$$

Therefore, the answer is:

$$\frac{1}{12}$$

Question 8. Find the rational number $𝐱$such that: $\frac{\mathrm{𝟓}}{\mathrm{𝟔}}(𝐱+\frac{\mathrm{𝟑}}{\mathrm{𝟓}})\mathbf{=}\frac{\mathrm{𝟓}}{\mathrm{𝟔}}𝐱\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟐}}$.

Answer:

$$\frac{5}{6}(x+\frac{3}{5})=\frac{5}{6}x+\frac{1}{2}$$

Using distributive property:

$$\frac{5}{6}x+\frac{5}{6}\times \frac{3}{5}=\frac{5}{6}x+\frac{1}{2}\frac{5}{6}x+\frac{3}{6}=\frac{5}{6}x+\frac{1}{2}\frac{5}{6}x+\frac{1}{2}=\frac{5}{6}x+\frac{1}{2}$$

This is true for every rational number $x$.

Therefore, any rational number can be the value of $x$.

Page 51 – Think and Reflect

Question. Try and represent $\frac{\mathrm{𝟖}}{\mathrm{𝟓}}$and $\mathbf{-}\frac{\mathrm{𝟕}}{\mathrm{𝟒}}$on a number line.

Answer:

$$\frac{8}{5}=1\frac{3}{5}$$

So, $\frac{8}{5}$lies between 1 and 2. Divide the interval between 1 and 2 into 5 equal parts and mark the third part after 1.

$$-\frac{7}{4}=-1\frac{3}{4}$$

So, $-\frac{7}{4}$lies between -2 and -1. Divide the interval between -2 and -1 into 4 equal parts and mark the first part to the right of -2, or the third part to the left of -1.

Page 52 – Exercise Set 3.4

Question 1. Represent the rational numbers $\frac{\mathrm{𝟐}}{\mathrm{𝟑}}$, $\mathbf{-}\frac{\mathrm{𝟓}}{\mathrm{𝟒}}$and $\frac{\mathrm{𝟏𝟏}}{\mathrm{𝟐}}$on a single number line.

Answer:

$$\frac{2}{3}$$

lies between 0 and 1. Divide the interval between 0 and 1 into 3 equal parts and mark the second part.

$$-\frac{5}{4}=-1\frac{1}{4}$$

lies between -2 and -1.

$$\frac{11}{2}=5\frac{1}{2}$$

lies between 5 and 6.

Therefore, on the same number line:

• $-\frac{5}{4}$is between -2 and -1

• $\frac{2}{3}$is between 0 and 1

• $\frac{11}{2}$is between 5 and 6

Question 2. Find three distinct rational numbers that lie strictly between $\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟐}}$and $\frac{\mathrm{𝟏}}{\mathrm{𝟒}}$.

Answer:

Convert the numbers to denominator 8:

$$-\frac{1}{2}=-\frac{4}{8}\frac{1}{4}=\frac{2}{8}$$

Three rational numbers between them are:

$$-\frac{3}{8},-\frac{1}{4},0$$

Therefore, three distinct rational numbers are:

$$-\frac{3}{8},-\frac{1}{4},0$$

Question 3. Simplify the expression: $\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟒}}\mathbf{+}\frac{\mathrm{𝟓}}{\mathrm{𝟏𝟐}}$.

Answer:

$$-\frac{1}{4}+\frac{5}{12}$$

LCM of 4 and 12 is 12.

$$-\frac{1}{4}=-\frac{3}{12}-\frac{3}{12}+\frac{5}{12}=\frac{2}{12}=\frac{1}{6}$$

Therefore, the answer is:

$$\frac{1}{6}$$

Question 4. A tailor has $\mathrm{𝟏𝟓}\frac{\mathrm{𝟑}}{\mathrm{𝟒}}$metres of fine silk. If making one kurta requires $\mathrm{𝟐}\frac{\mathrm{𝟏}}{\mathrm{𝟒}}$metres of silk, exactly how many kurtas can he make?

Answer:

Total silk:

$$15\frac{3}{4}=\frac{63}{4}$$

Silk needed for one kurta:

$$2\frac{1}{4}=\frac{9}{4}$$

Number of kurtas:

$$\frac{63}{4}÷\frac{9}{4}=\frac{63}{4}\times \frac{4}{9}=\frac{63}{9}=7$$

Therefore, the tailor can make 7 kurtas.

Question 5. Find three rational numbers between 3.1415 and 3.1416.

Answer:

Three rational numbers between 3.1415 and 3.1416 are:

$$3.14151,3.14155,3.14159$$

All these are terminating decimals, so they are rational numbers.

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

Answer:

Yes. There are many ways.

Method 1: Average Method

For two rational numbers $a$and $b$, one rational number between them is:

$$\frac{a+b}{2}$$

Method 2: Common Denominator Method

Convert both rational numbers to the same denominator, then choose a numerator between them.

Example:

Find a rational number between:

$$\frac{1}{3}\mathrm{\text{ and }}\frac{2}{3}$$

Convert to denominator 6:

$$\frac{1}{3}=\frac{2}{6}\frac{2}{3}=\frac{4}{6}$$

A rational number between them is:

$$\frac{3}{6}=\frac{1}{2}$$

Page 53 – Think and Reflect

Question. Can $\sqrt{\mathrm{𝟐}}$be written as a rational number $\frac{𝐩}{𝐪}$?

Answer:

No, $\sqrt{2}$cannot be written as a rational number.

It cannot be expressed in the form:

$$\frac{p}{q}$$

where $p$and $q$are integers and $q\ne 0$.

Therefore, $\sqrt{2}$is an irrational number.

Page 55 – Think and Reflect

Question. Try to prove the irrationality of $\sqrt{\mathrm{𝟑}}$using the approach of proof by contradiction. Will the same approach work for $\sqrt{\mathrm{𝟓}}$, $\sqrt{\mathrm{𝟕}}$, or $\sqrt{\mathrm{𝟏𝟎}}$?

Answer:

Assume that $\sqrt{3}$is rational.

So,

$$\sqrt{3}=\frac{p}{q}$$

where $p$and $q$are co-prime integers and $q\ne 0$.

Squaring both sides:

$$3=\frac{p^2}{q^2}{p}^2=3q^2$$

So, $p^2$is divisible by 3.

Therefore, $p$is divisible by 3.

Let:

$$p=3k$$

Substitute:

$$(3k{)}^2=3q^{2}9k^2=3q^2{q}^2=3k^2$$

So, $q^2$is divisible by 3.

Therefore, $q$is divisible by 3.

Now both $p$and $q$are divisible by 3, which contradicts the assumption that they are co-prime.

Therefore, $\sqrt{3}$is irrational.

The same method works for $\sqrt{5}$and $\sqrt{7}$, because 5 and 7 are prime numbers.

For $\sqrt{10}$, we can also prove it irrational because 10 is not a perfect square.

Page 55 – Think and Reflect

Question. We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?

Answer:

We can obtain line segments of irrational lengths using the Pythagoras theorem.

For example, to construct a line of length $\sqrt{2}$:

Draw a right triangle with perpendicular sides of length 1 unit and 1 unit.

Then the hypotenuse is:

$$\sqrt{1^2+1^2}=\sqrt{2}$$

So, the hypotenuse gives a line segment of irrational length $\sqrt{2}$.

Similarly, lengths like $\sqrt{3}$, $\sqrt{5}$, etc., can be constructed using right triangles.

Page 56 – Think and Reflect

Question. Try to extend this method for constructing line segments of lengths $\sqrt{\mathrm{𝟑}}$and $\sqrt{\mathrm{𝟓}}$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:

To construct $\sqrt{3}$:

First construct $\sqrt{2}$as the hypotenuse of a right triangle with sides 1 and 1.

Then use $\sqrt{2}$as one side and 1 as the other perpendicular side.

The new hypotenuse is:

$$\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{2+1}=\sqrt{3}$$

To construct $\sqrt{5}$:

First construct a line segment of length 2 units.

Draw a perpendicular of length 1 unit.

The hypotenuse will be:

$$\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$$

General method:

To construct $\sqrt{n}$, make a right triangle in such a way that:

$${\mathrm{\text{Hypotenuse}}}^2=n$$

Using the square root spiral method, each new right triangle has one side 1 unit and the previous hypotenuse as the other side. This gives:

$$\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},...$$

Page 57 – Think and Reflect

Question. Try to find the decimal expansions of $\frac{\mathrm{𝟏𝟎}}{\mathrm{𝟑}}$and $\frac{\mathrm{𝟏𝟏}}{\mathrm{𝟏𝟐}}$. What do you observe about the repetition of the digits after the decimal point?

Answer:

First:

$$\frac{10}{3}=3.3333...=3.\bar{3}$$

Here, the digit 3 repeats forever.

Now:

$$\frac{11}{12}=0.916666...=0.91\bar{6}$$

Here, after 91, the digit 6 repeats forever.

Observation:

Both are non-terminating repeating decimals. In $\frac{10}{3}$, repetition starts immediately after the decimal point. In $\frac{11}{12}$, repetition starts after some non-repeating digits.

Page 58 – Think and Reflect

Question. The decimal expansion of $\frac{𝐩}{𝐪}$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:

A decimal is based on powers of 10.

$$10=2\times 5$$

So,

$${10}^n=2^n\times 5^n$$

If the denominator $q$has only the prime factors 2 and/or 5, then it can be converted into a power of 10.

Example:

$$\frac{3}{20}20=2^2\times 5$$

We can make the denominator 100:

$$\frac{3}{20}=\frac{15}{100}=0.15$$

Therefore, the decimal terminates.

If $q$has any prime factor other than 2 or 5, the decimal will not terminate and will repeat.

Page 61–62 – Exercise Set 3.5

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

Answer:

$$\frac{\mathrm{𝟕}}{\mathrm{𝟐𝟎}}$$

$$20=2^2\times 5$$

The denominator has only 2 and 5 as prime factors.

So, it is terminating.

$$\frac{7}{20}=0.35$$

$$\frac{\mathrm{𝟒}}{\mathrm{𝟏𝟓}}$$

$$15=3\times 5$$

The denominator has 3 also.

So, it is non-terminating repeating.

$$\frac{4}{15}=0.2666...=0.2\bar{6}$$

$$\frac{\mathrm{𝟏𝟑}}{\mathrm{𝟐𝟓𝟎}}$$

$$250=2\times 5^3$$

The denominator has only 2 and 5.

So, it is terminating.

$$\frac{13}{250}=0.052$$

Final answers:

$$\frac{7}{20}=0.35\frac{4}{15}=0.2\bar{6}\frac{13}{250}=0.052$$

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

Answer:

$$\frac{1}{13}=0.076923076923...$$

Repeating block:

$$076923$$

Now:

$$\frac{2}{13}=0.153846153846...\frac{3}{13}=0.230769230769...\frac{4}{13}=0.307692307692...\frac{5}{13}=0.384615384615...\frac{6}{13}=0.461538461538...\frac{7}{13}=0.538461538461...\frac{8}{13}=0.615384615384...\frac{9}{13}=0.692307692307...\frac{10}{13}=0.769230769230...\frac{11}{13}=0.846153846153...\frac{12}{13}=0.923076923076...$$

Observation:

The decimal expansions repeat in blocks of 6 digits. The repeating blocks show cyclic-like patterns. Some blocks are cyclic shifts of each other.

Question 3. Classify the following numbers as rational or irrational:

Question 3(i). $\sqrt{\mathrm{𝟖𝟏}}$

Answer:

$$\sqrt{81}=9$$

Since 9 can be written as:

$$\frac{9}{1}$$

$\sqrt{81}$is rational.

Question 3(ii). $\sqrt{\mathrm{𝟏𝟐}}$

Answer:

$$\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$$

Since $\sqrt{3}$is irrational, $2\sqrt{3}$is also irrational.

Therefore, $\sqrt{12}$is irrational.

Question 3(iii). $\mathbf{0.33333}\mathbf{.}\mathbf{.}\mathbf{.}$

Answer:

$$0.33333...=0.\bar{3}=\frac{1}{3}$$

Therefore, it is rational.

Question 3(iv). $\mathbf{0.123451234512345}\mathbf{.}\mathbf{.}\mathbf{.}$

Answer:

The block 12345 repeats.

$$0.123451234512345...=0.\overline{12345}$$

So, it is rational.

Let:

$$x=0.\overline{12345}100000x=12345.\overline{12345}$$

Subtract:

$$100000x-x=1234599999x=12345x=\frac{12345}{99999}x=\frac{4115}{33333}$$

Therefore, it is rational.

Question 3(v). $\mathbf{1.01001000100001}\mathbf{.}\mathbf{.}\mathbf{.}$

Answer:

The number is:

$$1.01001000100001...$$

Here, the number of zeros keeps increasing. There is no fixed repeating block.

Therefore, it is irrational.

Question 3(vi). $\mathbf{23.560185612239874790120}$

Answer:

This is a terminating decimal.

Every terminating decimal is rational.

It can be written as:

$$23.560185612239874790120=\frac{23560185612239874790120}{1000000000000000000000}$$

So, it is rational.

Question 4. The number $\mathrm{𝟎}\mathbf{.}\overline{\mathrm{𝟗}}$is a rational number. Using algebra $𝐱\mathbf{=}\mathrm{𝟎}\mathbf{.}\overline{\mathrm{𝟗}}$, multiply by 10, and subtract, explain why $\mathrm{𝟎}\mathbf{.}\overline{\mathrm{𝟗}}$is exactly equal to 1.

Answer:

Let:

$$x=0.\bar{9}$$

Then:

$$10x=9.\bar{9}$$

Subtract the first equation from the second:

$$10x-x=9.\bar{9}-0.\bar{9}9x=9x=1$$

Therefore:

$$0.\bar{9}=1$$

Question 5. We have seen that the repeating block of $\frac{\mathrm{𝟏}}{\mathrm{𝟕}}$is a cyclic number. Try to find more numbers $𝐧$whose reciprocals $\frac{\mathrm{𝟏}}{𝐧}$produce decimals with repeating blocks that are cyclic.

Answer:

Some examples are:

$$\frac{1}{17}=0.\overline{0588235294117647}\frac{1}{19}=0.\overline{052631578947368421}$$

These repeating blocks show cyclic properties when multiplied by numbers less than the denominator.

Therefore, examples of such values of $n$are:

$$7,17,19$$

Page 64 – Think and Reflect

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

Answer:

There is no real number whose square is negative.

For example:

$$1^2=1(-1{)}^2=1$$

Both positive and negative real numbers give positive squares.

So, no real number satisfies:

$$x^2=-1$$

To solve this, mathematicians introduced the imaginary number:

$$i=\sqrt{-1}$$

Thus, $\sqrt{-1}$is not a real number. It is an imaginary number.

Page 64–66 – End-of-Chapter Exercises

Question 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:

Question 1(i). $\frac{\mathrm{𝟑}}{\mathrm{𝟓𝟎}}$

Answer:

$$\frac{3}{50}$$

Multiply numerator and denominator by 2:

$$\frac{3}{50}=\frac{6}{100}=0.06$$

Therefore,

$$\frac{3}{50}=0.06$$

It is a terminating decimal.

Question 1(ii). $\frac{\mathrm{𝟐}}{\mathrm{𝟗}}$

Answer:

$$\frac{2}{9}=0.2222...=0.\bar{2}$$

It is a non-terminating repeating decimal.

Question 2. Prove that $\sqrt{\mathrm{𝟓}}$is an irrational number.

Answer:

Assume that $\sqrt{5}$is rational.

So,

$$\sqrt{5}=\frac{p}{q}$$

where $p$and $q$are co-prime integers and $q\ne 0$.

Squaring both sides:

$$5=\frac{p^2}{q^2}{p}^2=5q^2$$

So, $p^2$is divisible by 5.

Therefore, $p$is divisible by 5.

Let:

$$p=5k$$

Substitute in the equation:

$$(5k{)}^2=5q^{2}25k^2=5q^2{q}^2=5k^2$$

So, $q^2$is divisible by 5.

Therefore, $q$is divisible by 5.

Thus, both $p$and $q$are divisible by 5. This contradicts the assumption that $p$and $q$are co-prime.

Therefore, $\sqrt{5}$is irrational.

Question 3. Convert the following decimal numbers in the form of $\frac{𝐩}{𝐪}$.

Question 3(i). $\mathbf{12.6}$

Answer:

$$12.6=\frac{126}{10}=\frac{63}{5}$$

Question 3(ii). $\mathbf{0.0120}$

Answer:

$$0.0120=\frac{120}{10000}=\frac{12}{1000}=\frac{3}{250}$$

Question 3(iii). $\mathrm{𝟑}\mathbf{.}\overline{\mathrm{𝟎𝟓𝟐}}$

Answer:

Let:

$$x=3.\overline{052}$$

Since 3 digits repeat, multiply by 1000:

$$1000x=3052.\overline{052}$$

Subtract:

$$1000x-x=3052.\overline{052}-3.\overline{052}999x=3049x=\frac{3049}{999}$$

Therefore,

$$3.\overline{052}=\frac{3049}{999}$$

Question 3(iv). $\mathrm{𝟏}\mathbf{.}\overline{\mathrm{𝟐𝟑𝟓}}$

Answer:

Let:

$$x=1.\overline{235}$$

Since 3 digits repeat, multiply by 1000:

$$1000x=1235.\overline{235}$$

Subtract:

$$1000x-x=1235.\overline{235}-1.\overline{235}999x=1234x=\frac{1234}{999}$$

Therefore,

$$1.\overline{235}=\frac{1234}{999}$$

Question 3(v). $\mathrm{𝟎}\mathbf{.}\overline{\mathrm{𝟐𝟑}}$

Answer:

Let:

$$x=0.\overline{23}$$

Since 2 digits repeat, multiply by 100:

$$100x=23.\overline{23}$$

Subtract:

$$100x-x=23.\overline{23}-0.\overline{23}99x=23x=\frac{23}{99}$$

Therefore,

$$0.\overline{23}=\frac{23}{99}$$

Question 3(vi). $\mathbf{2.0}\overline{\mathrm{𝟓}}$

Answer:

Let:

$$x=2.0\bar{5}$$

Here, 0 is non-repeating and 5 is repeating.

Multiply by 10:

$$10x=20.\bar{5}$$

Multiply again by 10:

$$100x=205.\bar{5}$$

Subtract:

$$100x-10x=205.\bar{5}-20.\bar{5}90x=185x=\frac{185}{90}x=\frac{37}{18}$$

Therefore,

$$2.0\bar{5}=\frac{37}{18}$$

Question 3(vii). $\mathbf{2.12}\overline{\mathrm{𝟓}}$

Answer:

Let:

$$x=2.12\bar{5}$$

Here, 12 is non-repeating and 5 is repeating.

Multiply by 100:

$$100x=212.\bar{5}$$

Multiply again by 10:

$$1000x=2125.\bar{5}$$

Subtract:

$$1000x-100x=2125.\bar{5}-212.\bar{5}900x=1913x=\frac{1913}{900}$$

Therefore,

$$2.12\bar{5}=\frac{1913}{900}$$

Question 3(viii). $\mathbf{3.12}\overline{\mathrm{𝟓}}$

Answer:

Let:

$$x=3.12\bar{5}$$

Here, 12 is non-repeating and 5 is repeating.

Multiply by 100:

$$100x=312.\bar{5}$$

Multiply again by 10:

$$1000x=3125.\bar{5}$$

Subtract:

$$1000x-100x=3125.\bar{5}-312.\bar{5}900x=2813x=\frac{2813}{900}$$

Therefore,

$$3.12\bar{5}=\frac{2813}{900}$$

Question 3(ix). $\mathbf{2.16}\overline{\mathrm{𝟐𝟓}}$

Answer:

Let:

$$x=2.16\overline{25}$$

Here, 16 is non-repeating and 25 is repeating.

Multiply by 100:

$$100x=216.\overline{25}$$

Since two digits repeat, multiply again by 100:

$$10000x=21625.\overline{25}$$

Subtract:

$$10000x-100x=21625.\overline{25}-216.\overline{25}9900x=21409x=\frac{21409}{9900}$$

Therefore,

$$2.16\overline{25}=\frac{21409}{9900}$$

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

Question 4(i). $\mathbf{0.532}$

Answer:

$$0.532=\frac{532}{1000}=\frac{133}{250}$$

So, $0.532$lies between 0 and 1.

To locate it, divide the interval between 0 and 1 into 250 equal parts and mark the 133rd part from 0.

Question 4(ii). $\mathrm{𝟏}\mathbf{.}\overline{\mathrm{𝟓}}$

Answer:

$$1.\bar{5}=1.5555...$$

Let:

$$x=1.\bar{5}10x=15.\bar{5}$$

Subtract:

$$10x-x=15.\bar{5}-1.\bar{5}9x=14x=\frac{14}{9}\frac{14}{9}=1\frac{5}{9}$$

So, $1.\bar{5}$lies between 1 and 2.

To locate it, divide the interval between 1 and 2 into 9 equal parts and mark the 5th part after 1.

Question 5. Find 6 rational numbers between 3 and 4.

Answer:

Write 3 and 4 with denominator 10:

$$3=\frac{30}{10}4=\frac{40}{10}$$

Six rational numbers between them are:

$$\frac{31}{10},\frac{32}{10},\frac{33}{10},\frac{34}{10},\frac{35}{10},\frac{36}{10}$$

That is:

$$3.1,3.2,3.3,3.4,3.5,3.6$$

Question 6. Find 5 rational numbers between $\frac{\mathrm{𝟐}}{\mathrm{𝟓}}$and $\frac{\mathrm{𝟑}}{\mathrm{𝟓}}$.

Answer:

Convert both numbers to denominator 60.

$$\frac{2}{5}=\frac{24}{60}\frac{3}{5}=\frac{36}{60}$$

Five rational numbers between them are:

$$\frac{25}{60},\frac{26}{60},\frac{27}{60},\frac{28}{60},\frac{29}{60}$$

Simplifying:

$$\frac{5}{12},\frac{13}{30},\frac{9}{20},\frac{7}{15},\frac{29}{60}$$

Question 7. Find 5 rational numbers between $\frac{\mathrm{𝟏}}{\mathrm{𝟔}}$and $\frac{\mathrm{𝟐}}{\mathrm{𝟓}}$.

Answer:

LCM of 6 and 5 is 30.

$$\frac{1}{6}=\frac{5}{30}\frac{2}{5}=\frac{12}{30}$$

Five rational numbers between them are:

$$\frac{6}{30},\frac{7}{30},\frac{8}{30},\frac{9}{30},\frac{10}{30}$$

Simplifying:

$$\frac{1}{5},\frac{7}{30},\frac{4}{15},\frac{3}{10},\frac{1}{3}$$

Question 8. If $\frac{𝐱}{\mathrm{𝟑}}\mathbf{+}\frac{𝐱}{\mathrm{𝟓}}\mathbf{=}\frac{\mathrm{𝟏𝟔}}{\mathrm{𝟏𝟓}}$, find the rational number $𝐱$.

Answer:

$$\frac{x}{3}+\frac{x}{5}=\frac{16}{15}$$

LCM of 3 and 5 is 15.

$$\frac{5x}{15}+\frac{3x}{15}=\frac{16}{15}\frac{8x}{15}=\frac{16}{15}$$

Multiply both sides by 15:

$$8x=16x=2$$

Therefore,

$$x=2$$

Question 9. Let $𝐚$and $𝐛$be two non-zero rational numbers such that $\frac{𝐚}{𝐛}\mathbf{+}\mathrm{𝟏}\mathbf{=}\mathrm{𝟎}$. Without assigning any numerical values, determine whether $𝐚𝐛$is positive or negative. Justify your answer.

Answer:

Given:

$$\frac{a}{b}+1=0\frac{a}{b}=-1$$

So,

$$a=-b$$

Now:

$$ab=(-b)bab=-b^2$$

Since $b$is non-zero, $b^2$is positive.

Therefore:

$$-b^2$$

is negative.

Hence, $ab$is negative.

Question 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 $\frac{𝐩}{{\mathrm{𝟏𝟎}}^{\mathrm{𝟒}}}$, 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 ${\mathrm{𝟐}}^{\mathrm{𝟒}}$or ${\mathrm{𝟓}}^{\mathrm{𝟒}}$? Give reasons.

Answer:

If the last non-zero digit occurs in the 4th decimal place, the number can be written with denominator:

$${10}^4$$

So, it can be written as:

$$\frac{p}{{10}^4}$$

where $p$is an integer.

Since the last digit at the 4th decimal place is non-zero, $p$cannot end in 0.

Therefore, $p$is not divisible by 10.

Now:

$${10}^4={(2\times 5)}^4=2^4\times 5^4$$

When the fraction is reduced to lowest form, common factors may cancel.

But since $p$is not divisible by 10, it cannot contain both 2 and 5 as factors together.

So, after cancellation, the denominator will still be divisible by at least one of:

$$2^4\mathrm{\text{ or }}{5}^4$$

Therefore, yes, in lowest form the denominator must be divisible by $2^4$or $5^4$.

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

Answer:

$$\frac{18}{125}$$

Prime factorisation:

$$125=5^3$$

The denominator has only 5 as prime factor.

Therefore, the decimal expansion is terminating.

To make the denominator a power of 10:

$$125\times 8=1000\frac{18}{125}=\frac{18\times 8}{125\times 8}=\frac{144}{1000}=0.144$$

Therefore, it terminates after 3 decimal places.

Question 12. A rational number in its lowest form has denominator ${\mathrm{𝟐}}^{\mathrm{𝟑}}\mathbf{\times }\mathrm{𝟓}$. How many decimal places will its decimal expansion have? Explain your answer.

Answer:

Denominator:

$$2^3\times 5$$

To make it a power of 10, we need equal powers of 2 and 5.

$$2^3\times 5=2^3\times 5^1$$

We need $5^3$, so multiply by $5^2$.

$$2^3\times 5^3={10}^3$$

Therefore, the denominator can be made 1000.

So, the decimal expansion will have 3 decimal places.

Question 13. Let $𝐚\mathbf{=}\frac{\mathrm{𝟕}}{\mathrm{𝟏𝟐}}$and $𝐛\mathbf{=}\frac{\mathrm{𝟓}}{\mathrm{𝟔}}$. Express both $𝐚$and $𝐛$in the form $\frac{{𝐤}_{\mathrm{𝟏}}}{𝐦}$and $\frac{{𝐤}_{\mathrm{𝟐}}}{𝐦}$where ${𝐤}_{\mathrm{𝟏}}$, ${𝐤}_{\mathrm{𝟐}}$and $𝐦$are integers and ${𝐤}_{\mathrm{𝟐}}\mathbf{-}{𝐤}_{\mathrm{𝟏}}\mathbf{>}\mathrm{𝟔}$. Using the same denominator $𝐦$, write exactly five distinct rational numbers lying between $𝐚$and $𝐛$keeping an integer numerator. Explain why the condition ${𝐤}_{\mathrm{𝟐}}\mathbf{-}{𝐤}_{\mathrm{𝟏}}\mathbf{>}𝐧\mathbf{+}\mathrm{𝟏}$is necessary to find $𝐧$such rational numbers between the two rational numbers $𝐚$and $𝐛$using this method.

Answer:

Given:

$$a=\frac{7}{12}b=\frac{5}{6}$$

Choose denominator:

$$m=48$$

Now:

$$\frac{7}{12}=\frac{28}{48}\frac{5}{6}=\frac{40}{48}$$

So,

$$k_1=28,k_2=40k_2-k_1=40-28=12$$

Since:

$$12>6$$

the condition is satisfied.

Five rational numbers between them are:

$$\frac{29}{48},\frac{30}{48},\frac{31}{48},\frac{32}{48},\frac{33}{48}$$

These all lie between:

$$\frac{28}{48}\mathrm{\text{ and }}\frac{40}{48}$$

Therefore, five rational numbers between $\frac{7}{12}$and $\frac{5}{6}$are:

$$\frac{29}{48},\frac{30}{48},\frac{31}{48},\frac{32}{48},\frac{33}{48}$$

Explanation:

Between $\frac{k_1}{m}$and $\frac{k_2}{m}$, the possible integer numerators are:

$$k_1+1,k_1+2,...,k_2-1$$

The number of such integers is:

$$k_2-k_1-1$$

To find $n$rational numbers, we need at least $n$integer numerators between $k_1$and $k_2$. So,

$$k_2-k_1-1\ge n{k}_2-k_1\ge n+1$$

Thus, the gap between $k_1$and $k_2$must be large enough to give $n$numerators between them.

Question 14. Three rational numbers $𝐱\mathbf{,}𝐲\mathbf{,}𝐳$satisfy $𝐱\mathbf{+}𝐲\mathbf{+}𝐳\mathbf{=}\mathrm{𝟎}$and $𝐱𝐲\mathbf{+}𝐲𝐳\mathbf{+}𝐳𝐱\mathbf{=}\mathrm{𝟎}$. Show that all the rational numbers $𝐱\mathbf{,}𝐲\mathbf{,}𝐳$must be simultaneously zero.

Answer:

Given:

$$x+y+z=0$$

and

$$xy+yz+zx=0$$

Now square the first equation:

$$(x+y+z{)}^2=0^2{x}^2+y^2+z^2+2xy+2yz+2zx=0x^2+y^2+z^2+2(xy+yz+zx)=0$$

But:

$$xy+yz+zx=0$$

So:

$$x^2+y^2+z^2+2\left(0\right)=0x^2+y^2+z^2=0$$

Now, square of any rational number is non-negative.

So:

$$x^2\ge 0,y^2\ge 0,z^2\ge 0$$

Their sum can be zero only when:

$$x^2=0,y^2=0,z^2=0$$

Therefore:

$$x=0,y=0,z=0$$

Hence, all three rational numbers must be zero.

Question 15. Show that the rational number $\frac{𝐚\mathbf{+}𝐛}{\mathrm{𝟐}}$lies between the rational numbers $𝐚$and $𝐛$.

Answer:

Assume:

$$a<b$$

We need to show:

$$a<\frac{a+b}{2}<b$$

First compare $a$and $\frac{a+b}{2}$:

$$a<\frac{a+b}{2}$$

Multiply both sides by 2:

$$2a<a+b$$

Subtract $a$from both sides:

$$a<b$$

This is true.

Now compare $\frac{a+b}{2}$and $b$:

$$\frac{a+b}{2}<b$$

Multiply both sides by 2:

$$a+b<2b$$

Subtract $b$from both sides:

$$a<b$$

This is also true.

Therefore:

$$a<\frac{a+b}{2}<b$$

Hence, $\frac{a+b}{2}$lies between $a$and $b$.

Question 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:

In a square root spiral, each new right triangle has one side of length 1 unit and the previous hypotenuse as the other side.

First triangle:

$$h_1=\sqrt{1^2+1^2}=\sqrt{2}$$

Second triangle:

$$h_2=\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{2+1}=\sqrt{3}$$

Third triangle:

$$h_3=\sqrt{(\sqrt{3}{)}^2+1^2}=\sqrt{3+1}=\sqrt{4}=2$$

Continuing in the same way, the hypotenuse lengths are:

$$\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},\sqrt{9},\sqrt{10},\sqrt{11}$$

So, the lengths of the hypotenuses are:

$$\boxed{\sqrt{2},\sqrt{3},2,\sqrt{5},\sqrt{6},\sqrt{7},2\sqrt{2},3,\sqrt{10},\sqrt{11}}$$