Mathematics solution NCERT
Class 9 - Chapter 3: The World of Numbers
Q1: 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) $$\frac{3}{50}$$
Solution:
Divide the numerator by the denominator.
$$ \begin{aligned} \frac{3}{50} &=3\div50\\ &=0.06 \end{aligned} $$Since the division terminates, the decimal expansion is a terminating decimal.
Answer:
$$ \boxed{\frac{3}{50}=0.06} $$(ii) $$\frac{2}{9}$$
Solution:
Divide the numerator by the denominator.
$$ \begin{aligned} \frac{2}{9} &=2\div9\\ &=0.22222\ldots \end{aligned} $$The digit 2 repeats indefinitely.
Hence, the decimal expansion is a non-terminating recurring decimal.
Answer:
$$ \boxed{\frac{2}{9}=0.22222\ldots} $$Question 2. Prove that $\sqrt{5}$ is an irrational number.
Solution:
Assume, on the contrary, that $$\sqrt{5}$$ is a rational number.
Then it can be written in the form
$$ \sqrt{5}=\frac{p}{q}, $$where $p$ and $q$ are coprime integers and $$q\neq0$$.
Squaring both sides, we get
$$ 5=\frac{p^2}{q^2} $$ $$ p^2=5q^2. $$Thus, $p^2$ is divisible by 5. Therefore, $p$ is also divisible by 5.
Let
$$ p=5k, $$where $$k$$ is an integer.
Substituting in the equation,
$$ (5k)^2=5q^2 $$ $$ 25k^2=5q^2 $$ $$ q^2=5k^2. $$Hence, $q$ is also divisible by 5.
This means both $p$ and $q$ are divisible by 5, which contradicts the fact that they are coprime.
Therefore, our assumption is false.
Answer:
$$ \boxed{\sqrt{5}\text{ is an irrational number.}} $$Question 3. Convert the following decimal numbers into the form of $\dfrac{p}{q}$.
(i) $12.6$
Solution:
Let the given decimal number be $x$.
$$ x=12.6 $$Since there is one digit after the decimal point, multiply both sides of the equation by $10$.
$$ \begin{aligned} 10x &= 10\times12.6\\ &=126 \end{aligned} $$Now divide both sides by $10$.
$$ \begin{aligned} x&=\frac{126}{10} \end{aligned} $$Simplify the fraction by dividing the numerator and denominator by $2$.
$$ \begin{aligned} x&=\frac{126\div2}{10\div2}\\ &=\frac{63}{5} \end{aligned} $$Hence, the given decimal number in the form of $\dfrac{p}{q}$ is
$$ \boxed{\frac{63}{5}} $$(ii) $0.0120$
Solution:
Let the given decimal number be $x$.
$$ x=0.0120 $$Since there are four digits after the decimal point, multiply both sides by $10000$.
$$ \begin{aligned} 10000x &=10000\times0.0120\\ &=120 \end{aligned} $$Now divide both sides by $10000$.
$$ \begin{aligned} x &=\frac{120}{10000} \end{aligned} $$Divide the numerator and denominator by $40$.
$$ \begin{aligned} x &=\frac{120\div40}{10000\div40}\\ &=\frac{3}{250} \end{aligned} $$Hence, the given decimal number in the form of $\dfrac{p}{q}$ is
$$ \boxed{\frac{3}{250}} $$(iii) $3.0\overline{52}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=3.05252525\ldots $$There is one non-recurring digit after the decimal point and two recurring digits. Therefore, multiply both sides by $10$.
$$ \begin{aligned} 10x &=30.5252525\ldots \end{aligned} $$Now multiply both sides by $100$ again so that the recurring parts line up.
$$ \begin{aligned} 1000x &=3052.5252525\ldots \end{aligned} $$Subtract the first equation from the second equation.
$$ \begin{aligned} 1000x-10x &=3052.525252\ldots-30.525252\ldots\\ 990x &=3022 \end{aligned} $$Now divide both sides by $990$.
$$ \begin{aligned} x &=\frac{3022}{990} \end{aligned} $$Simplify the fraction by dividing the numerator and denominator by $2$.
$$ \begin{aligned} x &=\frac{3022\div2}{990\div2}\\ &=\frac{1511}{495} \end{aligned} $$Hence, the given recurring decimal number in the form of $\dfrac{p}{q}$ is
$$ \boxed{\frac{1511}{495}} $$(iv) $1.\overline{235}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=1.235235235\ldots $$Since the repeating block contains three digits, multiply both sides by $1000$.
$$ \begin{aligned} 1000x &=1000\times1.235235235\ldots\\ &=1235.235235235\ldots \end{aligned} $$Subtract the original equation from the above equation.
$$ \begin{aligned} 1000x-x &=1235.235235235\ldots-1.235235235\ldots\\ 999x &=1234 \end{aligned} $$Now divide both sides by $999$.
$$ \begin{aligned} x &=\frac{1234}{999} \end{aligned} $$Since $1234$ and $999$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{1234}{999}} $$(v) $0.\overline{23}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=0.23232323\ldots $$Since the repeating block contains two digits, multiply both sides by $100$.
$$ \begin{aligned} 100x &=100\times0.23232323\ldots\\ &=23.23232323\ldots \end{aligned} $$Subtract the original equation from the above equation.
$$ \begin{aligned} 100x-x &=23.23232323\ldots-0.23232323\ldots\\ 99x &=23 \end{aligned} $$Now divide both sides by $99$.
$$ \begin{aligned} x &=\frac{23}{99} \end{aligned} $$Since $23$ and $99$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{23}{99}} $$(vi) $2.\overline{05}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=2.05050505\ldots $$Since the repeating block contains two digits, multiply both sides by $100$.
$$ \begin{aligned} 100x &=100\times2.05050505\ldots\\ &=205.05050505\ldots \end{aligned} $$Subtract the original equation from the above equation.
$$ \begin{aligned} 100x-x &=205.05050505\ldots-2.05050505\ldots\\ 99x &=203 \end{aligned} $$Now divide both sides by $99$.
$$ \begin{aligned} x &=\frac{203}{99} \end{aligned} $$Since $203$ and $99$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{203}{99}} $$(vii) $2.12\overline{5}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=2.1255555\ldots $$Since there are two non-recurring digits and one recurring digit after the decimal point, first multiply both sides by $100$.
$$ \begin{aligned} 100x &=100\times2.1255555\ldots\\ &=212.55555\ldots \end{aligned} $$Now multiply both sides by $10$ again so that the recurring parts line up.
$$ \begin{aligned} 1000x &=1000\times2.1255555\ldots\\ &=2125.55555\ldots \end{aligned} $$Subtract the first equation from the second equation.
$$ \begin{aligned} 1000x-100x &=2125.55555\ldots-212.55555\ldots\\ 900x &=1913 \end{aligned} $$Now divide both sides by $900$.
$$ \begin{aligned} x &=\frac{1913}{900} \end{aligned} $$Since $1913$ and $900$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{1913}{900}} $$(viii) $3.12\overline{5}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=3.1255555\ldots $$Since there are two non-recurring digits and one recurring digit after the decimal point, first multiply both sides by $100$.
$$ \begin{aligned} 100x &=100\times3.1255555\ldots\\ &=312.55555\ldots \end{aligned} $$Now multiply both sides by $10$ again.
$$ \begin{aligned} 1000x &=1000\times3.1255555\ldots\\ &=3125.55555\ldots \end{aligned} $$Subtract the first equation from the second equation.
$$ \begin{aligned} 1000x-100x &=3125.55555\ldots-312.55555\ldots\\ 900x &=2813 \end{aligned} $$Now divide both sides by $900$.
$$ \begin{aligned} x &=\frac{2813}{900} \end{aligned} $$Since $2813$ and $900$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{2813}{900}} $$(ix) $2.\overline{1625}$
Solution:
Let the given recurring decimal number be $x$.
$$ x=2.162516251625\ldots $$Since the repeating block contains four digits, multiply both sides by $10000$.
$$ \begin{aligned} 10000x &=10000\times2.162516251625\ldots\\ &=21625.162516251625\ldots \end{aligned} $$Subtract the original equation from the above equation.
$$ \begin{aligned} 10000x-x &=21625.162516251625\ldots-2.162516251625\ldots\\ 9999x &=21623 \end{aligned} $$Now divide both sides by $9999$.
$$ \begin{aligned} x &=\frac{21623}{9999} \end{aligned} $$Since $21623$ and $9999$ have no common factor other than $1$, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{21623}{9999}} $$Question 5. Find 6 rational numbers between $3$ and $4$.
Solution:
We have to find six rational numbers lying between $3$ and $4$.
First, express both numbers as fractions having the same denominator.
$$ 3=\frac{3}{1} \qquad\text{and}\qquad 4=\frac{4}{1} $$Since we need six rational numbers between them, multiply the numerator and denominator of both fractions by $10$.
$$ \begin{aligned} \frac31 &=\frac{3\times10}{1\times10}\\ &=\frac{30}{10} \end{aligned} $$ $$ \begin{aligned} \frac41 &=\frac{4\times10}{1\times10}\\ &=\frac{40}{10} \end{aligned} $$Now the fractions lying between $\frac{30}{10}$ and $\frac{40}{10}$ are
$$ \frac{31}{10}, \; \frac{32}{10}, \; \frac{33}{10}, \; \frac{34}{10}, \; \frac{35}{10}, \; \frac{36}{10}. $$Simplify the fractions wherever possible.
$$ \begin{aligned} \frac{32}{10}&=\frac{16}{5},\\[4pt] \frac{34}{10}&=\frac{17}{5},\\[4pt] \frac{35}{10}&=\frac72,\\[4pt] \frac{36}{10}&=\frac{18}{5}. \end{aligned} $$Hence, six rational numbers between $3$ and $4$ are
$$ \boxed{ \frac{31}{10}, \; \frac{16}{5}, \; \frac{33}{10}, \; \frac{17}{5}, \; \frac72, \; \frac{18}{5} } $$Question 6. Find 5 rational numbers between $\dfrac25$ and $\dfrac35$.
Solution:
We have to find five rational numbers between $\frac25$ and $\frac35$.
To obtain at least five fractions between them, multiply the numerator and denominator of each fraction by $6$.
$$ \begin{aligned} \frac25 &=\frac{2\times6}{5\times6}\\ &=\frac{12}{30} \end{aligned} $$ $$ \begin{aligned} \frac35 &=\frac{3\times6}{5\times6}\\ &=\frac{18}{30} \end{aligned} $$Now the fractions lying between $\frac{12}{30}$ and $\frac{18}{30}$ are
$$ \frac{13}{30}, \; \frac{14}{30}, \; \frac{15}{30}, \; \frac{16}{30}, \; \frac{17}{30}. $$Simplify the fractions wherever possible.
$$ \begin{aligned} \frac{14}{30} &=\frac{7}{15}, \\[4pt] \frac{15}{30} &=\frac12, \\[4pt] \frac{16}{30} &=\frac{8}{15}. \end{aligned} $$Therefore, five rational numbers between $\frac25$ and $\frac35$ are
$$ \boxed{ \frac{13}{30}, \; \frac{7}{15}, \; \frac12, \; \frac{8}{15}, \; \frac{17}{30} } $$Question 7. Find 5 rational numbers between $\dfrac16$ and $\dfrac25$.
Solution:
We have to find five rational numbers between $\frac16$ and $\frac25$.
First, express both fractions with the same denominator.
The LCM of $6$ and $5$ is $30$.
$$ \begin{aligned} \frac16 &=\frac{1\times5}{6\times5}\\ &=\frac5{30} \end{aligned} $$ $$ \begin{aligned} \frac25 &=\frac{2\times6}{5\times6}\\ &=\frac{12}{30} \end{aligned} $$There are only six integers between $5$ and $12$. To obtain at least five rational numbers conveniently, multiply the numerator and denominator of both fractions by $2$.
$$ \begin{aligned} \frac5{30} &=\frac{5\times2}{30\times2}\\ &=\frac{10}{60} \end{aligned} $$ $$ \begin{aligned} \frac{12}{30} &=\frac{12\times2}{30\times2}\\ &=\frac{24}{60} \end{aligned} $$Now the fractions lying between $\frac{10}{60}$ and $\frac{24}{60}$ are
$$ \frac{11}{60}, \; \frac{12}{60}, \; \frac{13}{60}, \; \frac{14}{60}, \; \frac{15}{60}. $$Simplify the fractions wherever possible.
$$ \begin{aligned} \frac{12}{60} &=\frac15, \\[4pt] \frac{14}{60} &=\frac7{30}, \\[4pt] \frac{15}{60} &=\frac14. \end{aligned} $$Hence, five rational numbers between $\frac16$ and $\frac25$ are
$$ \boxed{ \frac{11}{60}, \; \frac15, \; \frac{13}{60}, \; \frac7{30}, \; \frac14 } $$Question 8. If $\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{16}{15}$, find the rational number $x$.
Solution:
The given equation is
$$ \frac{x}{3}+\frac{x}{5}=\frac{16}{15}. $$Take $x$ common from the left-hand side.
$$ x\left(\frac13+\frac15\right)=\frac{16}{15} $$Add the fractions inside the brackets.
$$ \begin{aligned} \frac13+\frac15 &=\frac{5+3}{15}\\ &=\frac8{15} \end{aligned} $$Substitute this value into the equation.
$$ x\times\frac8{15}=\frac{16}{15} $$To find the value of $x$, multiply both sides by the reciprocal of $\frac8{15}$, which is $\frac{15}{8}$.
$$ \begin{aligned} x\times\frac8{15}\times\frac{15}{8} &=\frac{16}{15}\times\frac{15}{8} \end{aligned} $$Cancel the common factors.
$$ \begin{aligned} x &=\frac{16}{8}\\ &=2 \end{aligned} $$Verification:
Substitute $x=2$ into the given equation.
$$ \begin{aligned} \frac23+\frac25 &=\frac{10+6}{15}\\ &=\frac{16}{15} \end{aligned} $$The left-hand side is equal to the right-hand side. Hence, the value obtained is correct.
Answer:
$$ \boxed{x=2} $$Question 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.
Solution:
We are given that
$$ a+\frac1b=0. $$Subtract $\dfrac1b$ from both sides of the equation.
$$ \begin{aligned} a+\frac1b-\frac1b &=0-\frac1b\\ a &=-\frac1b \end{aligned} $$Now multiply both sides by $b$. Since $b\neq0$, this operation is valid.
$$ \begin{aligned} ab &=-\frac1b\times b\\ &=-1 \end{aligned} $$Since $ab=-1$, which is a negative number, the product $ab$ is always negative.
Answer:
$$ \boxed{ab=-1} $$Hence, $ab$ is a negative rational number.
Question 10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the fourth 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.
Solution:
Suppose the given terminating decimal number is
$$ x=a.bcde, $$where the digit $e$ is the last non-zero digit after the decimal point.
Since the last non-zero digit occurs in the fourth decimal place, multiplying the number by $10^4=10000$ removes the decimal point completely.
$$ 10000x=p, $$where $p$ is an integer.
Therefore,
$$ x=\frac{p}{10000} =\frac{p}{10^4}. $$The integer $p$ cannot be divisible by $10$. If $p$ were divisible by $10$, then both the numerator and denominator would have a common factor $10$. In that case, the decimal expansion would end before the fourth decimal place, which contradicts the given condition that the last non-zero digit occurs in the fourth decimal place.
Hence, $p$ is not divisible by $10$.
Now consider the fraction in its lowest form.
The denominator of a rational number in lowest form contains only the prime factors $2$ and $5$.
However, it is not necessary that the denominator must be divisible by $2^4$ or by $5^4$.
During simplification, common factors between the numerator and denominator may cancel.
For example,
$$ \frac{2500}{10000} = \frac14. $$Here, the denominator in the lowest form is
$$ 4=2^2, $$which is neither divisible by $2^4$ nor by $5^4$.
Similarly,
$$ \frac{2000}{10000} = \frac15, $$whose denominator is
$$ 5=5^1. $$Thus, after reducing the fraction to its lowest form, the denominator may contain powers of $2$ and $5$ smaller than $4$.
Answer:
$$ \boxed{ \text{The number can be written as } \frac{p}{10^4}, \text{ where }p\text{ is not divisible by }10. } $$ $$ \boxed{ \text{No, it is not necessary that the denominator in the lowest form is divisible by }2^4\text{ or }5^4. } $$Question 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.
Solution:
The given rational number is
$$ \frac{18}{125}. $$First, check whether the fraction is in its lowest form.
The HCF of $18$ and $125$ is $1$. Therefore, the fraction is already in its lowest form.
Now, write the denominator as the product of prime factors.
$$ \begin{aligned} 125 &=5\times5\times5\\ &=5^3 \end{aligned} $$Since the denominator contains only the prime factor $5$, the decimal expansion is terminating.
To determine the number of decimal places, make the powers of $2$ and $5$ equal by multiplying the numerator and denominator by $2^3=8$.
$$ \begin{aligned} \frac{18}{125} &=\frac{18\times8}{125\times8}\\ &=\frac{144}{1000} \end{aligned} $$Since
$$ 1000=10^3, $$the decimal expansion has three decimal places.
Answer:
$$ \boxed{\text{The decimal expansion is terminating and has 3 decimal places.}} $$Question 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.
Solution:
The denominator of the given rational number is
$$ 2^3\times5. $$To determine the number of decimal places, express the denominator as a power of $10$.
The powers of $2$ and $5$ must be equal. Here, the denominator contains $2^3$ but only one factor of $5$.
Multiply both the numerator and denominator by $5^2=25$.
$$ \begin{aligned} 2^3\times5\times5^2 &=2^3\times5^3\\ &=(2\times5)^3\\ &=10^3\\ &=1000 \end{aligned} $$Thus, after multiplying by $25$, the denominator becomes $1000$.
Since
$$ 1000=10^3, $$the decimal expansion will have three decimal places.
Answer:
$$ \boxed{\text{The decimal expansion has 3 decimal places.}} $$Question 13. Let $a=\dfrac{7}{12}$ and $b=\dfrac{5}{6}$. 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 the two rational numbers using this method.
Solution:
The given rational numbers are
$$ a=\frac{7}{12} \qquad\text{and}\qquad b=\frac56. $$First, express both fractions with the same denominator.
The LCM of $12$ and $6$ is $12$.
$$ \frac56=\frac{10}{12}. $$Thus,
$$ a=\frac7{12}, \qquad b=\frac{10}{12}. $$Here,
$$ k_1=7,\qquad k_2=10,\qquad m=12. $$Since
$$ k_2-k_1=10-7=3, $$the condition $k_2-k_1>6$ is not satisfied. Therefore, we must choose a larger common denominator.
Multiply the numerator and denominator of both fractions by $3$.
$$ \begin{aligned} \frac7{12} &=\frac{7\times3}{12\times3}\\ &=\frac{21}{36} \end{aligned} $$ $$ \begin{aligned} \frac{10}{12} &=\frac{10\times3}{12\times3}\\ &=\frac{30}{36} \end{aligned} $$Now,
$$ k_1=21,\qquad k_2=30,\qquad m=36. $$Hence,
$$ k_2-k_1=30-21=9>6. $$Therefore, the required condition is satisfied.
The rational numbers lying between $\dfrac{21}{36}$ and $\dfrac{30}{36}$ are
$$ \frac{22}{36}, \; \frac{23}{36}, \; \frac{24}{36}, \; \frac{25}{36}, \; \frac{26}{36}, \; \frac{27}{36}, \; \frac{28}{36}, \; \frac{29}{36}. $$Any five of these rational numbers may be chosen.
Hence, one possible set is
$$ \boxed{ \frac{22}{36}, \; \frac{23}{36}, \; \frac{24}{36}, \; \frac{25}{36}, \; \frac{26}{36} } $$Explanation of the condition $k_2-k_1>n+1$:
Suppose we want to find exactly $n$ rational numbers between
$$ \frac{k_1}{m} \qquad\text{and}\qquad \frac{k_2}{m}, $$where both fractions have the same denominator.
The possible numerators lying strictly between $k_1$ and $k_2$ are
$$ k_1+1,\; k_1+2,\; \ldots,\; k_2-1. $$The total number of integers between $k_1$ and $k_2$ is
$$ k_2-k_1-1. $$To obtain at least $n$ rational numbers, we must have
$$ \begin{aligned} k_2-k_1-1 &\ge n. \end{aligned} $$Adding $1$ to both sides, we get
$$ \boxed{ k_2-k_1\ge n+1. } $$Therefore, the condition $k_2-k_1>n+1$ (or equivalently, at least $n+1$ difference between the numerators) ensures that there are enough integer numerators available to obtain $n$ distinct rational numbers between the given rational numbers.
Question 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.
Solution:
We are given that
$$ x+y+z=0 \tag{1} $$and
$$ xy+yz+zx=0. \tag{2} $$We have to prove that
$$ x=y=z=0. $$Square both sides of equation (1).
$$ (x+y+z)^2=0^2 $$Expand the left-hand side using the identity
$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. $$Therefore,
$$ x^2+y^2+z^2+2xy+2yz+2zx=0. $$Take $2$ common from the last three terms.
$$ x^2+y^2+z^2+2(xy+yz+zx)=0. $$From equation (2),
$$ xy+yz+zx=0. $$Substitute this value into the above equation.
$$ \begin{aligned} x^2+y^2+z^2+2(0) &=0\\ x^2+y^2+z^2 &=0. \end{aligned} $$Now observe that $x^2$, $y^2$, and $z^2$ are squares of rational numbers.
The square of any rational number is always non-negative.
Therefore, each of the numbers
$$ x^2,\qquad y^2,\qquad z^2 $$is greater than or equal to zero.
The sum of three non-negative numbers is zero only when each number is zero.
Hence,
$$ x^2=0,\qquad y^2=0,\qquad z^2=0. $$Taking the square root of both sides, we obtain
$$ x=0,\qquad y=0,\qquad z=0. $$Therefore, all three rational numbers are simultaneously zero.
Answer:
$$ \boxed{x=y=z=0} $$Hence proved.
Question 15. Show that the rational number $\dfrac{a+b}{2}$ lies between the rational numbers $a$ and $b$.
Solution:
Let $a$ and $b$ be two rational numbers such that
$$ aSince $a Now divide both sides by $2$. Since $2$ is a positive number, the inequality remains unchanged. Thus, Step 2: Show that $\dfrac{a+b}{2} Again, since $a Divide both sides by $2$. Step 3: Combine the two results. From Step 1 and Step 2, we get Hence, the rational number $\dfrac{a+b}{2}$ lies between the rational numbers $a$ and $b$. Solution: In the given square root spiral, each new right triangle is formed by taking: Using the Pythagoras Theorem, we find the hypotenuse of each triangle. First Triangle The two perpendicular sides are $1$ unit and $1$ unit. Second Triangle The perpendicular sides are $\sqrt2$ units and $1$ unit. Third Triangle The perpendicular sides are $\sqrt3$ units and $1$ unit. Fourth Triangle The perpendicular sides are $2$ units and $1$ unit. Fifth Triangle The perpendicular sides are $\sqrt5$ units and $1$ unit. Sixth Triangle The perpendicular sides are $\sqrt6$ units and $1$ unit. Seventh Triangle The perpendicular sides are $\sqrt7$ units and $1$ unit. Eighth Triangle The perpendicular sides are $\sqrt8$ units and $1$ unit. Ninth Triangle The perpendicular sides are $3$ units and $1$ unit. Summary Therefore, the lengths of the hypotenuses of all the right triangles are or equivalently,
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.