Examples (Revised) - Chapter 1 - Number Systems - Ncert Solutions class 9 - Maths

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Chapter 1 - Number Systems | NCERT Solutions Class 9 Maths

Example 1 :

Are the following statements true or false? Give reasons for your answers.
(i) Every whole number is a natural number.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.

Solution :

(i) False, because zero is a whole number but not a natural number.
(ii) True, because every integer $m$ can be expressed in the form $\frac{m}{1}$, and so it is a rational number.

(iii) False, because $\frac{3}{5}$ is not an integer.

Example 2 :

Find five rational numbers between 1 and 2.
We can approach this problem in at least two ways.
Solution 1 :

Recall that to find a rational number between $r$ and $s$, you can add $r$ and $s$ and divide the sum by 2 , that is $\frac{r+s}{2}$ lies between $r$ and $s$. So, $\frac{3}{2}$ is a number between 1 and 2. You can proceed in this manner to find four more rational numbers between 1 and 2 . These four numbers are $\frac{5}{4}, \frac{11}{8}, \frac{13}{8}$ and $\frac{7}{4}$.
Solution 2

: The other option is to find all the five rational numbers in one step. Since we want five numbers, we write 1 and 2 as rational numbers with denominator $5+1$, i.e., $1=\frac{6}{6}$ and $2=\frac{12}{6}$. Then you can check that $\frac{7}{6}, \frac{8}{6}, \frac{9}{6}, \frac{10}{6}$ and $\frac{11}{6}$ are all rational numbers between 1 and 2. So, the five numbers are $\frac{7}{6}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}$ and $\frac{11}{6}$.

Let us see how we can locate some of the irrational numbers on the number line.
Example 3

: Locate $\sqrt{2}$ on the number line.
Solution : It is easy to see how the Greeks might have discovered $\sqrt{2}$. Consider a square OABC, with each side 1 unit in length (see Fig. 1.6). Then you can see by the Pythagoras theorem that

Fig. 1.6 $\mathrm{OB}=\sqrt{1^2+1^2}=\sqrt{2}$. How do we represent $\sqrt{2}$ on the number line?
This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex $\mathrm{O}$ coincides with zero (see Fig. 1.7).

We have just seen that $O B=\sqrt{2}$. Using a compass with centre $O$ and radius $O B$, draw an arc intersecting the number line at the point $P$. Then $P$ corresponds to $\sqrt{2}$ on the number line.

Example 4 :

Locate $\sqrt{3}$ on the number line.
Solution :

Let us return to Fig. 1.7.

Construct $\mathrm{BD}$ of unit length perpendicular to $\mathrm{OB}$ (as in Fig. 1.8). Then using the Pythagoras theorem, we see that $\mathrm{OD}=\sqrt{(\sqrt{2})^2+1^2}=\sqrt{3}$. Using a compass, with centre $\mathrm{O}$ and radius $\mathrm{OD}$, draw an arc which intersects the number line at the point $\mathrm{Q}$. Then $\mathrm{Q}$ corresponds to $\sqrt{3}$.

In the same way, you can locate $\sqrt{n}$ for any positive integer $n$, after $\sqrt{n-1}$ has been located.

Example 5 :

Find the decimal expansions of $\frac{10}{3}, \frac{7}{8}$ and $\frac{1}{7}$.
Solution :

What have you noticed? You should have noticed at least three things:
(i) The remainders either become 0 after a certain stage, or start repeating themselves.
(ii) The number of entries in the repeating string of remainders is less than the divisor (in $\frac{10}{3}$ one number repeats itself and the divisor is 3 , in $\frac{1}{7}$ there are six entries 326451 in the repeating string of remainders and 7 is the divisor).
(iii) If the remainders repeat, then we get a repeating block of digits in the quotient (for $\frac{10}{3}, 3$ repeats in the quotient and for $\frac{1}{7}$, we get the repeating block 142857 in the quotient).

Although we have noticed this pattern using only the examples above, it is true for all rationals of the form $\frac{p}{q}(q \neq 0)$. On division of $p$ by $q$, two main things happen - either the remainder becomes zero or never becomes zero and we get a repeating string of remainders. Let us look at each case separately.

Case (i) : The remainder becomes zero
In the example of $\frac{7}{8}$, we found that the remainder becomes zero after some steps and the decimal expansion of $\frac{7}{8}=0.875$. Other examples are $\frac{1}{2}=0.5, \frac{639}{250}=2.556$. In all these cases, the decimal expansion terminates or ends after a finite number of steps. We call the decimal expansion of such numbers terminating.
Case (ii) : The remainder never becomes zero
In the examples of $\frac{10}{3}$ and $\frac{1}{7}$, we notice that the remainders repeat after a certain stage forcing the decimal expansion to go on for ever. In other words, we have a repeating block of digits in the quotient. We say that this expansion is non-terminating recurring. For example, $\frac{10}{3}=3.3333 \ldots$ and $\frac{1}{7}=0.142857142857142857 \ldots$

The usual way of showing that 3 repeats in the quotient of $\frac{10}{3}$ is to write it as $3 . \overline{3}$. Similarly, since the block of digits 142857 repeats in the quotient of $\frac{1}{7}$, we write $\frac{1}{7}$ as $0 . \overline{142857}$, where the bar above the digits indicates the block of digits that repeats. Also $3.57272 \ldots$ can be written as $3.5 \overline{72}$. So, all these examples give us non-terminating recurring (repeating) decimal expansions.
Thus, we see that the decimal expansion of rational numbers have only two choices: either they are terminating or non-terminating recurring.

Now suppose, on the other hand, on your walk on the number line, you come across a number like 3.142678 whose decimal expansion is terminating or a number like $1.272727 \ldots$ that is, $1 . \overline{27}$, whose decimal expansion is non-terminating recurring, can you conclude that it is a rational number? The answer is yes!

We will not prove it but illustrate this fact with a few examples. The terminating cases are easy.

Example 6 :

Show that 3.142678 is a rational number. In other words, express 3.142678 in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Solution :

We have $3.142678=\frac{3142678}{1000000}$, and hence is a rational number.
Now, let us consider the case when the decimal expansion is non-terminating recurring.

Example 7 :

Show that $0.3333 \ldots=0 . \overline{3}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Solution :

Since we do not know what $0 . \overline{3}$ is, let us call it ' $x$ ' and so
$
x=0.3333 \ldots
$

Now here is where the trick comes in. Look at
$
10 x=10 \times(0.333 \ldots)=3.333 \ldots
$

Now,
Therefore,
Solving for $x$, we get
$
9 x=3 \text {, i.e., } x=\frac{1}{3}
$

Example 8 :

Show that $1.272727 \ldots=1 . \overline{27}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
Solution :

Let $x=1.272727 \ldots$ Since two digits are repeating, we multiply $x$ by 100 to get
$
100 x=127.2727 \ldots
$

So,
$
\begin{aligned}
100 x & =126+1.272727 \ldots=126+x \\
100 x-x & =126, \text { i.e., } 99 x=126
\end{aligned}
$

i.e.,
$
x=\frac{126}{99}=\frac{14}{11}
$

You can check the reverse that $\frac{14}{11}=1 . \overline{27}$.

Example 9 :

Show that $0.2353535 \ldots=0.2 \overline{35}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
Solution :

Let $x=0.2 \overline{35}$. Over here, note that 2 does not repeat, but the block 35 repeats. Since two digits are repeating, we multiply $x$ by 100 to get

So,
$
100 x=23.53535 \ldots
$

Therefore,
$
\begin{aligned}
100 x & =23.3+0.23535 \ldots=23.3+x \\
99 x & =23.3
\end{aligned}
$
i.e.,
$
99 x=\frac{233}{10}, \text { which gives } x=\frac{233}{990}
$

You can also check the reverse that $\frac{233}{990}=0.2 \overline{35}$.
So, every number with a non-terminating recurring decimal expansion can be expressed in the form $\frac{p}{q}(q \neq 0)$, where $p$ and $q$ are integers. Let us summarise our results in the

following form:
The decimal expansion of a rational number is either terminating or nonterminating recurring. Moreover; a number whose decimal expansion is terminating or non-terminating recurring is rational.
So, now we know what the decimal expansion of a rational number can be. What about the decimal expansion of irrational numbers? Because of the property above, we can conclude that their decimal expansions are non-terminating non-recurning.
So, the property for irrational numbers, similar to the property stated above for rational numbers, is
The decimal expansion of an irrational number is non-terminating non-recurring. Moreover; a number whose decimal expansion is non-terminating non-recurring is irrational.

Recall $s=0.10110111011110 \ldots$ from the previous section. Notice that it is nonterminating and non-recurring. Therefore, from the property above, it is irrational. Moreover, notice that you can generate infinitely many irrationals similar to $s$.
What about the famous irrationals $\sqrt{2}$ and $\pi$ ? Here are their decimal expansions up to a certain stage.
$
\begin{aligned}
\sqrt{2} & =1.4142135623730950488016887242096 \ldots \\
\pi & =3.14159265358979323846264338327950 \ldots
\end{aligned}
$
(Note that, we often take $\frac{22}{7}$ as an approximate value for $\pi$, but $\pi \neq \frac{22}{7}$.)
Over the years, mathematicians have developed various techniques to produce more and more digits in the decimal expansions of irrational numbers. For example, you might have learnt to find digits in the decimal expansion of $\sqrt{2}$ by the division method. Interestingly, in the Sulbasutras (rules of chord), a mathematical treatise of the Vedic period ( $800 \mathrm{BC}-500 \mathrm{BC}$ ), you find an approximation of $\sqrt{2}$ as follows:
$
\sqrt{2}=1+\frac{1}{3}+\left(\frac{1}{4} \times \frac{1}{3}\right)-\left(\frac{1}{34} \times \frac{1}{4} \times \frac{1}{3}\right)=1.4142156
$

Notice that it is the same as the one given above for the first five decimal places. The history of the hunt for digits in the decimal expansion of $\pi$ is very interesting.

Now, let us see how to obtain irrational numbers.
Example 10 :

Find an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$.
Solution :

We saw that $\frac{1}{7}=0 . \overline{142857}$. So, you can easily calculate $\frac{2}{7}=0 . \overline{285714}$.
To find an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$, we find a number which is

non-terminating non-recurring lying between them. Of course, you can find infinitely many such numbers.
An example of such a number is $0.150150015000150000 \ldots$

Example 11 :

Check whether $7 \sqrt{5}, \frac{7}{\sqrt{5}}, \sqrt{2}+21, \pi-2$ are irrational numbers or not.

Solution :

$\sqrt{5}=2.236 \ldots, \sqrt{2}=1.4142 \ldots, \pi=3.1415 \ldots$
Then $7 \sqrt{5}=15.652 \ldots, \frac{7}{\sqrt{5}}=\frac{7 \sqrt{5}}{\sqrt{5} \sqrt{5}}=\frac{7 \sqrt{5}}{5}=3.1304 \ldots$
$
\sqrt{2}+21=22.4142 \ldots, \pi-2=1.1415 \ldots
$

All these are non-terminating non-recurring decimals. So, all these are irrational numbers.
Now, let us see what generally happens if we add, subtract, multiply, divide, take square roots and even $n$th roots of these irrational numbers, where $n$ is any natural number. Let us look at some examples.

Example 12 :

Add $2 \sqrt{2}+5 \sqrt{3}$ and $\sqrt{2}-3 \sqrt{3}$.
Solution :
$
\begin{aligned}
(2 \sqrt{2}+5 \sqrt{3})+(\sqrt{2}-3 \sqrt{3}) & =(2 \sqrt{2}+\sqrt{2})+(5 \sqrt{3}-3 \sqrt{3}) \\
& =(2+1) \sqrt{2}+(5-3) \sqrt{3}=3 \sqrt{2}+2 \sqrt{3}
\end{aligned}
$

Example 13 :

Multiply $6 \sqrt{5}$ by $2 \sqrt{5}$.
Solution :

$6 \sqrt{5} \times 2 \sqrt{5}=6 \times 2 \times \sqrt{5} \times \sqrt{5}=12 \times 5=60$
Example 14 :

Divide $8 \sqrt{15}$ by $2 \sqrt{3}$.
Solution :

$8 \sqrt{15} \div 2 \sqrt{3}=\frac{8 \sqrt{3} \times \sqrt{5}}{2 \sqrt{3}}=4 \sqrt{5}$
These examples may lead you to expect the following facts, which are true:
(i) The sum or difference of a rational number and an irrational number is irrational.
(ii) The product or quotient of a non-zero rational number with an irrational number is irrational.
(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational.
We now turn our attention to the operation of taking square roots of real numbers. Recall that, if $a$ is a natural number, then $\sqrt{a}=b$ means $b^2=a$ and $b>0$. The same definition can be extended for positive real numbers.
Let $a>0$ be a real number. Then $\sqrt{a}=b$ means $b^2=a$ and $b>0$.
In Section 1.2, we saw how to represent $\sqrt{n}$ for any positive integer $n$ on the number line. We now show how to find $\sqrt{x}$ for any given positive real number $x$ geometrically.
For example, let us find it for $x=3.5$, i.e., we find $\sqrt{3.5}$ geometrically.

Mark the distance 3.5 units from a fixed point $A$ on a given line to obtain a point $B$ such that $\mathrm{AB}=3.5$ units (see Fig. 1.11). From B, mark a distance of 1 unit and mark the new point as $\mathrm{C}$. Find the mid-point of $\mathrm{AC}$ and mark that point as $\mathrm{O}$. Draw a semicircle with centre $\mathrm{O}$ and radius $\mathrm{OC}$. Draw a line perpendicular to $\mathrm{AC}$ passing through $\mathrm{B}$ and intersecting the semicircle at $\mathrm{D}$. Then, $\mathrm{BD}=\sqrt{3.5}$.

More generally, to find $\sqrt{x}$, for any positive real number $x$, we mark $\mathrm{B}$ so that $\mathrm{AB}=x$ units, and, as in Fig. 1.12, mark $\mathrm{C}$ so that $\mathrm{BC}=1$ unit. Then, as we have done for the case $x=3.5$, we find $\mathrm{BD}=\sqrt{x}$ (see Fig. 1.12). We can prove this result using the Pythagoras Theorem.

Notice that, in Fig. 1.12, $\Delta$ OBD is a right-angled triangle. Also, the radius of the circle is $\frac{x+1}{2}$ units.
Therefore, $\mathrm{OC}=\mathrm{OD}=\mathrm{OA}=\frac{x+1}{2}$ units.
Now, OB $=x-\left(\frac{x+1}{2}\right)=\frac{x-1}{2}$.
So, by the Pythagoras Theorem, we have
$
\mathrm{BD}^2=\mathrm{OD}^2-\mathrm{OB}^2=\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2=\frac{4 x}{4}=x .
$

This shows that $\mathrm{BD}=\sqrt{x}$.
This construction gives us a visual, and geometric way of showing that $\sqrt{x}$ exists for all real numbers $x>0$. If you want to know the position of $\sqrt{x}$ on the number line, then let us treat the line $\mathrm{BC}$ as the number line, with $\mathrm{B}$ as zero, $\mathrm{C}$ as 1 , and so on. Draw an arc with centre $\mathrm{B}$ and radius $\mathrm{BD}$, which intersects the number line in $\mathrm{E}$ (see Fig. 1.13). Then, E represents $\sqrt{x}$.

We would like to now extend the idea of square roots to cube roots, fourth roots, and in general $n$th roots, where $n$ is a positive integer. Recall your understanding of square roots and cube roots from earlier classes.

What is $\sqrt[3]{8}$ ? Well, we know it has to be some positive number whose cube is 8 , and you must have guessed $\sqrt[3]{8}=2$. Let us try $\sqrt[5]{243}$. Do you know some number $b$ such that $b^5=243$ ? The answer is 3 . Therefore, $\sqrt[5]{243}=3$.

From these examples, can you define $\sqrt[n]{a}$ for a real number $a>0$ and a positive integer $n$ ?
Let $a>0$ be a real number and $n$ be a positive integer. Then $\sqrt[n]{a}=b$, if $b^n=a$ and $b>0$. Note that the symbol ' $\sqrt{ }$, used in $\sqrt{2}, \sqrt[3]{8}, \sqrt[n]{a}$, etc. is called the radical sign.

We now list some identities relating to square roots, which are useful in various ways. You are already familiar with some of these from your earlier classes. The remaining ones follow from the distributive law of multiplication over addition of real numbers, and from the identity $(x+y)(x-y)=x^2-y^2$, for any real numbers $x$ and $y$.
Let $a$ and $b$ be positive real numbers. Then
(i) $\sqrt{a b}=\sqrt{a} \sqrt{b}$
(ii) $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
(iii) $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$
(iv) $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$
(v) $(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{d})=\sqrt{a c}+\sqrt{a d}+\sqrt{b c}+\sqrt{b d}$
(vi) $(\sqrt{a}+\sqrt{b})^2=a+2 \sqrt{a b}+b$

Let us look at some particular cases of these identities.

Example 15 :

Simplify the following expressions:
(i) $(5+\sqrt{7})(2+\sqrt{5})$
(ii) $(5+\sqrt{5})(5-\sqrt{5})$
(iii) $(\sqrt{3}+\sqrt{7})^2$
(iv) $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$

Solution :
(i) $(5+\sqrt{7})(2+\sqrt{5})=10+5 \sqrt{5}+2 \sqrt{7}+\sqrt{35}$
(ii) $(5+\sqrt{5})(5-\sqrt{5})=5^2-(\sqrt{5})^2=25-5=20$
(iii) $(\sqrt{3}+\sqrt{7})^2=(\sqrt{3})^2+2 \sqrt{3} \sqrt{7}+(\sqrt{7})^2=3+2 \sqrt{21}+7=10+2 \sqrt{21}$
(iv) $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})=(\sqrt{11})^2-(\sqrt{7})^2=11-7=4$

Example 16 :

Rationalise the denominator of $\frac{1}{\sqrt{2}}$.

Solution :

We want to write $\frac{1}{\sqrt{2}}$ as an equivalent expression in which the denominator is a rational number. We know that $\sqrt{2} \cdot \sqrt{2}$ is rational. We also know that multiplying $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ will give us an equivalent expression, since $\frac{\sqrt{2}}{\sqrt{2}}=1$. So, we put these two facts together to get
$
\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}
$

In this form, it is easy to locate $\frac{1}{\sqrt{2}}$ on the number line. It is half way between 0 and $\sqrt{2}$.

Example 17 :

Rationalise the denominator of $\frac{1}{2+\sqrt{3}}$.
Solution :

We use the Identity (iv) given earlier. Multiply and divide $\frac{1}{2+\sqrt{3}}$ by $2-\sqrt{3}$ to get $\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$.

Example 18

Rationalise the denominator of $\frac{5}{\sqrt{3}-\sqrt{5}}$.
Solution :

Here we use the Identity (iii) given earlier.
So, $\frac{5}{\sqrt{3}-\sqrt{5}}=\frac{5}{\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}=\frac{5(\sqrt{3}+\sqrt{5})}{3-5}=\left(\frac{-5}{2}\right)(\sqrt{3}+\sqrt{5})$
Example 19 : Rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$.
Solution : $\frac{1}{7+3 \sqrt{2}}=\frac{1}{7+3 \sqrt{2}} \times\left(\frac{7-3 \sqrt{2}}{7-3 \sqrt{2}}\right)=\frac{7-3 \sqrt{2}}{49-18}=\frac{7-3 \sqrt{2}}{31}$
So, when the denominator of an expression contains a term with a square root (or a number under a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator.

Example 20 :

Simplify (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}}$
(ii) $\left(3^{\frac{1}{5}}\right)^4$
(iii) $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}$
(iv) $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}$

Solution :
(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}}=2^{\left(\frac{2}{3}+\frac{1}{3}\right)}=2^{\frac{3}{3}}=2^1=2$
(ii) $\left(3^{\frac{1}{5}}\right)^4=3^{\frac{4}{5}}$
(iii) $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}=7^{\left(\frac{1}{5}-\frac{1}{3}\right)}=7^{\frac{3-5}{15}}=7^{\frac{-2}{15}}$
(iv) $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}=(13 \times 17)^{\frac{1}{5}}=221^{\frac{1}{5}}$