Examples - Chapter 1 Real Numbers class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) - Chapter 1 - Real Numbers - Ncert Solutions class 10 - Maths

Example 1:

Consider the numbers $4^n$, where $n$ is a natural number. Check whether there is any value of $n$ for which $4^n$ ends with the digit zero.
Solution :

If the number $4^n$, for any $n$, were to end with the digit zero, then it would be divisible by 5 . That is, the prime factorisation of $4^n$ would contain the prime 5 . This is not possible because $4^n=(2)^{2 n}$; so the only prime in the factorisation of $4^n$ is 2 . So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of $4^n$. So, there is no natural number $n$ for which $4^n$ ends with the digit zero.

You have already learnt how to find the $\mathrm{HCF}$ and $\mathrm{LCM}$ of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! This method is also called the prime factorisation method. Let us recall this method through an example.

Example 2 :

Find the LCM and HCF of 6 and 20 by the prime factorisation method.

Solution :

We have : $\quad 6=2^1 \times 3^1$ and $20=2 \times 2 \times 5=2^2 \times 5^1$.
You can find $\operatorname{HCF}(6,20)=2$ and $\operatorname{LCM}(6,20)=2 \times 2 \times 3 \times 5=60$, as done in your earlier classes.

Note that $\operatorname{HCF}(6,20)=2^1=$ Product of the smallest power of each common prime factor in the numbers.
$\operatorname{LCM}(6,20)=2^2 \times 3^1 \times 5^1=$ Product of the greatest power of each prime factor, involved in the numbers.
From the example above, you might have noticed that $\operatorname{HCF}(6,20) \times \operatorname{LCM}(6,20)$ $=6 \times 20$. In fact, we can verify that for any two positive integers $a$ and $b$, $\operatorname{HCF}(a, b) \times \operatorname{LCM}(a, b)=a \times b$. We can use this result to find the LCM of two positive integers, if we have already found the $\mathrm{HCF}$ of the two positive integers.

Example 3:

Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.

Solution :

The prime factorisation of 96 and 404 gives:
$
96=2^5 \times 3,404=2^2 \times 101
$

Therefore, the HCF of these two integers is $2^2=4$.

Also,
$
\operatorname{LCM}(96,404)=\frac{96 \times 404}{\operatorname{HCF}(96,404)}=\frac{96 \times 404}{4}=9696
$

Example 4 :

Find the HCF and LCM of 6, 72 and 120 , using the prime factorisation method.

Solution :

We have :
$
6=2 \times 3,72=2^3 \times 3^2, 120=2^3 \times 3 \times 5
$

Here, $2^1$ and $3^1$ are the smallest powers of the common factors 2 and 3 , respectively.
So,
$
\operatorname{HCF}(6,72,120)=2^1 \times 3^1=2 \times 3=6
$
$2^3, 3^2$ and $5^1$ are the greatest powers of the prime factors 2,3 and 5 respectively involved in the three numbers.
So,
$
\operatorname{LCM}(6,72,120)=2^3 \times 3^2 \times 5^1=360
$

Remark:

Notice, $6 \times 72 \times 120 \neq \operatorname{HCF}(6,72,120) \times \operatorname{LCM}(6,72,120)$. So, the product of three numbers is not equal to the product of their HCF and LCM.

Example 5 :

Prove that $\sqrt{3}$ is irrational.
Solution :

Let us assume, to the contrary, that $\sqrt{3}$ is rational.
That is, we can find integers $a$ and $b(\neq 0)$ such that $\sqrt{3}=\frac{a}{b}$.
Suppose $a$ and $b$ have a common factor other than 1 , then we can divide by the common factor, and assume that $a$ and $b$ are coprime.
So, $b \sqrt{3}=a$.
Squaring on both sides, and rearranging, we get $3 b^2=a^2$.
Therefore, $a^2$ is divisible by 3 , and by Theorem 1.3 , it follows that $a$ is also divisible by 3 .
So, we can write $a=3 c$ for some integer $c$.
Substituting for $a$, we get $3 b^2=9 c^2$, that is, $b^2=3 c^2$.
This means that $b^2$ is divisible by 3 , and so $b$ is also divisible by 3 (using Theorem 1.3 with $p=3$ ).

Therefore, $a$ and $b$ have at least 3 as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.
So, we conclude that $\sqrt{3}$ is irrational.
In Class IX, we mentioned that :
- the sum or difference of a rational and an irrational number is irrational and
- the product and quotient of a non-zero rational and irrational number is irrational.
We prove some particular cases here.

Example 6 :

Show that $5-\sqrt{3}$ is irrational.
Solution :

Let us assume, to the contrary, that $5-\sqrt{3}$ is rational.
That is, we can find coprime $a$ and $b(b \neq 0)$ such that $5-\sqrt{3}=\frac{a}{b}$.
Therefore, $5-\frac{a}{b}=\sqrt{3}$.
Rearranging this equation, we get $\sqrt{3}=5-\frac{a}{b}=\frac{5 b-a}{b}$.
Since $a$ and $b$ are integers, we get $5-\frac{a}{b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.

So, we conclude that $5-\sqrt{3}$ is irrational.

Example 7 :

Show that $3 \sqrt{2}$ is irrational.
Solution :

Let us assume, to the contrary, that $3 \sqrt{2}$ is rational.
That is, we can find coprime $a$ and $b(b \neq 0)$ such that $3 \sqrt{2}=\frac{a}{b}$.
Rearranging, we get $\sqrt{2}=\frac{a}{3 b}$.
Since $3, a$ and $b$ are integers, $\frac{a}{3 b}$ is rational, and so $\sqrt{2}$ is rational.

But this contradicts the fact that $\sqrt{2}$ is irrational.
So, we conclude that $3 \sqrt{2}$ is irrational.