NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Exercise 1.1 

1. Use Euclid’s division algorithm to find the HCF of:

(i) $135$ and $225$ 

Ans: We have to find the HCF of $135$ and $225$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a > b$ is found as :

• First find the values of $q$ and $r$, where $a=bq+r$, $0\le r < b$.

• If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

• Continue steps till the remainder is zero. When we get the remainder zero, divisor will be the HCF.

Let $a=225$ and $b=135$.

Since, $a > b$ 

Using division algorithm, we get

$a=bq+r$

$\Rightarrow 225=135\times 1+90$ 

Here,

$\Rightarrow b=135$ 

$\Rightarrow q=1$ 

$\Rightarrow r=90$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new remainder). We get

$\Rightarrow 135=90\times 1+45$ 

Here, 

$\Rightarrow b=90$ 

$\Rightarrow q=1$ 

$\Rightarrow r=45$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 90=2\times 45+0$ 

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, divisor will be the HCF.

Therefore, the HCF of $135$ and $225$ is $45$.

(ii) $196$ and $38220$

Ans: We have to find the HCF of $196$ and $38220$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a > b$ is found as :

• First find the values of $q$ and $r$, where $a=bq+r$, $0\le r < b$.

• If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

• Continue steps till the remainder is zero. When we get the remainder zero, divisor will be the HCF.

Let $a=38220$ and $b=196$.

Since, $a > b$ 

Using division algorithm, we get

$a=bq+r$

$\Rightarrow 38220=196\times 195+0$ 

Here, 

$\Rightarrow b=196$ 

$\Rightarrow q=195$ 

$\Rightarrow r=0$ 

Since, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, divisor will be the HCF.

Therefore, the HCF of $196$ and $38220$ is $196$.

(iii) $867$ and $255$ 

Ans: We have to find the HCF of $867$ and $255$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a > b$ is found as :

• First find the values of $q$ and $r$, where $a=bq+r$, $0\le r < b$.

• If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

• Continue steps till the remainder is zero. When we get the remainder zero, divisor will be the HCF.

Let $a=867$ and $b=255$.

Since, $a > b$ 

Using division algorithm, we get

$a=bq+r$

\[\Rightarrow 867=255\times 3+102\] 

Here,

$\Rightarrow b=255$ 

$\Rightarrow q=3$ 

$\Rightarrow r=102$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new remainder). We get

$\Rightarrow 255=102\times 2+51$ 

Here, 

$\Rightarrow b=102$ 

$\Rightarrow q=2$ 

$\Rightarrow r=51$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 102=51\times 2+0$ 

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, divisor will be the HCF.

Therefore, the HCF of $867$ and $255$ is $51$.

2. Show that any positive odd integer is of the form, $6q+1$ or $6q+3$, or $6q+5$, where $q$ is some integer.

Ans: Let $a$ be any positive integer and $b=6$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r < b$.

Here, $0\le r < 6$.

Substitute the values, we get

$\Rightarrow a=6q+r$

If $r=0$, we get

$\Rightarrow a=6q+0$

$\Rightarrow a=6q$

If $r=1$, we get

$\Rightarrow a=6q+1$

If $r=2$, we get

$\Rightarrow a=6q+2$ and so on

Therefore, $a=6q$ or $6q+1$ or $6q+2$ or $6q+3$ or $6q+4$ or $6q+5$.

We can write the obtained expressions as 

\[6q+1=2\times 3q+1\]

$\Rightarrow 6q+1=2{{k}_{1}}+1$ 

Where, ${{k}_{1}}$ is an integer.

\[6q+3=6q+2+1\]

$\Rightarrow 6q+3=2\left( 3q+1 \right)+1$ 

$\Rightarrow 6q+3=2{{k}_{2}}+1$ 

Where, ${{k}_{2}}$ is an integer.

\[6q+5=6q+4+1\]

$\Rightarrow 6q+5=2\left( 3q+2 \right)+1$ 

$\Rightarrow 6q+5=2{{k}_{3}}+1$  

Where, ${{k}_{3}}$ is an integer.

Thus, $6q+1,6q+3,6q+5$ are of the form $2k+1$ and are not exactly divisible by $2$.

Also, all these expressions are of odd numbers.

Therefore, any positive odd integer can be expressed in the form, $6q+1$ or $6q+3$, or $6q+5$, where $q$ is some integer.

3. An army contingent of $616$ members is to march behind an army band of $32$ members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?  

Ans: Given that an army contingent of $616$ members is to march behind an army band of $32$ members in a parade. The two groups are to march in the same number of columns.

We have to find the maximum number of columns in which they can march.

We need to find the HCF of $616$ and $32$ to find the maximum number of columns.

We will use the Euclid’s division algorithm to find the HCF.

Let $a=616$ and $b=32$.

Since, $a > b$ 

Using division algorithm, we get

$a=bq+r$

\[\Rightarrow 616=32\times 19+8\] 

Here,

$\Rightarrow b=32$ 

$\Rightarrow q=19$ 

$\Rightarrow r=8$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new remainder). We get

$\Rightarrow 32=8\times 4+0$ 

Since, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, divisor will be the HCF.

Therefore, the HCF of $616$ and $32$ is $8$.

Therefore, in $8$ columns members can march.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of form $3m$ or $3m+1$ for some integer $m$.

(Hint: Let $x$ be any positive integer than it is of the form $3q,3q+1$ or $3q+2$. Now, square each of these and show that they can be rewritten in the form $3m$ or $3m+1$.)

Ans: Let $a$ be any positive integer and $b=3$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r < b$.

Here, $0\le r < 3$.

Substitute the values, we get

$\Rightarrow a=3q+r$

If $r=0$, we get

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

If $r=1$, we get

$\Rightarrow a=3q+1$

If $r=2$, we get

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, squaring each of these, we get

$\Rightarrow {{a}^{2}}={{\left( 3q \right)}^{2}}$ or ${{\left( 3q+1 \right)}^{2}}$ or ${{\left( 3q+2 \right)}^{2}}$

Now, applying the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$

Thus, we get

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}$

$\Rightarrow {{a}^{2}}=3m$, where, $m=3{{q}^{2}}$ 

And, 

${{a}^{2}}=3\times 3{{q}^{2}}+3\times 2q+1$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+2q \right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+2q$ 

And, 

${{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+4$

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+3+1$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q+1 \right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+4q+1$

Therefore, we can say that the square of any positive integer is either of form $3m$ or $3m+1$ for some integer $m$.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Ans: Let $a$ be any positive integer and $b=3$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r < b$.

Here, $0\le r < 3$.

Substitute the values, we get

$\Rightarrow a=3q+r$

If $r=0$, we get

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

If $r=1$, we get

$\Rightarrow a=3q+1$

If $r=2$, we get

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, consider $a=3q$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q \right)}^{3}}$ 

$\Rightarrow {{a}^{3}}=27{{q}^{3}}$ 

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}} \right)$

$\Rightarrow {{a}^{3}}=9m$, where, $m=3{{q}^{3}}$ 

Now, consider $a=3q+1$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+1 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$

$\Rightarrow {{a}^{3}}=9m+1$, where, \[m=3{{q}^{3}}+3{{q}^{2}}+q\]  

Now, consider $a=3q+2$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+2 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$

$\Rightarrow {{a}^{3}}=9m+8$, where, \[m=3{{q}^{3}}+6{{q}^{2}}+4q\]  

Therefore, we can say that cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Exercise 1.2

1. Express each number as product of its prime factors:

(i) $140$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 140=2\times 2\times 5\times 7$ 

$\therefore 140={{2}^{2}}\times 5\times 7$ 

Therefore, the prime factors of $140$ are $2,5,7$.

(ii) $156$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 156=2\times 2\times 3\times 13$ 

$\therefore 156={{2}^{2}}\times 3\times 13$ 

Therefore, the prime factors of $156$ are $2,3,13$.

(iii) $3825$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 3825=3\times 3\times 5\times 5\times 17$ 

$\therefore 3825={{3}^{2}}\times {{5}^{2}}\times 17$ 

Therefore, the prime factors of $3825$ are $3,5,17$.

(iv) $5005$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 5005=5\times 7\times 11\times 13$ 

$\therefore 5005=5\times 7\times 11\times 13$ 

Therefore, the prime factors of $5005$ are $5,7,11,13$.

(v) $7429$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 7429=17\times 19\times 23$ 

$\therefore 7429=17\times 19\times 23$ 

Therefore, the prime factors of $7429$ are $17,19,23$.

2. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=\text{Product of two numbers}$.

(i) $26$ and $91$ 

Ans: First we write the prime factors of $26$ and $91$. We get

$\Rightarrow 26=2\times 13$ and 

$\Rightarrow 91=7\times 13$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $26$ and $91$ is $13$.

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $26$ and $91$ will be

$\Rightarrow 2\times 7\times 13=182$ 

Therefore, the LCM of $26$ and $91$ is $182$.

Now, the product of two numbers is 

$\Rightarrow 26\times 91=2366$ 

Product of LCM and HCF is

$\Rightarrow 13\times 182=2366$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(ii) $510$ and $92$

Ans: First we write the prime factors of $510$ and $92$. We get

$\Rightarrow 510=2\times 3\times 5\times 17$ and 

$\Rightarrow 92=2\times 2\times 23$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $510$ and $92$ is $2$.

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $510$ and $92$ will be

$\Rightarrow 2\times 2\times 3\times 5\times 17\times 23=23460$ 

Therefore, the LCM of $510$ and $92$ is $23460$.

Now, the product of two numbers is 

$\Rightarrow 510\times 92=46920$ 

Product of LCM and HCF is

$\Rightarrow 2\times 23460=46920$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(iii) $336$ and $54$

Ans: First we write the prime factors of $336$ and $54$. We get

\[\Rightarrow 336=2\times 2\times 2\times 2\times 3\times 7\] 

and 

$\Rightarrow 54=2\times 3\times 3\times 3$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $336$ and $54$ is $2\times 3=6$.

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $336$ and $54$ will be

$\Rightarrow 2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024$ 

Therefore, the LCM of $336$ and $54$ is $3024$.

Now, the product of two numbers is 

$\Rightarrow 336\times 54=18144$ 

Product of LCM and HCF is

$\Rightarrow 6\times 3024=18144$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) $12,15$ and $21$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $12,15$ and $21$ are as follows:

\[\Rightarrow 12=2\times 2\times 3\] 

\[\Rightarrow 15=3\times 5\] and 

$\Rightarrow 21=3\times 7$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $12,15$ and $21$ is $3$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12,15$ and $21$ will be

$\Rightarrow 2\times 2\times 3\times 5\times 7=420$

Therefore, the LCM of $12,15$ and $21$ is $420$.

(ii) $17,23$ and $29$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $17,23$ and $29$ are as follows:

\[\Rightarrow 17=17\times 1\] 

$\Rightarrow 23=23\times 1$ and 

$\Rightarrow 29=29\times 1$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $17,23$ and $29$ is $1$.

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $17,23$ and $29$ will be

$\Rightarrow 17\times 23\times 29=11339$ 

Therefore, the LCM of $17,23$ and $29$ is $11339$.

(iii) $8,9$ and $25$ 

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $8,9$ and $25$ are as follows:

\[\Rightarrow 8=2\times 2\times 2\] 

\[\Rightarrow 9=3\times 3\] and 

$\Rightarrow 25=5\times 5$

Now, we know that HCF is the highest factor, among the common factors of two numbers. as there is no common factor.

Therefore, the HCF of $8,9$ and $25$ is $1$.

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $8,9$ and $25$ will be

$\Rightarrow 2\times 2\times 2\times 3\times 3\times 5\times 5=1800$ 

Therefore, the LCM of $8,9$ and $25$ is $1800$.

4. Given that HCF $\left( 306,657 \right)=9$, find LCM $\left( 306,657 \right)$.

Ans: We have been given the HCF of two numbers $\left( 306,657 \right)=9$.

We have to find the LCM of $\left( 306,657 \right)$.

Now, we know that $LCM\times HCF=\text{Product of two numbers}$

Substitute the values, we get

$LCM\times 9=306\times 657$

$\Rightarrow LCM=\dfrac{306\times 657}{9}$ 

$\therefore LCM=22338$ 

Therefore, the LCM of $\left( 306,657 \right)=22338$.

5. Check whether ${{6}^{n}}$ can end with the digit $0$ for any natural number $n$.

Ans: We have to check whether ${{6}^{n}}$ can end with the digit $0$ for any natural number $n$.

By divisibility rule we know that if any number ends with the digit $0$, it is divisible by $2$ and $5$.

Thus, the prime factors of ${{6}^{n}}$ is

$\Rightarrow {{6}^{n}}={{\left( 2\times 3 \right)}^{n}}$

Now, we will observe that for any value of $n$, ${{6}^{n}}$ is not divisible by $5$.

Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.

6. Explain why $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are composite numbers.

Ans: The given numbers are $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$.

We can rewrite the given numbers as

$\Rightarrow 7\times 11\times 13+13=13\times \left( 7\times 11+1 \right)$

$\Rightarrow 7\times 11\times 13+13=13\times \left( 77+1 \right)$ 

$\Rightarrow 7\times 11\times 13+13=13\times 78$

$\Rightarrow 7\times 11\times 13+13=13\times 13\times 6$

And, 

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 7\times 6\times 4\times 3\times 2\times 1+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 1008+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times 1009$

Here, we can observe that the given expressions has its factors other than $1$ and the number itself.

A composite number have factors other than $1$ and the number itself.

Therefore, $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are composite numbers.

7. There is a circular path around a sports field. Sonia takes $18$ minutes to drive one round of the field, while Ravi takes $12$ minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Ans: It can be observed that Ravi takes lesser time than Sonia for completing $1$ round of the circular path. Both are going in the same direction, they will meet again when Ravi will have completed $1$ round of that circular path with respect to Sonia. 

The total time taken for completing this $1$ round of circular path will be the LCM of time taken by Sonia and Ravi for ending $1$ round of circular path respectively i.e., LCM of $18$ minutes and $12$ minutes.

The prime factors of $12$ and $18$ are as follows:

\[\Rightarrow 12=2\times 2\times 3\]  and 

$\Rightarrow 18=2\times 3\times 3$

Now, we know that LCM is least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12$ and $18$ will be

$\Rightarrow 2\times 2\times 3\times 3=36$

Therefore, Ravi and Sonia meet again at the starting point after $36$ minutes.

Exercise 1.3 

1. Prove that $\sqrt{5}$ is irrational.

Ans: We have to prove that $\sqrt{5}$ is irrational.

We will use contradiction method to prove it.

Let $\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\sqrt{5}=\dfrac{a}{b}$

Now, squaring both sides, we get

${{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}$

\[\Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}}\] 

$\Rightarrow {{a}^{2}}=5{{b}^{2}}$ …….(1)

If ${{a}^{2}}$ is divisible by $5$ than $a$ is also divisible by $5$.

Let $a=5k$, where, $k$ is any integer.

Again squaring both sides, we get

\[\Rightarrow {{a}^{2}}={{\left( 5k \right)}^{2}}\] 

Substitute the value in eq. (1), we get

\[\Rightarrow {{\left( 5k \right)}^{2}}=5{{b}^{2}}\]

$\Rightarrow {{b}^{2}}=5{{k}^{2}}$ …..(2)

If ${{b}^{2}}$ is divisible by $5$ than $b$ is also divisible by $5$.

From, eq. (1) and (2), we can conclude that $a$ and $b$ have $5$ as a common factor.

This contradicts our assumption.

Therefore, we can say that $\sqrt{5}$ is irrational.

Hence proved.

2. Prove that $3+2\sqrt{5}$ is irrational.

Ans: We have to prove that $3+2\sqrt{5}$ is irrational.

We will use contradiction method to prove it.

Let $3+2\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $3+2\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow 2\sqrt{5}=\dfrac{a}{b}-3$ 

$\Rightarrow \sqrt{5}=\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ ……..(1)

From eq. (1) we can say that $\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $3+2\sqrt{5}$ is irrational.

Hence proved.

3. Prove that following are irrationals:

(i) $\dfrac{1}{\sqrt{2}}$ 

Ans: We have to prove that $\dfrac{1}{\sqrt{2}}$ is irrational.

We will use contradiction method to prove it.

Let $\dfrac{1}{\sqrt{2}}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{b}{a}$ ………..(1)

From eq. (1) we can say that $\dfrac{b}{a}$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $\dfrac{1}{\sqrt{2}}$ is irrational.

Hence proved.

(ii) $7\sqrt{5}$ 

Ans: We have to prove that $7\sqrt{5}$ is irrational.

We will use contradiction method to prove it.

Let $7\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $7\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow \sqrt{5}=\dfrac{a}{7b}$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{7b}$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $7\sqrt{5}$ is irrational.

Hence proved.

(iii) $6+\sqrt{2}$ 

Ans: We have to prove that $6+\sqrt{2}$ is irrational.

We will use contradiction method to prove it.

Let $6+\sqrt{2}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $6+\sqrt{2}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{a}{b}-6$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{b}-6$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $6+\sqrt{2}$ is irrational.

Hence proved.

Exercise 1.4 

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) $\dfrac{13}{3125}$

Ans: Given a rational number $\dfrac{13}{3125}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $3125$.

Then, factors of $3125$ are

$\Rightarrow 3125=5\times 5\times 5\times 5\times 5$ 

$\Rightarrow 3125={{5}^{5}}$

Here, the factors of denominator are of the form ${{5}^{m}}$.

Therefore, $\dfrac{13}{3125}$ has terminating decimal expansion.

(ii) $\dfrac{17}{8}$

Ans: Given a rational number $\dfrac{17}{8}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $8$.

Then, factors of $8$ are

$\Rightarrow 8=2\times 2\times 2$ 

$\Rightarrow 8={{2}^{3}}$

Here, the factors of denominator are of the form ${{2}^{n}}$.

Therefore, $\dfrac{17}{8}$ has terminating decimal expansion.

(iii) $\dfrac{64}{455}$

Ans: Given a rational number $\dfrac{64}{455}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $455$.

Then, factors of $455$ are

$\Rightarrow 455=5\times 7\times 13$ 

Here, the factors of denominator are not in the form ${{2}^{n}}{{5}^{m}}$. The denominator has factors other than $2$ and $5$.

Therefore, $\dfrac{64}{455}$ has non-terminating repeating decimal expansion.

(iv) $\dfrac{15}{1600}$

Ans: Given a rational number $\dfrac{15}{1600}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $1600$.

Then, factors of $1600$ are

$\Rightarrow 1600=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5$ 

$\Rightarrow 1600={{2}^{6}}\times {{5}^{5}}$

Here, the factors of denominator are of the form ${{2}^{n}}{{5}^{m}}$.

Therefore, $\dfrac{15}{1600}$ has terminating decimal expansion.

(v) $\dfrac{29}{343}$

Ans: Given a rational number $\dfrac{29}{343}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $343$.

Then, factors of $343$ are

$\Rightarrow 343=7\times 7\times 7$ 

$\Rightarrow 343={{7}^{3}}$

Here, the factors of denominator are not in the form ${{2}^{n}}{{5}^{m}}$. The denominator has factors other than $2$ and $5$.

Therefore, $\dfrac{29}{343}$ has non-terminating repeating decimal expansion.

(vi) $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$

Ans: Given a rational number $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is ${{2}^{3}}{{5}^{2}}$.

Here, the denominator is of the form ${{2}^{n}}{{5}^{m}}$.

Therefore, $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ has terminating decimal expansion.

(vii) $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$

Ans: Given a rational number $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is ${{2}^{2}}{{5}^{7}}{{7}^{5}}$.

Here, the denominator is of the form ${{2}^{n}}{{5}^{m}}$ but also has factors other than $2$ and $5$.

Therefore, $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$ has non-terminating repeating decimal expansion.

(viii) $\dfrac{6}{15}$

Ans: Given a rational number $\dfrac{6}{15}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $15$.

Then, factors of $3125$ are

$\Rightarrow 15=3\times 5$ 

But we can write the numerator of the given number as

$\dfrac{6}{15}=\dfrac{2\times 3}{3\times 5}=\dfrac{2}{5}$ 

Here, the factors of denominator are of the form ${{5}^{m}}$.

Therefore, $\dfrac{6}{15}$ has terminating decimal expansion.

(ix) $\dfrac{35}{50}$

Ans: Given a rational number $\dfrac{35}{50}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $50$.

Then, factors of $50$ are

$\Rightarrow 50=10\times 5$ 

But we can write the numerator of the given number as

$\dfrac{35}{50}=\dfrac{7\times 5}{10\times 5}=\dfrac{7}{10}$ 

$\Rightarrow 10=2\times 5$

Here, the factors of denominator are of the form ${{2}^{n}}{{5}^{m}}$.

Therefore, $\dfrac{35}{50}$ has terminating decimal expansion.

(x) $\dfrac{77}{210}$ 

Ans: Given a rational number $\dfrac{77}{210}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $210$.

Then, factors of $210$ are

$\Rightarrow 210=2\times 3\times 5\times 7$ 

Here, the denominator has factors other than $2$ and $5$.

Therefore, $\dfrac{77}{210}$ has non-terminating repeating decimal expansion.

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) $\dfrac{13}{3125}$

Ans: To find the decimal expansion of $\dfrac{13}{3125}$, we will divide the numerator of the number by denominator using long division method. We get

$3125\overset{0.00416}{\overline{\left){\begin{align}  & 13.00000 \\ & 0 \\ & \overline{130\text{    }} \\ & \text{    0} \\ & \overline{\text{13000}} \\ & \text{12500} \\ & \overline{\text{  5000  }} \\ & \text{  3125} \\ & \overline{\text{  18750  }} \\ & \underline{\text{  18750  }} \\ & \text{   0} \\ \end{align}}\right.}}$ 

Therefore, the decimal expansion of $\dfrac{13}{3125}$ is $0.00416$. 

(ii) $\dfrac{17}{8}$

Ans: To find the decimal expansion of $\dfrac{17}{8}$, we will divide the numerator of the number by denominator using long division method. We get

$8\overset{2.125}{\overline{\left){\begin{align}  & 17 \\ & 16 \\ & \overline{\text{  10    }} \\ & \text{   8} \\ & \overline{\text{    20  }} \\ & \text{    16} \\ & \overline{\text{     40  }} \\ & \text{     }\underline{\text{40}} \\ & \text{      0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{17}{8}$ is $2.125$. 

(iii) $\dfrac{15}{1600}$ 

Ans: To find the decimal expansion of $\dfrac{15}{1600}$, we will divide the numerator of the number by denominator using long division method. We get

$1600\overset{0.009375}{\overline{\left){\begin{align}  & 15.000000 \\ & 0 \\ & \overline{150\text{    }} \\  & \text{    0} \\ & \overline{\text{1500}} \\ & \text{0} \\ & \overline{\text{15000  }} \\ & \text{14400} \\ & \overline{\text{  6000  }} \\ & \underline{\text{  4800  }} \\ & \text{  12000} \\ & \underline{\text{  11200}} \\ & \text{     8000} \\& \text{     }\underline{\text{8000}} \\ & \text{      0} \\ \end{align}}\right.}}$ 

Therefore, the decimal expansion of $\dfrac{15}{1600}$ is $0.009375$. 

(iv) $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$

Ans: To find the decimal expansion of $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$, we will divide the numerator of the number by denominator using long division method. We get

$\dfrac{23}{{{2}^{3}}{{5}^{2}}}=\dfrac{23}{200}$

$200\overset{00.115}{\overline{\left){\begin{align} & 23.000 \\ & 0 \\ & \overline{23\text{    }} \\ & \text{ 0} \\  & \overline{\text{230   }} \\ & 200 \\ & \overline{\text{  300  }} \\  & \text{  200} \\ & \overline{\text{  1000  }} \\ & \underline{\text{  1000  }} \\  & \text{   0} \\ \end{align}}\right.}}$ 

Therefore, the decimal expansion of $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ is $0.115$.

(v) $\dfrac{6}{15}$

Ans: To find the decimal expansion of $\dfrac{6}{15}$, we will divide the numerator of the number by denominator using long division method. We get

$\dfrac{6}{15}=\dfrac{2\times 3}{3\times 5}=\dfrac{2}{5}$ 

$5\overset{0.4}{\overline{\left){\begin{align} & 2.0 \\ & 0 \\ & \overline{20\text{    }} \\ & \underline{20\text{    }} \\ & 0 \\ \end{align}}\right.}}$ 

Therefore, the decimal expansion of $\dfrac{6}{15}$ is $0.4$.

(vi) $\dfrac{35}{50}$

Ans: To find the decimal expansion of $\dfrac{35}{50}$, we will divide the numerator of the number by denominator using long division method. We get

$50\overset{0.7}{\overline{\left){\begin{align} & 35.0 \\ & 0 \\ & \overline{\text{350 }} \\ & \underline{350\text{ }} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{35}{50}$ is $0.7$.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form $\dfrac{p}{q}$, what can you say about the prime factors of $q$?

(i) $43.123456789$ 

Ans: Given a decimal expansion $43.123456789$.

The given number has terminating expansion, we can write the number as $\dfrac{43123456789}{1000000000}$, which is of the form $\dfrac{p}{q}$.

Therefore, the number $43.123456789$ is a rational number.

Since the number has terminating decimal expansion, the factors of $q$ must be of the form ${{2}^{n}}{{5}^{m}}$.

(ii) $0.120120012000120000......$ 

Ans: Given a decimal expansion $0.120120012000120000......$.

When we observe the given expansion we can say that the number has non-terminating and non-repeating decimal expansion. Hence we cannot express it in the form of $\dfrac{p}{q}$.

Therefore, the number is irrational.

(iii) $43.\overline{123456789}$ 

Ans: Given the decimal expansion $43.\overline{123456789}$.

The given number has non-terminating but repeating decimal expansion. So the number will be of the form $\dfrac{p}{q}$. 

Therefore, the number $43.\overline{123456789}$ is a rational number.

But the factors of denominator are not of the form ${{2}^{n}}{{5}^{m}}$. Denominator also has factors other than $2$ and $5$.

Chapter 1 - Real Numbers

You can opt for Chapter 1 - Real Numbers NCERT Solutions for Class 10 Maths PDF for upcoming Exams  and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 10 Maths

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solutions for Class 10 Maths Chapter 1 All Exercises

Important Topics under NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

The first chapter of the class 10 maths syllabus includes Real Numbers, which is an important chapter in maths covered in class 10. The chapter on Real Numbers is divided into 4 major parts. The following is a list of the important topics covered under the chapter Real Numbers. It is recommended that students read through these topics carefully to be able to learn, use, and master this chapter and be able to solve problems based on Real Numbers.

• Introduction to Real Numbers

• Euclid’s Division Lemma

• Fundamental Theorem of Arithmetic (H.C.F. and L.C.M.) 

• Revisiting Irrational Numbers

1.1 Introduction

In the introduction part of ch 1 Maths Class 10 students will be reminded of what they learned in class IX about real numbers and irrational numbers. This section gives a glimpse of what students would learn about positive numbers in the later sections of Chapter 1 Maths Class 10, i.e. Euclid’s division algorithm and the fundamental theorem of arithmetic. The Fundamental Theorem of Arithmetic is based on the fact that a composite number can be expressed as a product of prime numbers, in distinct ways. This theorem has deep and significant applications in mathematics.

1.2 Euclid’s Division Lemma

You will learn about Euclid’s division lemma and algorithm in this section of NCERT Solutions Class 10 Maths Chapter 1. A lemma is a statement that is proven and acts as a stepping stone to prove other statements. Euclid’s Division Lemma states the usual division system in mathematics in a formal way, i.e. for every pair of positive numbers x and y; there are two unique whole numbers a and b that satisfy the equation:

• X = ya + b where 0 <= b <= y and x is a dividend, y is a divisor.

• A is the quotient.

• B is the remainder. 

In other words: dividend = (quotient * divisor) + remainder.

You would also learn Euclid’s Division Algorithm in this portion of ch 1 Maths Class 10 NCERT Solutions which is based on the lemma. An algorithm is a set of well-defined steps that can procedurally solve a problem. Euclid’s Division Algorithm is used in calculating the HCF(highest common factor) of two positive integers.

1.3 The Fundamental Theorem of Arithmetic

According to this theorem, every composite number can be factorized as a product of some prime numbers. It is a unique prime factorization of natural numbers as the order of the factors does not matter. We will understand this with an example that is based on the following fundamentals:

• HCF - The highest common factor of two or more integers is the greatest integer that can exactly divide all the given integers. For example, HCF of 60 and 75 is 15.

• LCM - The Least Common Multiple of two or more integers is the smallest integer that is exactly divisible by all the given integers. For example, LCm of 2, 4, and 5 is 20.

• For two positive integers a and b; HCF(a,b) * LCM (a, b) = a * b

So, going by this theorem we can express any natural number as a multiplication of prime number, for example, 253 = 11 * 23, 4 = 2 * 2, etc.

1.4  Revisiting Irrational Numbers

In this section of NCERT Solutions for Class 10th Maths Chapter 1, you will remember the definition of Irrational numbers learned in earlier classes and then prove p is an irrational number, where p is a prime number.

If a number “n” can not be written in the form x/y, then it is called an irrational number. Here x and y are integers and n <> 0. Few examples of irrational numbers are 2,3, etc.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1

The Class 10 Maths NCERT Solutions Chapter 1 prepared by the scholars of Vedantu is one of the most reliable online resources. The key takeaways of these NCERT Solutions for Class 10 Maths ch 1 are:

• Students will get answers to all the questions in Maths Class 10 NCERT Solutions Chapter 1 and no question is left out.

• The Class 10 Maths Chapter 1 Solutions are also available for download in a PDF format which makes revisions very quick and easy during stressful exam times.

• You will find that going through Class 10 Maths Chapter 1 NCERT Solutions is effortless as they are written in a simple and easily comprehensible manner which is apt for the understanding level of class 10 students.