Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25

Very Short Answer Questions                                                                        (1 Mark)

1. 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 then 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, then we can write $a=3q+r$ for some integer $q\ge 0$  …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.    ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.    ……(3)

Squaring both sides of equation (3) we get,

${{a}^{2}}={{(3q)}^{2}}$ or ${{(3q+1)}^{2}}$ or ${{(3q+2)}^{2}}$ 

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

Taking $3$ common from LHS of equation (4) we get,

${{a}^{2}}=3\times (3{{q}^{2}})$ or $3(3{{q}^{2}}+2q)+1$ or $3(3{{q}^{2}}+4q+1)+1$.   …..(5)

Equation (5) could be written as 

${{a}^{2}}=3{{k}_{1}}$ or $3{{k}_{2}}+1$ or $(3{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $3m$ or $3m+1$.

2. Express each number as product of its prime factors.

(i). $140$ 

Ans: The product of prime factors of number $140$ is \[140=2\cdot 2\cdot 5\cdot 7={{2}^{2}}\cdot 5\cdot 7\].

(ii). $156$ 

Ans: The product of prime factors of number $156$ is $156=2\cdot 2\cdot 3\cdot 13={{2}^{2}}\cdot 3\cdot 13$.

(iii). \[\mathbf{382}5\] 

Ans: The product of prime factors of number $3825$ is $3825=3\cdot 3\cdot 5\cdot 5\cdot 17={{3}^{2}}\cdot {{5}^{2}}\cdot 17$.

(iv). \[5\mathbf{005}\] 

Ans: The product of prime factors of number $5005$ is $5005=5\cdot 7\cdot 11\cdot 13$.

(v). \[\mathbf{7429}\] 

Ans: The product of prime factors of number $7429$ is $7429=17\cdot 19\cdot 23$.

3. Given that HCF $(306,657)=9$, find LCM $(306,657)$.                  

Ans: It is given that the HCF of the two numbers $306,657$ is $9$. We have to find its LCM.

We know that, LCM$\times $HCF = Product of two numbers. Therefore,

LCM$\times $HCF $=306\times 657$ 

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

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

$\therefore LCM=22338$

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

Ans: If any number ends with the digit $0$, it should be divisible by $10$.

Let us check this for natural number $n=2$. 

Clearly, ${{6}^{2}}=36$, which is not divisible by $10$. Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.

5. Prove that $\left( 3+2\sqrt{5} \right)$ is irrational.    

Ans: We will prove this by contradiction. To solve this problem, suppose that $\left( 3+2\sqrt{5} \right)$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=3+2\sqrt{5}$     …..(1)

Subtracting $3$ from both sides of the equation (1) and simplifying it further we get,

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

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

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

Since we know that $\sqrt{5}$ is irrational. Therefore, from (2) $\dfrac{a-3b}{2b}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $(3+2\sqrt{5})$ cannot be rational.

Hence, $(3+2\sqrt{5})$ is irrational.

6. The number $7\times 11\times 13\times 15+15$ is a

1. Composite Number

2. Whole Number

3. Prime Number

4. None of these

Ans: (a) and (b) both

7. For what least value of $'n'$ a natural number, ${{(24)}^{n}}$ is divisible by $8$?

1.  $0$ 

2. $-1$ 

3. $1$ 

4. No value of $'n'$ is possible

Ans: (c) 1

8. The sum of a rational and an irrational number is

1. Rational

2. Irrational

3. Both (a) & (b)

4. Either (a) & (b)

Ans: (b) Irrational

9. HCF of two numbers is $113,$ their LCM is $56952$. If one number is $904$, then other number is:

1. $7719$

2. $7119$

3. $7791$

4. $7911$

Ans: (b) $7119$

10. A lemma is an axiom used for providing

1. other statement

2. no statement

3. contradictory statement

4. none of these

Ans: (a) other statement

11. If HCF of two numbers is $1$, the two numbers are called relatively ______ or _______.

1. prime, co-prime

2. composite, prime

3. Both (a) and (b)

4. None of these

Ans: (a) prime, co-prime

12. The number $2.\overline{35}$ is

1. a terminating decimal number

2. a rational number 

3. an irrational number

4. Both (a) and (b)

Ans: (b) a rational number

13. The number $2.13113111311113......$ is        

1. a rational number

2. a non-terminating decimal number

3. an irrational number

4. Both (a) & (c)

Ans: (b) a non-terminating decimal number

14. The smallest composite number is      

1.  $1$

2. $2$

3. $3$

4. $4$

Ans: (d) $4$ 

15. The number $1.23\overline{48}$ is

1. an integer

2. an irrational number

3. a rational number

4. None if these

Ans: (c) a rational number

16. The number $\pi $ is

1. a rational number

2. an irrational number

3. Both (a) and (b)

4. neither rational nor irrational

Ans: (b) an irrational number

17. The number $(2+\sqrt{5})$ is

1. a rational number

2. an irrational number

3. an integer

4. not real number

Ans: (b) an irrational number

Short Answer Questions                                                                               (2 Marks)

1. 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 odd positive integer, then by Euclid’s algorithm we can write $a=6q+r$ for some integer $q\ge 0$.        …..(1)

Clearly, in the expression (1) we are dividing $a$ by $6$ with quotient $q$ and remainder $r$, 

$r=1,3,5$ because $0\le r  < 6$ and $a$ is odd.    …..(2)

Therefore from (1) and (2) we can get, 

$a=6q+1$ or $6q+3$ or $6q+5$ 

And therefore, any odd integer can be expressed in the form $6q+1,6q+3,$ or $6q+5$.

2. An army contingent of $616$ members are to march behind a 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: To solve this problem, we have to find the HCF of two numbers $(616,32)$.

To find the HCF, we can use Euclid’s algorithm which states that if there are any two integers \[a\] and \[b\], then there exists \[q\] and \[r\] such that it satisfies the condition \[a=bq+r\] where \[0\le r  < b\].

For $a=616$, $616=32\times 19+8$       …..(1)

For $a=32$, $32=8\times 4+0$       …..(2) 

Therefore, from (1) and (2), the HCF $(616,32)$ is $8$ and hence they can march in $8$ columns each.

3. 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, then by Euclid’s algorithm we can write $a=3q+r,$ for some integer $q\ge 0$.    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.     …..(2)

Therefore from (1) and (2) we can get, 

$a=3q$ or $3q+1$ or $3q+2$    …..(3)

Cubing both sides of equation (3) we get,

Case 1: When $a=3q$,

${{a}^{3}}=27{{q}^{3}}$

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

$\Rightarrow {{a}^{3}}=9m$ 

Where $m$ is an integer such that $m=3{{q}^{3}}$

Case 2: When $a=3q+1$

${{a}^{3}}={{(3q+1)}^{3}}$

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

$\begin{align} & \Rightarrow {{a}^{3}}=9\left( {{q}^{3}}+{{q}^{2}}+q \right)+1 \\ & \Rightarrow {{a}^{3}}=9m+1 \\ \end{align}$

Where $m$ is an integer such that $m=(3{{q}^{3}}+3{{q}^{2}}+q)$

Case 3: When $a=3q+2$

${{a}^{3}}={{(3q+2)}^{3}}$

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

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

$\Rightarrow {{a}^{3}}=9m+8$ 

Where $m$ is an integer such that $m=(3{{q}^{3}}+6{{q}_{2}}+4q)$

Therefore, from case 1,2 and 3 we conclude that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$.

4. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=$product of two numbers.                       

(i). $26\text{ and }91$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

$26=2\times 13$ and $91=7\times 13$.       …..(1) 

Since $13$ is common in the prime factorization of both of the numbers we get $\text{HCF}=13$.      …..(2)

From (1), $\text{LCM = 2}\times \text{7}\times \text{13=182}$     …..(3) 

Product of two numbers, $26\times 91=2366$   …..(4)

From (2) and (3), $\text{HCF}\times \text{LCM =13}\times \text{182=2366}$.    …..(5) 

Hence from (4) and (5), product of two numbers \[=HCF\times LCM\]. 

(ii). $510\text{ and }92$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

$510=2\times 3\times 5\times 17$ and $92=2\times 2\times 23$.   …..(1) 

Since $2$ is common in the prime factorization of both of the numbers we get $HCF=2$.         …..(2)

From (1),$LCM=2\times 2\times 3\times 5\times 17\times 23=23460$    …..(3) 

Product of two numbers, $510\times 92=46920$    …..(4)

From (2) and (3),$HCF\times LCM=2\times 23460=46920$.  …..(5) 

Hence from (4) and (5), product of two numbers \[=HCF\times LCM\].

(iii). $336\text{ and }54$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

$336=2\times 2\times 2\times 2\times 3\times 7={{2}^{4}}\times 3\times 7$and$54=2\times 3\times 3\times 3=2\times {{3}^{3}}$.     …..(1) 

Since $2,3$ are common in the prime factorization of the numbers, $HCF=2\times 3=6$.      .....(2)

From (1),$LCM={{2}^{4}}\times {{3}^{3}}\times 7=3024$      …..(3) 

Product of two numbers $=336\times 54=18144$      …..(4)

From (2) and (3), $HCF\times LCM=6\times 3024=18144$.      …..(5) 

Hence from (4) and (5), product of two numbers \[=HCF\times LCM\].

5. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i). $12,15\text{ and }221$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

$12={{2}^{2}}\times 3$, $15=3\times 5$ and$21=3\times 7$.    …..(1) 

Since $3$ is common in the prime factorization of all the numbers we get $HCF=3$.                                   

From (1), $LCM={{2}^{2}}\times 3\times 5\times 7=420$. 

(ii). $17,23\text{ and }29$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

$17=1\times 17$, $23=1\times 23$ and $29=1\times 29$.   …..(1) 

Since $1$ is common in the prime factorization of all the numbers we get $HCF=1$.                                   

From (1), $LCM=17\times 23\times 29=11339$. 

(iii). $8,9\text{ and }25$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

$8=2\times 2\times 2={{2}^{3}}$, $9=3\times 3={{3}^{2}}$ and $25=5\times 5={{5}^{2}}$.    …..(1) 

Since $1$ is common in the prime factorization of all the numbers we get $HCF=1$.                                   

From (1), $LCM={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}=1800$. 

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: Numbers are of two types – prime and composite.                  

Prime numbers can be divided by $1$ and only itself, whereas composite numbers have factors other than $1$ and itself.

It can be observed that $7\times 11\times 13+13=13\times (7\times 11+1)$

The given expression has $13$ as one of the factors. Therefore, it is a composite number.

Also, $7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times (7\times 6\times 4\times 3\times 2\times 1+1)$.

The given expression has $5$ as one of the factors. Therefore, it is a composite number.

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 one round of the circular path. As they are going in the same direction, they will meet again at the time that will be the $LCM$ of time taken by Sonia and Ravi for completing $1$ round of circular path respectively i.e., $LCM$ of $18$ minutes and $12$ minutes.

Let us now calculate $LCM$ of $18$ minutes and $12$ minutes.

$18=2\times 3\times 3$ and $12=2\times 2\times 3$    …..(1)

$LCM\left( 12\text{,}18 \right)=2\times 2\times 3\times 3=36$

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

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

Ans: We will prove this by contradiction.

Let us suppose that $\sqrt{5}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow b\sqrt{5}=a$   …..(1)

Squaring both sides of (1), we get $5b={{a}^{2}}$    …..(2)

From (2) we can conclude that $5$ is factor of ${{a}^{2}}$, this means that $5$ is also factor of $a$.     …..(3)

From (3) we can write $a=5c$ for some integer $c$.   …..(4)

Substituting value of (4) in (2) we get,

$\begin{align} & 5{{b}^{2}}=25{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=5{{c}^{2}} \\ \end{align}$

It means that $5$ is factor of ${{b}^{2}}$, this means that $5$ is also factor of $b$.   …..(5)

From $(3)\text{,}(5)$, we can say that $5$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $\sqrt{5}$ is irrational.

9. Write down the decimal expansion of those rational numbers in Question 1 which have terminated decimal expansion.        

Ans: 

(i).The decimal expansion of $\dfrac{13}{3125}$ is $\dfrac{13}{{{5}^{5}}}=\dfrac{13\times {{2}^{5}}}{{{5}^{5}}\times {{2}^{5}}}=\dfrac{13\times {{2}^{5}}}{{{10}^{5}}}=\dfrac{416}{{{10}^{5}}}=0.00416$ 

(ii) The decimal expansion of $\dfrac{17}{8}$ is $\dfrac{17}{{{2}^{3}}}=\dfrac{17\times {{5}^{3}}}{{{2}^{3}}\times {{5}^{3}}}=\dfrac{17\times {{5}^{3}}}{{{10}^{3}}}=\dfrac{2125}{{{10}^{3}}}=2.215$ 

(iii) The decimal expansion of $\dfrac{15}{1600}$ is $\dfrac{15}{{{2}^{6}}\times {{5}^{2}}}=\dfrac{15\times {{5}^{4}}}{2\times {{5}^{2}}\times {{5}^{4}}}=\dfrac{15\times {{5}^{4}}}{{{10}^{6}}}=\dfrac{9375}{{{10}^{6}}}=0.009375$ 

(iv) The decimal expansion of $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$ is $\dfrac{23\times {{5}^{1}}}{{{2}^{3}}\times {{5}^{2}}\times {{5}^{1}}}=\dfrac{23\times {{5}^{1}}}{{{10}^{3}}}=\dfrac{115}{{{10}^{3}}}=0.115$ 

(v) The decimal expansion of $\dfrac{6}{15}$ is $\dfrac{2}{5}=\dfrac{2\times 2}{5\times 2}=\dfrac{4}{10}=0.4$ 

(vi) The decimal expansion of $\dfrac{35}{50}$ is $\dfrac{7}{10}=0.7$

10. 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 form $\dfrac{p}{q}$, what can you say about the prime factors of $q$?                                                                               

(i). Number - $43.123456789$

Ans: Since the decimal expansion is terminating so it is a rational number and therefore it can be expressed in $\dfrac{p}{q}$ form.

 \[43.123456789=\dfrac{43123456789}{{{10}^{9}}}\]

Hence, the factors of $q$ are of the form ${{2}^{n}}\times {{5}^{m}}$ where $n\text{ and }m$ are non-negative integers.

(ii). Number - $0.1201120012000120000...$

Ans: Since the decimal expansion is neither terminating nor non-terminating repeating so it is an irrational number.

(iii). Number - $43.\overline{123456789}$

Ans: Since the decimal expansion is non-terminating repeating so it is a rational number and therefore it can be expressed in $\dfrac{p}{q}$ form.

\[43.\overline{123456789}=43\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot ...\]

Hence, the factors of $q$ are of the form ${{2}^{n}}\times {{5}^{m}}$ where $n\text{ and }m$ are non-negative integers.

11. Show that every positive integer is of the form $2q$ and that every positive odd integer is of the form $2q+1$ for some integer $q$.                               

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=2q+r,$ for some integer $q\ge 0$.   …..(1)

Clearly, in the expression (1) we are dividing $a$ by $2$ with quotient $q$ and remainder $r$, $r=0,1$ because $0\le r  < 2$.  …..(2)

Therefore from (1) and (2) we can get, 

$a=2q$ or $2q+1$   …..(3)

If $a=2q$ (which is even)

If $a=2q+1$ (which is odd)

So, every positive even integer is of the form $2q$ and odd integer is of the form $2q+1$.

12. Show that any number of the form ${{4}^{n}}$, $n\in N$ can never end with the digit $0$.  

Ans: The prime factorization of $10$ is $2\times 5$. Therefore, for a number to end with $0$ it should have both 2 and 5 as a factor.

For the given number of the form ${{4}^{n}}={{[{{2}^{2}}]}^{n}}={{2}^{2n}}$, we can observe that it has only $2$ as a factor and not $5$. So ${{4}^{n}},n\in N$ can never end with the digit $0.$

13. Use Euclid’s Division Algorithm to find the $HCF$ of $4052\text{ and }12576$.

Ans: Step 1: Since$12576 >4052$ , we will apply Euclid’s division lemma such that $12576=4052\times 3+420$   …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[4052\] and \[420\], such that $4052=420\times 9+272$   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $420$ and $272$, such that $420=272\times 1+148$.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $148$ and $272$, such that $272=148\times 1+124$.   …..(4) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to $148$ and $124$, such that $148=124\times 1+24$.  …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to $24$ and $124$, such that $124=24\times 5+4$.  …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to $24$ and $4$, such that $24=4\times 6+0$.  …..(7) 

Now the remainder in equation (7) is zero. Therefore, $HCF$ of $12576$ and $4052$ is $'4'.$

14. Given that $HCF$ of two numbers is $23$ and there $LCM$ is $1449$. If one of the numbers is $161$, find the other.                             

Ans: It is given that the HCF of the two numbers is $23$ and their LCM is $1449$. We have to find the other number if one of the numbers is $161$. Let the other number be $a$. 

We know that, LCM$\times $HCF = Product of two numbers. Therefore,

$1449\times 23=a\times 161$ 

$\Rightarrow a=\dfrac{1449\times 23}{161}$

$\therefore a=207$

15. Show that every positive odd integer is of the form $(4q+1)$ or $(4q+3)$ for some integer $q$.

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=4q+r,$ for some integer $q\ge 0$. …..(1)

Clearly, in the expression (1) we are dividing $a$ by $4$ with quotient $q$ and remainder $r$, $r=0,1,2,3$ because $0\le r  < 4$.     …..(2)

Therefore from (1) and (2) we can get, 

$a=4q$ or $4q+1$ or $4q+2$ or $4q+3$     …..(3)

If $a=4q=2\left( 2q \right)$ (which is even)

If $a=4q+1=2\left( 2q \right)+1$ (which is odd)

If $a=4q+2=2(2q+1)$ (which is even)

If $a=4q+3=2(2q+1)+1$ (which is odd) 

So, every positive odd integer is of the form $(4q+1)$ or $(4q+3)$.

16. Show that any number of the form ${{6}^{x}},x\in N$ can never end with the digit $0.$       

Ans: The prime factorization of $10$ is $2\times 5$. Therefore, for a number to end with $0$ it should have both 2 and 5 as a factor.

For the given number of the form ${{6}^{n}}={{(2\times 3)}^{n}}={{2}^{n}}\times 3{}^{n}$., we can observe that it has only $2$ as a factor and not $5$. So ${{6}^{x}},x\in N$ can never end with the digit $0.$

17. Find $HCF$ and $LCM$ of $18$ and $24$ by the prime factorization method.   

Ans: The prime factorization of $18$ is $18=2\times 3\times 3=2\times {{3}^{2}}$  …..(1)

The prime factorization of $24$ is $24=2\times 2\times 2\times 3={{2}^{3}}\times 3$   …..(2)

From (1) and (2) we can see that $18,24$ has prime factors $2,3$ in common. 

$\therefore HCF=2\times 3=6$

From (1) and (2) we can see that $3$ appears twice in prime factorization of $18$ and $2$ appears thrice in prime factorization of $24$.

 $\therefore LCM=\left( {{3}^{2}} \right)\times \left( {{2}^{3}} \right)=72$

18. The $HCF$ of two numbers is $23$ and their $LCM$ is $1449$. If one of the numbers is $161,$ find the other.

Ans: It is given that the HCF of the two numbers is $23$ and their LCM is $1449$. We have to find the other number if one of the numbers is $161$. Let the other number be $a$. 

We know that, LCM$\times $HCF = Product of two numbers. Therefore,

$1449\times 23=a\times 161$ 

$\Rightarrow a=\dfrac{1449\times 23}{161}$

$\therefore a=207$

19. Prove that the square of any positive integer of the form $5g+1$ is of the same form.

Ans: Let $a$ be any positive integer, of the form $a=5g+1$      …..(1)

Squaring both sides of equation (1) we get,

\[{{a}^{2}}={{(5g+1)}^{2}}\]

\[\Rightarrow {{a}^{2}}=25{{g}^{2}}+10g+1\] 

\[\Rightarrow {{a}^{2}}=5(5{{g}^{2}}+2g)+1\]

\[\Rightarrow {{a}^{2}}=5m+1\] 

Hence, the square of any positive integer of the form $5g+1$ is of the same form.

20. Use Euclid’s Division Algorithm to find the $HCF$ of $4052$ and $12576$.      

Ans: Step 1: Since$12576 >4052$ , we will apply Euclid’s division lemma such that $12576=4052\times 3+420$    …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[4052\] and \[420\], such that $4052=420\times 9+272$   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $420$ and $272$, such that $420=272\times 1+148$.    …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $148$ and $272$, such that $272=148\times 1+124$.   …..(4) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to $148$ and $124$, such that $148=124\times 1+24$.     …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to $24$ and $124$, such that $124=24\times 5+4$.    …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to $24$ and $4$, such that $24=4\times 6+0$.   …..(7) 

Now the remainder equation (7) is zero. Therefore, $HCF$ of $12576$ and $4052$ is $'4'.$

21. Find the largest number which divides $245$ and $1029$ leaving remainder $5$ in each case.

Ans. Since we want the remainder to be $5$, therefore the required number is the $HCF$ of $(245-5)$ and $(1029-5)$ i.e., $240$ and $1024$.

Step 1: Since$1024 >240$ , we will apply Euclid’s division lemma such that $1024=240\times 4+64$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[240\] and \[64\], such that $240=64\times 3+48$  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $64$ and $48$, such that$64=48\times 1+16$.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $48$ and $16$, such that$48=16\times 3+0$.    …..(4) 

Now the remainder in equation (4) is zero. Therefore, $HCF$ of $1024$ and $240$ is $'16'$.

The largest number which divides $245$ and $1029$ leaving remainder $5$ in each case is $16$. 

22. A shopkeeper has $120$ litres of petrol, $180$ litres of diesel and $240$ litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin?                                                                         

Ans: The required greatest capacity is the $HCF$ of $120,180$ and $240$. First let us find the HCF of $240,180$ by Euclid’s division algorithm.

Step 1: Since$240 >180$ , we will apply Euclid’s division lemma such that $240=180\times 1+60$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[180\] and \[60\], such that $180=60\times 3+0$     …..(2)

Now the remainder in equation (2) is zero. Therefore, $HCF$ of $180$ and $240$ is $60$.

Let us now find the HCF of $60,120$. Clearly, $120=60\times 2+0$.

\[\therefore HCF\] of $120,180$ and $240$ is $60$ and hence the required capacity is $60$ litres.

Short Answer Questions                                                                               (3 Marks)

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

(i). $135$ and $225$

Ans: Step 1: Since $225 >135$, we will apply Euclid’s division lemma such that $225=135\times 1+90$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[135\] and \[90\], such that $135=90\times 1+45$ …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $90$ and $45$, such that $90=2\times 45+0$.   …..(3) 

Now the remainder in equation (3) is zero. Therefore, the $HCF$ of $135$ and $225$ is $45$.

(ii). $196$ and $38220$

Ans: Since $38220 >196$, we will apply Euclid’s division lemma such that $38220=196\times 195+0$                     …..(1)

Since, the remainder in (1) is zero. Therefore, the $HCF$ of $38220$ and $196$ is $196$.

(iii). $867$ and $255$

Ans: Step 1: Since $867 >255,$, we will apply Euclid’s division lemma such that $867=255\times 3+102$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to $255$ and $102$, such that $255=102\times 2+51$   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $102$ and $51$, such that $102=51\times 2+0$.   …..(3) 

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

2. Find the greatest number of $6$ digits exactly divisible by $24,15$ and $36.$

Ans: To find the greatest number of $6$ digits exactly divisible by $24,15$ and $36$, first we have to find the LCM of $24,15$ and $36$.

The prime factorization of $24,15$ and $36$ are:

$24=2\times 2\times 2\times 3$    …..(1)

$15=5\times 3$    …..(2)

$36=2\times 2\times 3\times 3$   …..(3)

Hence, from (1), (2) and (3), $LCM\left( 15,24,36 \right)=2\times 2\times 2\times 5\times 3\times 3=360$.  .….(4)

Now, we know that the greatest number of $6$ digits is $999999.$  …..(5)

From (4) and (5), using Euclid’s division lemma we get, $999999=360\times 2777+279$.

$\Rightarrow 999999-279=360\times 2777$ 

$\Rightarrow 999720=360\times 2777$

Hence, the greatest number of $6$ digits exactly divisible by $24,15$ and $36$ is $999720$.  

3. Prove that the square of any positive integer is of the form $4m$ or $4m+1$ for some integer $m$.

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=4q+r,$ for some integer $q\ge 0$.      …..(1)

Clearly, in the expression (1) we are dividing $a$ by $4$ with quotient $q$ and remainder $r$, $r=0,1,2,3$ because $0\le r  < 4$.   …..(2)

Therefore from (1) and (2) we can get, 

$a=4q$ or $4q+1$ or $4q+2$ or $4q+3$   …..(3)

Case 1: If $a=4q$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}} \right)$ …..(4)

Equation (4) could be written as \[{{a}^{2}}=4m\] for some integer $m$ ….. (5)

Case 2: If $a=4q+1$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+8q+1$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}}+2q \right)+1$ …..(6)

Equation (6) could be written as \[{{a}^{2}}=4m+1\] for some integer $m$ ….. (7)

Case 3: If $a=4q+2$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+16q+4$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}}+4q+1 \right)$ …..(8)

Equation (8) could be written as \[{{a}^{2}}=4m\] for some integer $m$ ….. (9)

Case 4: If $a=4q+3$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+24q+9$

${{a}^{2}}=4\left( 4{{q}^{2}}+6q+2 \right)+1$ …..(10)

Equation (10) could be written as \[{{a}^{2}}=4m+1\] for some integer $m$ ….. (11)

Therefore, from equations (5), (7), (9) and (11) it can be said that the square of any positive integer is either of the form $4m$ or $4m+1$ for some integer $m$.

4. There are $144$ cartons of coke can and $90$ cartons of Pepsi can to be stacked in a canteen. If each stack is of the same height and is to contain cartoons of the same drink. What would be the greater number of cartons each stack would have?     

Ans: Given $144$ cartons of coke can and $90$ cartons of Pepsi can are to be stacked in a canteen. For stack to be of the same height and is to contain cartoons of the same drink, we have to find the $HCF$ of $144$ and $90$.

Using Euclid’s division algorithm,

Step 1: Since $144 >90,$, we will apply Euclid’s division lemma such that $144=90\times 1+54$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to $90$ and $54$, such that $90=54\times 1+36$  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $54$ and $36$, such that $54=36\times 1+18$   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $18$ and $36$, such that$36=18\times 2+0$.      …..(4)

Now the remainder in (4) is zero. Therefore, the $HCF$ of $144$ and $90$ is $18$.

So, greatest number of cartoons is $18.$

5. Prove that product of three consecutive positive integers is divisible by $6$.   

Ans: Let us consider three consecutive positive integers $x,(x+1)$ and $(x+2)$.

Using Euclid’s lemma, let us write $x=6q+r$   …..(1)

Clearly, in the expression (1) we are dividing $a$ by $6$ with quotient $q$ and remainder $r$, $r=0,1,2,3,4,5$ because $0\le r  < 6$.     …..(2)

Therefore from (1) and (2) we can get,                                                                                       

$x=6q,6q+1,6q+2,6q+3,6q+4,6q+5$

Case 1: If $x=6q$ then product 

$x(x+1)(x+2)=6q(6q+1)(6q+2)$, which is divisible by $6$.   …..(3)

Case 2: If $x=6q+1$ then product 

$x(x+1)(x+2)=(6q+1)(6q+2)(6q+3)$

$\Rightarrow x(x+1)(x+2)=2(3q+1).3(2q+1)(6q+1)$ 

$\Rightarrow x(x+1)(x+2)=6(3q+1).(2q+1)(6q+1)$ 

 which is divisible by $6$.    …..(4)

Case 3: If $x=6q+2$ then product 

$x(x+1)(x+2)=(6q+2)(6q+3)(6q+4)$

$\Rightarrow x(x+1)(x+2)=3(2q+1).2(3q+1)(6q+4)$ 

$\Rightarrow x(x+1)(x+2)=6(2q+1)(3q+1)(6q+4)$ 

 which is divisible by $6$.  …..(5)

Case 4: If $x=6q+3$ then product 

$x(x+1)(x+2)=(6q+3)(6q+4)(6q+5)$

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

 which is divisible by $6$.   …..(6)

Case 5: If $x=6q+4$ then product 

$x(x+1)(x+2)=(6q+4)(6q+5)(6q+6)$

$\Rightarrow x(x+1)(x+2)=6(6q+4)(6q+5)(q+1)$ 

 which is divisible by $6$.     …..(7)

Case 6: If $x=6q+5$ then product 

$x(x+1)(x+2)=(6q+5)(6q+6)(6q+7)$

$\Rightarrow x(x+1)(x+2)=6(6q+5)(q+1)(6q+7)$ 

 which is divisible by $6$.   …..(8)

From equations (3) to (8) we can conclude that the product of three consecutive positive integers is divisible by $6$.

6. Prove that $\left( 3-\sqrt{5} \right)$ is an irrational number.

Ans: We will prove this by contradiction. To solve this problem, suppose that $\left( 3-\sqrt{5} \right)$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=3-\sqrt{5}$      …..(1)

Subtracting $3$ from both sides of the equation (1) and simplifying it further we get,

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

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

Since we know that $\sqrt{5}$ is irrational. Therefore, from (2) $\dfrac{3b-a}{b}$should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $\left( 3-\sqrt{5} \right)$ cannot be rational.

7. Prove that if $x$ and $y$ are odd positive integers, then ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by $4$.

Ans: Let $x=2p+1$ and $y=2q+1$ be two odd positive integers $\forall p,q\in \mathbb{N}$.       …..(1)

Squaring $x,y$ from (1) and adding them we get, 

\[{{x}^{2}}+{{y}^{2}}={{(2p+1)}^{2}}+{{(2q+1)}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\left( 4{{p}^{2}}+4p+1 \right)+\left( 4{{q}^{2}}+4q+1 \right)\]

$\Rightarrow {{x}^{2}}+{{y}^{2}}=2(2{{p}^{2}}+2{{q}^{2}}+2p+2q+1)$

$\Rightarrow {{x}^{2}}+{{y}^{2}}=2m$    …..(2)

Where \[m=(2{{p}^{2}}+2{{q}^{2}}+2p+2q+1)\] 

From (2), it is clear that ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by $4$.

8. Show that one and only one out of $n,(n+2)$ or $(n+4)$ is divisible by $3$, where $n\in N.$

Ans: Let $n$ be a number such that $n=3q+r$.   …..(1)

Clearly, in the expression (1) we are dividing $n$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.    …..(2)

Therefore from (1) and (2) we can get, $n=3q,3q+1,3q+2$

Case 1: If $n=3q$ then, numbers are $3q,(3q+2)(3q+4)$ out of these $3q$ is divisible by $3$.     …..(1)

Case 2: If $n=3q+1$ then, numbers are $(3q+1),(3q+3),(3q+5)$ out of these $(3q+3)$ is divisible by $3$.     …..(2)

Case 3: If $n=3q+2$ then, numbers are $(3q+2),(3q+4),(3q+6)$ out of these $(3q+6)$ is divisible by $3$.     …..(3)

$\therefore $ from (1), (2) and (3) we can conclude that out of $n,(n+2)$ and $(n+4)$ only one is divisible by $3$.

9. Use Euclid’s Division Lemma to show that the square of any positive integer of the form $3m$ or $(3m+1)$ for some integer $q$.

Ans: Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.     ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.     ……(3)

Squaring both sides of equation (3) we get,

${{a}^{2}}={{(3q)}^{2}}$ or ${{(3q+1)}^{2}}$ or ${{(3q+2)}^{2}}$ 

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

Taking $3$ common from LHS of equation (4) we get,

${{a}^{2}}=3\times (3{{q}^{2}})$ or $3(3{{q}^{2}}+2q)+1$ or $3(3{{q}^{2}}+4q+1)+1$.    …..(5)

Equation (5) could be written as 

${{a}^{2}}=3{{k}_{1}}$ or $3{{k}_{2}}+1$ or $(3{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $3m$ or $3m+1$.

10. Prove that if $\sqrt{n}$ is not a rational number, if $n$ is not a perfect square. 

Ans: We will prove this by contradiction.

Let us suppose that $\sqrt{n}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\sqrt{n}=\dfrac{a}{b}$

$\Rightarrow b\sqrt{n}=a$   …..(1)

Squaring both sides of (1), we get $n{{b}^{2}}={{a}^{2}}$   …..(2)

From (2) we can conclude that $n$ is factor of ${{a}^{2}}$, this means that $n$ is also factor of $a$.    …..(3)

From (3) we can write $a=nc$ for some integer $c$.   …..(4)

Substituting value of (4) in (2) we get,

$\begin{align} & n{{b}^{2}}={{n}^{2}}{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=n{{c}^{2}} \\ \end{align}$

It means that $n$ is factor of ${{b}^{2}}$, this means that $n$ is also factor of $b$.    …..(5)

From $(3)\text{,}(5)$, we can say that $n$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $\sqrt{n}$ is irrational.

11. Prove that the difference and quotient of $(3+2\sqrt{3})$ and $(3-2\sqrt{3})$ are irrational. 

Ans: Difference of $(3+2\sqrt{3})$ and $(3-2\sqrt{3})$ is $\left( 3+2\sqrt{3} \right)-\left( 3-2\sqrt{3} \right)=4\sqrt{3}$.

We will prove this by contradiction.

Let us suppose that $4\sqrt{3}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $4\sqrt{3}=\dfrac{a}{b}$

$\Rightarrow 4b\sqrt{3}=a$   …..(1)

Squaring both sides of (1), we get $48{{b}^{2}}={{a}^{2}}$  …..(2)

From (2) we can conclude that $48$ is factor of ${{a}^{2}}$, this means that $48$ is also factor of $a$.   …..(3)

From (3) we can write $a=48c$ for some integer $c$.   …..(4)

Substituting value of (4) in (2) we get,

$\begin{align} & 4\sqrt{3}{{b}^{2}}=48{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=4\sqrt{3}{{c}^{2}} \\ \end{align}$

It means that $48$ is factor of ${{b}^{2}}$, this means that $48$ is also factor of $b$.   …..(5)

From $(3)\text{,}(5)$, we can say that $48$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $4\sqrt{3}$ is irrational.

Now, the quotient of $\left( 3+2\sqrt{3} \right),\left( 3-2\sqrt{3} \right)$ is

$\dfrac{3+2\sqrt{3}}{3-2\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3-2\sqrt{3}}\times \dfrac{3+2\sqrt{3}}{3+2\sqrt{3}}$

\[\Rightarrow \dfrac{3+2\sqrt{3}}{3-2\sqrt{3}}=-\left( 5+4\sqrt{3} \right)\] 

Similarly, using the approach used above, we can prove that \[-\left( 5+4\sqrt{3} \right)\] is also irrational. 

12. Show that $({{n}^{2}}-1)$ is divisible by $8$, if $n$ is an odd positive integer.

Ans: Let $n=4q+1$ be an odd positive integer.   …..(1)

Then from (1),

${{n}^{2}}-1={{(4q+1)}^{2}}-1$

$\Rightarrow {{n}^{2}}-1=16{{q}^{2}}+8q$    …..(2)

Taking $8$ common from RHS of equation (2) we get,

${{n}^{2}}-1=8\left( 2{{q}^{2}}+q \right)$ 

\[\Rightarrow {{n}^{2}}-1=8m\] where $m=2{{q}^{2}}+q$.   …..(3)

Hence, from (3) we can conclude that $({{n}^{2}}-1)$ is divisible by $8$, if $n$ is an odd positive integer.

13. Use Euclid Division Lemma to show that cube of any positive integer is either of the form $9m,(9m+1)$ or $(9m+8)$.

Ans: Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r <3$.    ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.    ……(3)

Cubing both sides of equation (3) we get,

${{a}^{3}}={{(3q)}^{3}}$ or ${{(3q+1)}^{3}}$ or ${{(3q+2)}^{3}}$ 

$\Rightarrow {{a}^{3}}=27{{q}^{3}}$ or $27{{q}^{3}}+27{{q}^{2}}+9q+1$ or $27{{q}^{3}}+54{{q}^{2}}+36q+8$    …..(4)

Taking $9$ common from RHS of equation (4) we get,

${{a}^{3}}=9\left( 3{{q}^{3}} \right)$ or $9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$ or $9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$    …..(5)

Equation (5) could be written as 

${{a}^{3}}=9{{k}_{1}}$ or $9{{k}_{2}}+1$ or $(9{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $9m,9m+1$ or $3m+8$.

Long Answer Questions                                                                                (4 Marks)

1. Prove that the following are irrationals.       

(i). $\dfrac{1}{\sqrt{2}}$

Ans: (i) We will prove this by contradiction. Let us suppose that $\dfrac{1}{\sqrt{2}}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

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

Since we know that $\sqrt{2}$ is irrational. Therefore, from (1) $\dfrac{b}{a}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $\dfrac{1}{\sqrt{2}}$ cannot be rational.

(ii). $7\sqrt{5}$

Ans: We will prove this by contradiction. Let us suppose that $7\sqrt{5}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $7\sqrt{5}=\dfrac{a}{b}$  

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

Since we know that $\sqrt{5}$ is irrational. Therefore, from (1) $\dfrac{a}{7b}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $7\sqrt{5}$ cannot be rational.

(iii). $6+\sqrt{2}$

Ans: We will prove this by contradiction. To solve this problem, suppose that $6+\sqrt{2}$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=6+\sqrt{2}$        …..(1)

Subtracting $6$ from both sides of the equation (1) and simplifying it further we get,

$\dfrac{a}{b}-6=\sqrt{2}$

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

Since we know that $\sqrt{2}$ is irrational. Therefore, from (2) $\dfrac{a-6b}{b}$should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $6+\sqrt{2}$ cannot be rational.

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

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{13}{3125}$, the prime factorization of $q$ is $q=3125=5\times 5\times 5\times 5\times 5={{5}^{5}}$.                       …..(2)

From (2) we can see that the denominator $q=3125$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=5$ and $n=0$.

Therefore, it follows from theorem (1), the rational number $\dfrac{13}{3125}$ has a terminating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{17}{8}$, the prime factorization of $q$ is 

$q=8=2\times 2\times 2={{2}^{3}}$.    …..(2)

From (2) we can see that the denominator $q=8$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where \[m=0\] and $n=3$.

Therefore, it follows from theorem (1), the rational number $\dfrac{17}{8}$ has a terminating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{64}{455}$, the prime factorization of $q$ is 

$q=455=5\times 91$.   …..(2)

From (2) we can see that the denominator $q=455$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{64}{455}$ has a non-terminating repeating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{15}{1600}$, the prime factorization of $q$ is 

$q=320=2\times 2\times 2\times 2\times 2\times 2\times 5={{2}^{6}}\times 5$.    …..(2)

From (2) we can see that the denominator $q=8$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=6$.

Therefore, it follows from theorem (1), the rational number $\dfrac{15}{1600}$ has a terminating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{29}{343}$, the prime factorization of $q$ is 

$q=343=7\times 7\times 7$.   …..(2)

From (2) we can see that the denominator $q=343$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{29}{343}$ has a non-terminating repeating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$, the prime factorization of $q$ is 

$q={{2}^{3}}\times {{5}^{2}}$.   …..(2)

From (2) we can see that the denominator $q={{2}^{3}}\times {{5}^{2}}$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=2$ and $n=3$.

Therefore, it follows from theorem (1), the rational number $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$ has a terminating decimal expansion.

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

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}$, the prime factorization of $q$ is 

$q={{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}$.    …..(2)

From (2) we can see that the denominator $q={{2}^{3}}\times {{5}^{2}}$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}$ has a non-terminating repeating decimal expansion.

(viii). $\dfrac{6}{15}=\dfrac{2}{5}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative  integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{6}{15}=\dfrac{2}{5}$, the prime factorization of $q$ is $q=5={{5}^{1}}$.    …..(2)

From (2) we can see that the denominator $q=5$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=0$. Therefore, it follows from theorem (1), the rational number $\dfrac{6}{15}=\dfrac{2}{5}$ has a terminating decimal expansion.

(ix). $\dfrac{35}{50}=\dfrac{7}{10}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.   …..(1)

For rational number $\dfrac{35}{50}=\dfrac{7}{10}$, the prime factorization of $q$ is $q=10=2\times 5={{2}^{1}}\times {{5}^{1}}$.  …..(2)

From (2) we can see that the denominator $q=5$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=1$. Therefore, it follows from theorem (1), the rational number $\dfrac{35}{50}$ has a terminating decimal expansion.

(x). $\dfrac{77}{210}=\dfrac{11}{30}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{77}{210}=\dfrac{11}{30}$, the prime factorization of $q$ is 

$q=30=2\times 3\times 5$.  …..(2)

From (2) we can see that the denominator $q={{2}^{3}}\times {{5}^{2}}$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{77}{210}=\dfrac{11}{30}$ has a non-terminating repeating decimal expansion.

Maths Chapter 1- Real Numbers

Introduction

All numbers can be classified into two categories -- Real and Imaginary. A real number can be actually perceived and can be represented in a number line. It includes all integers -- positive and negative, all natural numbers (1, 2, 3, ….., ) all whole numbers (0, 1, 2, 3, ….., ), fractions, rational and irrational numbers.

• An imaginary number on the other hand cannot be perceived. Square root of all negative numbers is imaginary.

• A real number can be represented as an infinite decimal expansion. Ex: 5 = 5.0000….., ⅓ = 0.3333……

• A real number can be rational or irrational.

• All rational numbers and all irrational numbers together make a group of real numbers.

(i) A group of rational numbers is denoted by Q.

(ii) A group of irrational numbers is denoted by Q’.

(iii) A group of real numbers is denoted by R.

• A rational number an be written in the form pq, both p and q being integers (q 0) while an irrational number can not be impressed in such a way.

• The decimal expansion of a rational number say x can terminate after a finite number of digits.

Ex: ⅝ = 0.625 or eventually begins to repeat the same finite sequence of digits continuously. 

Ex: 56/99 = 0.565656……….

• On the other hand in case of an irrational number, the decimal expansion continues without repeating. Ex: 1.7320508….,  𝜋= 3.141592….)

Important Questions For CBSE Class 10 Math :

• Important Questions For CBSE Class 10 Math Chapter 2 Polynomials

• Important Questions For CBSE Class 10 Math Chapter 3 Pair of Linear Equations In Two Variables

• Important Questions For CBSE Class 10 Math Chapter 4 Quadratic Equations

• Important Questions For CBSE Class 10 Math Chapter 5 Arithmetic Progression

• Important Questions For CBSE Class 10 Math Chapter 6 Triangles

• Important Questions For CBSE Class 10 Math Chapter 7 Coordinate Geometry

• Important Questions For CBSE Class 10 Math Chapter 8 Introduction To Trigonometry

• Important Questions For CBSE Class 10 Math Chapter 9 Some Application To Trigonometry

• Important Questions For CBSE Class 10 Math Chapter 10 Circles

• Important Questions For CBSE Class 10 Math Chapter 11 Constructions

• Important Questions For CBSE Class 10 Math Chapter 12 Area Related To Circles

• Important Questions For CBSE Class 10 Math Chapter 13 Surface Area And Volume

• Important Questions For CBSE Class 10 Math Chapter 14 Statistics

• Important Questions For CBSE Class 10 Math Chapter 15 Probability

Euclid’s Division Lemma and Algorithm

First, let us understand the meaning of Lemma and Algorithms.

A Lemma is a minor proven statement which is used to prove other statements. The Euclid’s division Lemma is actually another statement of the common long division process.

For example, on dividing 36 by 5, we get quotient = 7 and remainder = 1 which is less than 5.

If there are positive integers a and b, then there are unique integers existing q and r that satisfy a = bq + r, 0 rb.

Example: If a =47 and b = 9, then we can write 

  47 = 9 x 5 + 2, where q = 4 and r = 15 b

Note: 

(i) The unique characteristics of qand r are nothing but the quotient and remainder respectively.

(ii) Although Euclid’s division algorithm is expressed for only positive integers but it can be extended for all integers except 0.

(iii) When aand bare positive integers, then qand rcan take values only from whole numbers.

An algorithm means a series of well defined steps, which provides  a procedure of calculations repeatedly successfully on the results of earlier steps. With the help of defined procedure the desired result can be obtained. 

Euclid’s Division Algorithm

The Euclid’s Division Algorithm is also an algorithm to determine the H.C.F (Highest Common Factor) of the given positive integers. 

Ordinarily, if we want to find the HCF of two positive integers, we first find the factors of the two integers. 

For ex: For two integers 24 and 42, the common factors are 1, 2, 3 and 6. As 6 is the highest of other numbers, 6 is the HCF of 24 and 42. 

In Euclid’s Division Algorithm, the largest integer is divided by the smaller one and a remainder is obtained. 

Next the smaller integer is divided by the above remainder and the process is repeated till zero is obtained. The remainder in the last but one division is the required HCF.

The Fundamental Theorem of Arithmetic

According to this theorem every positive integer is either a prime number or it can be expressed as the product of primes. It is also known as Unique Factorization Theorem. Thus, any integer that is greater than 1 can be either a prime number or can be expressed as a product of prime factors.

Example: 8 = 2 x 2 x 2, where 2 is a prime factor. 

Note: 

(i) When a number has no factors other than 1 and the number itself, the number is called a prime number.

(ii) 1 is neither prime nor composite.

(iii) A number is a composite number if it has at least one factor other than 1 and the number itself.

(iv) 2 is the smallest prime number and an even number. It is the only number that is both prime and even.

(v) Two numbers are co-prime if they have no common factors other than 1, i.e, their HCF = 1.

This theorem is termed as ‘Fundamental’ because of its importance in the development of number theory. 

Factor Tree

A chain of factors that is demonstrated in the form of a tree, is called a factor tree. 

HCF and LCM using Prime Factorization

HCF is the product of the smallest power of each common prime factor present in the numbers.

The product of the largest power of each prime factor present in the numbers is the LCM. 

Important Note: 

(i) LCM of two or more numbers is the smallest number which is the smallest number and divisible by all the given numbers.

(ii) If there is no common prime factor, then the HCF of the given number is 1.

Relationship Between Numbers and Their HCF and LCM

• For given positive integers aand b, the relation between these numbers and their HCF and LCM is 

HCF (a, b) = \[\frac{a \times b}{LCM(a,b)}\] or LCM (a, b) = \[\frac{a          \times b}{HCF(a,b)}\]

• For given three positive integers a, b and c, the relation between these numbers and their HCF and LCM is  HCF (a, b, c) = \[\frac{a \times b \times c \times LCM(a, b, c) }{LCM(a,b) \times LCM(b, c) \times LCM(c, a)}\] or LCM (a, b,c) = \[\frac{a \times b \times c \times HCF(a, b, c) }{HCF(a,b) \times HCF(b, c) \times HCF(c, a)}\]

Method of Proving Irrationality of Numbers

We use the method of contradiction in order to prove irrationality of the numbers. i.e., first we assume that the given number is rational and after reaching a contradiction we prove that the given number is irrational. 

Theorem

If a prime number  p divides a2, then p divides a, where ais a positive integer.

Let us understand how to write the decimal expansion of those rational numbers with terminating decimal expansion

Let a rational number in the lowest form be  p/q such that the prime factorization of q is of the form 2mx 5n, where m, n are non-negative integers. To write decimal expansion of p/q, convert p/q to an equivalent rational number of the form c/d, where dis a power of 10. 

The above notes will be a huge benefit for solving the Important Questions Of Chapter 1 of Maths for Class 10, Real Numbers. You can now master the Euclid’s Division Lemma and Algorithm. The important questions for this chapter is basically a compilation of higher and more advanced techniques a student can expect in Class 10 examinations. The questions of the chapter and the notes related to the topic provided by Vedantu will not only help you to understand the concept better but also solve the questions successfully. If you harbour any doubts then you can get answers to all your queries with our experienced teachers by registering with Vedantu and gain expertise in the subject. 

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