NCERT Exemplar for Class 10 Maths Chapter 1 - Real Numbers (Book Solutions)

SOLVED EXAMPLE

1. The decimal expansion of the rational number $\frac{33}{2^2.5}$ will terminate after 

(A) One decimal place  

(B) Two decimal place   

(C) Three decimal place 

(D) More than 3 decimal places  

Ans: B 

Express the denominator as the multiple of 10,

If we divide both numerator and denominator by $5,$ denominator can be changed to $100$,

$\frac{33}{2^2\cdot 5}=\frac{33\times 5}{4\times 5\times 5}$

$\Rightarrow \frac{33\times 5}{100}=\frac{165}{100}$

Convert the obtained fraction into decimal. 

$\frac{165}{100}=1\cdot 65$ 

2. Euclid’s division Lemma states that for two positive integers $a$ and $b$ there exist unique integers $q$ and $r$ such that $a=bq+r$, where $r$ must satisfy.

(A) $1<r<b$

(B) $0<r⩽b$

(C) $0⩽r<b$

(D) $0<r<b$

Ans: C

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that $a=bq+r$, where $0⩽r<b$.

EXERCISE 1.1

1.  For some integer $m$, every even integer is of the form 

(A) $m$  

(B)$m+1$   

(C) $2m$

(D) $2m+1$

Ans: C

Every Even integer is a multiple of $2$.

Let,

$m=\text{Integer}=Z$

$\Rightarrow m=....-3,-2,-1,0,1,2,3....$ 

$\Rightarrow 2m=....-6,-4,-2,0,2,4,6....$(all even numbers)  

Hence, 

Even numbers can be written in the form of 2m.

2. For some integer $q$, every odd integer is of the form  

(A) $q$  

(B) $q+1$   

(C) $2q$

(D) $2q+1$ 

Ans: D 

Every Even integer is a multiple of 2.

Let,

$q=\text{Integer}=Z$

$\Rightarrow q=....-3,-2,-1,0,1,2,3....$ 

$\Rightarrow 2q=....-6,-4,-2,0,2,4,6....$(all are even numbers) 

$\Rightarrow 2q+1=....-5,-3,-1,1,3,5,7....$(all are odd numbers) 

Hence, every odd integer can be written in the form of $2q+1.$

3. $n^2-1$ is divisible by 8, if $n$ is 

(A) an integer

(B)a natural number   

(C) an odd integer

(D) More than 3 decimal places

Ans: C

Let $x=n^2-1$

Since, ‘n’ is any number. Therefore, it can be even or odd.

Condition I,

When ‘n’ is an even integer.

Let, $n=2k$  (where ${}^{\prime }{k}^{\prime }$ is an integer) 

$\Rightarrow x=n^2-1$

$\Rightarrow x={\left(2k\right)}^2-1$

$\Rightarrow x=4k^2-1$ 

When,

$k=-1$

$\Rightarrow x=4k^2-1$

$\Rightarrow x=4{\left(-1\right)}^2-1=3$(which is not divisible by $8$) 

$k=0$

$\Rightarrow x=4k^2-1$ 

$\Rightarrow x=4{\left(0\right)}^2-1=-1$ (which is not divisible by $8$) 

Condition II,

When ‘n’ is an odd integer.

$n=2k+1$ (where ${}^{\prime }{k}^{\prime }$ is an integer) 

$\Rightarrow x={\left(2k+1\right)}^2-1$

$\Rightarrow x=4k^2+1+4k-1$

$\Rightarrow x=4k^2+4k$

$\Rightarrow x=4k\left(k+1\right)$ 

When,

$k=-1$

$\Rightarrow x=4k\left(k+1\right)$

$\Rightarrow x=4\left(-1\right)\left(-1+1\right)=0$ (which is divisible by $8$)

$k=0$

$\Rightarrow x=4k\left(k+1\right)$

$\Rightarrow x=4\left(0\right)\left(0+1\right)=0$ (which is divisible by $8$)

$k=1$

$\Rightarrow x=4k\left(k+1\right)$

$\Rightarrow x=4\left(1\right)\left(1+1\right)=8$ (which is divisible by $8$) 

Hence, we can conclude from the above two cases that if n is odd, then $n^2-1$ is divisible by $8$.

4. If the HCF of $65$ and $117$ is expressible in the form $65m--117$, then the value of $m$ is 

(A)$4$  

(B) $2$  

(C) $1$

(D) $3$

Ans: B 

HCF of $65$ and $117$ by prime factorisation method.

$65=5\times 13$

$117=3\times 3\times 13$ 

As, $13$ is the common factor between $65$ and $117$.

Therefore, $\text{HCF}\left(65,117\right)=13$     

$\because 65m-117=\text{HCF}$

$\therefore 65m-117=13$

$\Rightarrow 65m=13+117$

$\Rightarrow 65m=130$

$\Rightarrow m=\frac{130}{65}=2$ 

5. The largest number which divides $70and125$, leaving remainder $5and8$ respectively is 

(A) 13  

(B) 65

(C) 875

(D) 1750

Ans: A

Since, 5 and 8 are the remainder of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers $65=\left(70-5\right)$ and$117=\left(125--8\right)$, which is divisible by the required number.

The number to be obtained $\text{ = HCF}\left(\text{70,}\text{125}\right)$

$\Rightarrow 117=65\times 1+52$  ($\therefore$ Dividend $=$ divisor $\times$ quotient $+$remainder)

$\Rightarrow 65=52\times 1+13$

$\Rightarrow 52=13\times 4+0$ 

Hence, 13 is the highest common factor of 70 and 125, leaving 5 and 8 as remainder receptively.

6. If two positive integers a and b are written as $a=x^3{y}^2$ and $b=x{y}^3;x,y$are prime numbers, then HCF $\left(a,b\right)$ is 

(A) $xy$  

(B) $x{y}^2$   

(C) $x^3{y}^3$

(D) $x^2{y}^2$

Ans: B

HCF is defined as the highest common factor between the two numbers and for variables it’s the smallest exponent of every common variable.

We can write,

$a=x\times x\times x\times y\times y$

$b=x\times y\times y\times y$

So,

$x$ and $y$ are both common variables in $a$ and $b$, with lowest exponents as 1 and 2 respectively.

Therefore, the HCF will be $x{y}^2$.

7. If two positive integer $p$and $q$ can be expressed as $p=a{b}^2$ and $q=a^{3}b;a,b$ being prime number then LCM $\left(p,q\right)$ is 

(A) $ab$  

(B) $a^2{b}^2$   

(C) $a^3{b}^2$

(D) $a^3{b}^3$

Ans: C

LCM is defined as the least common multiple of integers $a$ and $b$. For variables it’s the highest exponent of every common variable.

We can write,

$p=a\times b\times b$

$q=a\times a\times a\times b$

So,

a and b are both common variables in p and q, with highest exponent as 3 and 2 respectively.

Therefore, the LCM will be $a^3{b}^2$.

8.  The product of a non-zero rational and an irrational no. is 

(A) Always irrational 

(B) Always rational Two decimal place   

(C) Rational or Irrational

(D) One 

Ans: A

Irrational number can be defined as a non- repeating, non-terminating decimal number unlike the case of rational number which is a terminating number. 

When a non-repeating and non-terminating number is multiplied by an irrational number. The result number is always an irrational number.

Example:

$\frac{2}{5}\times \sqrt{3}=\frac{2\sqrt{3}}{5}$

9. The least number that is divisible by all the numbers from 1-10 (both inclusive) is 

(A) 10 

(B) 100  

(C) 504 

(D)2520

Ans: D

Since, the required number is divisible from all the numbers from 1 to 10. Therefore, we need to find the LCM.

Factors of numbers from 1 to 10.

$1=1$ 

$2=1\times 2$ 

$3=1\times 3$ 

$4=2\times 2$ 

$5=1\times 5$ 

$6=2\times 3$ 

$7=1\times 7$ 

$8=2\times 2\times 2$ 

$9=3\times 3$ 

$10=2\times 5$ 

We know that:

LCM is defined as the least common multiple of integers $a$ and $b$. For variables it’s the highest exponent of every common variable.

Therefore,

$\text{LCM}\left(1,2,3,4,5,6,7,8,9,10\right)=2\times 2\times 2\times 3\times 3\times 5\times 7=2520$

10. The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:

(A) One decimal place  

(B) Two decimal place   

(C) Three decimal place 

(D) More than 3 decimal places

Ans: D

Rational number = $\frac{14587}{1250}$

Multiply the numerator and the denominator with $2^3$ to make the denominator as a multiple of 10.

$\frac{14587}{1250}=\frac{14587}{2^1\times 5^4}\times \frac{2^3}{2^3}$

$\Rightarrow =\frac{14587\times 8}{2^4\times 5^4}$

$\Rightarrow =\frac{116696}{{\left(10\right)}^4}=\frac{116696}{10000}$

$\Rightarrow =11.6696$ 

Hence, a given rational number will terminate after four decimal places.

SOLVED EXAMPLE

1. The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer.

Ans: The given statement “The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only.” is False.

According to Euclid's division Lemma, 

$a=bq+r,a=3q+r$

where b = 3

$0⩽r<b$

$\Rightarrow 0⩽r<3$ 

Therefore, r can have 0, 1 and 2.

2. Can the number$6^n$, n being a natural number end with the digits 5 ? Give reasons.

Ans: Consider the number $6^n$

On prime factorisation of 6,

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

$6^n=2^n\times 3^n$ 

Since, on factoring $6^n$ we get $2^n\times 3^n$, which does not contain 5. Therefore, $6^n$ cannot end with the digit 5.

EXERCISE 1.2

1. Write whether every positive integer can be of the form $4q+2$, where q is an integer. Justify your answer

Ans: Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that $a=bq+r$, where $0⩽r<b$.

Where,

a = dividend

b = divisor

q = quotient

r = remainder

According to question,

‘a’ is any positive integer.

$b=4$

$r=0,1,2,3$$0⩽r<b;0⩽r<4$

Therefore, ‘a’ can be in the form of $4q,4q+1,4q+2,4q+3$ i.e., $a=bq+r$.

Hence, it is not necessary that every positive integer is in the form of $4q+2$. 

2. “The product of two consecutive positive integers is divisible by 2”. is this statement true or false? Give reasons.

Ans: Two consecutive positive integers are always an even and an odd number. 

The product of even and an odd number is always an even number.

Since, the even number is divisible by 2. 

Therefore, the product of two consecutive positive integers is also divisible by 2.

Hence, the given statement “The product of two consecutive positive integers is divisible by 2”. is true. 

3. “The product of three consecutive positive integers is divisible by $6$. Is this statement true or false”? Justify your answer

Ans: We can verify it by taking different random groups of three consecutive positive integers.

a) Let take 2, 3, 4 as 3 consecutive positive integers

So, $2\times 3\times 4=24$ which is divisible by 6

b) Let take 5, 6, 7 as 3 consecutive positive integers

So, $5\times 6\times 7=210;$ which is divisible by 6

Hence, The statement “The product of three consecutive positive integers is divisible by $6$.”

4. Write whether the square of any positive integer can be of the form $3m+2$, where $m$ is a natural number. Justify your answer.

Ans: No

Consider any random positive integer say ‘a’.

According to Euclid’s division lemma; a can be written as 

$a=bq+r;0⩽r<b$

Let b = 3 then,

$a=3q+r\text{ where }0⩽r<3$

So, r can have the following values 0, 1, 2.

Since, any positive integer can have the following forms.

$r=0\Rightarrow a=3q$ 

$r=1\Rightarrow a=3q+1$ 

$r=2\Rightarrow a=3q+2$ 

When $r=0,a=3q$

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

$\Rightarrow 9q^2=3\left(3q^2\right)=3m$(where $m=3q^2$)

 When $r=1,a=3q+1$

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

$\Rightarrow 9q^2+6q+1=3\left(3q^2+2q\right)+1=3m+1$ (where $m=3q^2+2q$)

When $r=2,a=3q+2$

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

$\Rightarrow 9q^2+12q+4=3\left(3q^2+4q+3\right)+1=3m+1$(where $m=3q^2+4q+3$)

Hence, squares of positive integers cannot be expressed in the form $3m+2$.

5. A positive integer is of the form $3q+1,q$ being a natural number. Can you write its square in any form other than$3m+1$, i.e., $3m$ or $3m+2$ for some integer m? Justify your answer.

Ans: According to Euclid’s division lemma,

Put $b=3$

$a=bq+r$, where all of a, b, q, r are positive integers and $0⩽r<b$ i.e., $0⩽r<3$.

$\Rightarrow a=3q,3q+1\text{ or }3q+2$.    

Now,

When $a=3q$,

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

$=9q^2=3.3q^2$(here, $m=3q^2$)

When $a=3q+1$,

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

$=9q^2+6q+1=3\left(3q^2+2q\right)+1$(here, $m=3q^2+2q$)$$

When $a=3q+2$,

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

$=9q^2+12q+4$ 

$=3\left(3q^2+4q+1\right)+1$(here, $m=3q^2+4q+1$)

Therefore, the square of a positive integer can be expressed in the form of $3m+1$.

6. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.

Ans: The common factors of $3000$ and $525$ is $75.$

By Euclid’s division lemma,

$3000=525\times 5+375$ 

$525=375\times 1+150$ 

375 = 150 \times 2 + 75$ 

150 = 75 \times 2 + 0$ 

By Euclid’s method, we get the HCF (525, 3000) as 75.

7. Explain why $3\times 5\times 7+7$ is a composite number.

Ans: A composite number can be expressed as product of primes 

Considering the given number,

$=\left(3\times 5\times 7\right)+7$

$\Rightarrow 105+7=112$ 

On prime factorisation of 112, 

$112=2\times 2\times 2\times 2\times 7$

$=2^4\times 7$ 

Here 2 and 7 both are prime numbers.

We know that:

The product of two prime numbers is always a composite number. 

Therefore, $3\times 5\times 7+7$ is a composite number.

8. Can two numbers have $18$ as their HCF and $380$ as their LCM? Give reasons.

Ans: HCF of two numbers means highest common factor and LCM of two numbers means their lowest common multiple. So, as per definition LCM of two numbers is always divisible by the HCF of two numbers. HCF is the factor of LCM.

Since, 18 is not the factor of 380. Therefore, no numbers can have 18 and 380 as their HCF and LCM respectively.

9. Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

Ans: Let $\frac{987}{10500}$

On simplifying the given number,

$\Rightarrow \frac{987}{10500}=\frac{3\times 7\times 47}{3\times 7\times 500}$

$=\frac{47}{500}$

$=\frac{47}{2^2\times 5^3}$ 

Here, the denominator is in the form $2^m\times 5^n$.

If the denominator is in the form of $2^m\times 5^n$ then its terminating rational number.

Hence, the number$\frac{987}{10500}$ is a terminating rational number.

10. A rational number in its decimal expansion is $327.7081$. What can you say about the prime factors of $q$, when this number is expressed in the form$\frac{p}{q}$? Give reasons.

Ans:

$327.7081$ can be written as,

$\frac{3277081}{10000}=\frac{32777081}{104}=\frac{p}{q}$

$q=10000$

$={10}^4$

$={\left(2\times 5\right)}^4$ 

Hence, the denominator can be expressed in the form of $2^m\times 5^n$.

Sample Questions

1. Using Euclid’s division algorithm, find which of the following pairs of numbers are co-prime:

(i) $231,396$ 

Ans: Let's calculate the HCF of each number pair.

Here, Co-primes are two numbers that have only $1$ as a common factor.

Then, by Euclid's lemma $a=bq+r,0⩽r<b$

(i) $231,396$ 

So, HCF of $231,396$is 

$\Rightarrow 396=1\times 231+165$

$\Rightarrow 231=1\times 165+66$

$\Rightarrow 165=2\times 66+33$

$\Rightarrow 66=2\times 33+0$

The HCF of $231,396$ is $33$ which is not equal to$1$, 

Therefore, $231,396$are not co-prime.

(ii) $847,2160$ 

Let's calculate the HCF of each number pair.

Here, Co-primes are two numbers that have only $1$ as a common factor.

Then, by Euclid's lemma $a=bq+r,0⩽r<b$

So, HCF of $847$ and $2160$ is 

$2160=847\times 2+466$

$\Rightarrow 847=466\times 1+381$

$\Rightarrow 466=381\times 1+85$

$\Rightarrow 85=41\times 2+3$

$\Rightarrow 41=3\times 13+2$

$\Rightarrow 3=2\times 1+1$

$\Rightarrow 2=1\times 2+0$

Since, The HCF of $847$ and $2160$is $1$.

Therefore, $847$ and $2160$  are co-primes.

2.  Show that the square of an odd positive integer is of the form $8m+1$, for some whole number$m$.

Ans: Let us take $2p+1$ is any odd positive integer for which$p\in I$

$\Rightarrow (2p+1{)}^2=4p^2+1+4p$

$\Rightarrow (2p+1{)}^2=4p^2+4p+1$

$\Rightarrow (2p+1{)}^2=4p(p+1)+1$ 

$p(p+1)$is even and also a product of two consecutive integers .

Let $p(p+1)=2m$ for some whole number $m\in \text{W}$

$\Rightarrow (2p+1{)}^2$

$\Rightarrow {\left(2p\right)}^2+1+4p$

$\Rightarrow 4p^2+4p+1$

$\Rightarrow 4p\left(p+1\right)+1$

$\Rightarrow 4(2m)+1$

$\Rightarrow 8m+1$ 

Therefore, the square of an odd integer is of the form $8\text{m}+1$ for some whole number $m\in \text{W}$.

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

Ans: Let us assume $\sqrt{2}+\sqrt{3}$is a rational number.

So it can be represented in the form of $a/b$ 

$\Rightarrow \sqrt{2}+\sqrt{3}=\frac{a}{b}$

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

Now, squaring on both sides, we get

$\Rightarrow {\left(\sqrt{2}\right)}^2={\left(\frac{a}{b}-\sqrt{3}\right)}^2$

Expand  ${\left(\frac{a}{b}-\sqrt{3}\right)}^2={\frac{a}{b^2}}^2+3-2\left(\frac{a}{b}\right)\sqrt{3}$

$\Rightarrow 2=\frac{a^2}{b^2}+3-2\sqrt{3}\left(\frac{a}{b}\right)$

$\Rightarrow \frac{a^2}{b^2}+3-2=2\sqrt{3}\frac{a}{b}$

$\Rightarrow \frac{a^2}{b^2}+1=2\times \sqrt{3}\left(\frac{a}{b}\right)$

$\Rightarrow \frac{\left(a^2+b^2\right)}{b^2}\times \frac{b}{2a}=\sqrt{3}$

$\Rightarrow \frac{\left(a^2+b^2\right)}{2ab}=\sqrt{3}$

$\frac{\left(a^2+b^2\right)}{2ab}$ which is a contradiction.

Since, $a,b$ are integers. Therefore,$\frac{\left(a^2+b^2\right)}{2ab}$  is a rational number but $\sqrt{3}$ is an irrational number. Therefore, it contradicts the fact that $\sqrt{2}+\sqrt{3}$ is rational.

Hence, it is an irrational number.

EXERCISE 1.3:

1. Show that the square of any positive integer is either of the form $4q$ or $4q+1$ for some integer$q$.

Ans: Let us consider $a$ be any positive integer 

Using Euclid's division lemma $a=bm+r$

$a=4q+r$,$0⩽r<4$when$b=4$

So, $r=0,1,2,3$  

Now, squaring on both sides the equation $a=4q+r$ in every case 

Case. i) when $r=0$

$a=4m$

$a^2=(4m{)}^2$

$a^2=4\left(4m^2\right)$

$a^2=4q$

where $q=4m^2$

When $r=1$

$a=4m+1$

$a^2=(4m+1{)}^2$

$a^2=16m^2+1+8m$

$a^2=4\left(4m^2+2m\right)+1$

$a^2=4q+1$, where $q=4m^2+2m$

Case. ii) when $r=2$

$a=4m+2$

$a^2=(4m+2{)}^2$

$a^2=16m^2+4+16m$

$a^2=4\left(4m^2+4m+1\right)$

$a^2=4q$, Where $q=4m^2+4m+1$

Case. iii) when $r=3$

$a=4m+3$

$a^2=(4m+3{)}^2$

$a^2=16m^2+9+24m$

$a^2=16m^2+24m+8+1$

$a^2=4\left(4m^2+6m+2\right)+1$

$a^2=4q+1$, where $q=4m^2+6m+2$

Therefore, Square of any positive integer is in the form of $4q$ or $4q+1$, where $q$ is any integer.

2. Show that cube of any positive integer is of the form $4m,4m+1$ or $4m+3$, for some integer 

Ans: Let us consider $a$ be any positive integer 

Using Euclid's division lemma $a=bq+r$

$a=4q+r$ ,$0⩽r<4$when$b=4$

So, $r=0,1,2,3$  

Now, cubing on both sides of $a=4q+r$in every case

Case 1: When $r=0$ 

$a=4q$ 

$(a{)}^3=(4q{)}^3$

$(a{)}^3=64q^3=4\left(16q^3\right)$

$a^3=4m$ , where $m=16q^3$

Case. 2) when  $r=1$ 

$a=4q+1$ 

$a^3=(4q+1{)}^3$

Since, $(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

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

$a^3=64q^3+1+3\times 16q^2+12q$

$a^3=64q^3+1+48q^2+12q$

$a^3=4\left(16q^3+12q^3+3q\right)+1$

$a^3=4m+1$, where $m=16q^3+12q^2+3q$

Case. 3) when $r=2$ 

$a=4q+2$

$a^3=(4q+2{)}^3$

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

$a^3=64q^3+8+3\times 16q^2\times 2+3\times 4q\times 4$

$a^3=64q^3+8+96q^2+48q$

$a^3=4\left(16q^3+2+24q^3+12q\right)$

$a^3=4m$, Where $m=16q^3+2+24q^3+12q$

Case. 4) when $r=3$ 

$a=4q+3$

$a^3=(4q+3{)}^3$

$a^3=\left(64q^3+24+144q^2+108q\right)+3$

$a^3=4\left(16q^3+6+36q^2+27q\right)+3$ 

$a^3=4m+3$, where $m=16q^3+6+36q^2+27q$

Therefore,  $a^3$ is of the form of $4m,4m+1,4m+3$.

Hence, the cube of any positive integer is of the form $4m,4m+1$ or $4m+3$, for some integer $m$.

3. Show that the square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any integer $q$.

Ans: Let us consider $a$ be any positive integer 

Using Euclid's division lemma $a=bm+r$

$a=5m+r$, $0⩽r<5$ when $b=5$

So, $r=0,1,2,3,4$  

Now, Squaring both sides of an equation  $a=6m+r$in every case 

Case 1: when$r=0$ 

$a=5m$ 

$a^2=25m^2$

$a^2=5\left(5m^2\right)$

$a^2=5q,\text{ where q = }5m^2$

Case 2: when $r=1$

$a=5m+1$

$a^2=(5m+1{)}^2$

$a^2=25m^2+10m+1$ 

$a^2=5\left(5m^2+2m\right)+1$

$=5q+1$ ,  where $q=5m^2+2m$

Case 3: when $r=2$

$a=5m+2$

$a^2=25m^2+20m+4$

$a^2=5\left(5m^2+4m\right)+4$

$a^2=5q+4$,  where $q=5m^2+4m$

Case 4: when $r=3$

$a=5m+3$

$a^2=25m^2+30m+9$

$=25m^2+30m+5+4$

$=5\left(5m^2+6m+1\right)+4$

$=5q+4$,  where $q=5m^2+6m+1$

Case 5: when $r=4$

$a=5m+4$

$a^2=25m^2+40m+16$

$a^2=25m^2+40m+15+1$ 

$=5\left(5m^2+8m+3\right)+1$

$=5q+1$ ,  Where $q=5m^2+8m+3$

From the above cases, $(5q+2)$ and, $(5q+3)$ are not perfect squares for some value of $q$.
Hence, the square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any integer $q$.

4. Show that the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer $m$.

Ans: Let us consider $a$ be any positive integer.

Using Euclid's division lemma $a=bm+r$

$a=6q+r$, $0⩽r<6$ when $b=6$

So that $r=0,1,2,3,4,5$ 

Now, squaring both sides of an equation  $a=6q+r$ in every case. 

Case 1: when $r=0$

$a^2=6\left[6q^2+2q\times 0\right]+0^2$

$\Rightarrow a^2=36q^2$

$\Rightarrow a^2=6m$, where $m=6q^2$

Case 2: when $r=1$

$a^2=6\left[6q^2+2q\times 1\right]+1^2$ 

$\Rightarrow a^2=6\left[6q^2+2q\right]+1$

$\Rightarrow a^2=6m+1$, where $m=6q^2+2q$

Case 3: When $r=2$

$a^2=6\left[6q^2+2q\cdot 2\right]+2^2$ 

$\Rightarrow a^2=6m+4$, where $m=\left(6q^2+4q\right)$

Case 4: When $r=3$

$a^2=6\left[6q^2+2q\cdot 3\right]+3^2$ 

$\Rightarrow a^2=6\left[6q^2+6q\right]+6+3$

$\Rightarrow a^2=6\left[6q^2+6q+1\right]+3$

$\Rightarrow a^2=6m+3$, where $m=6q^2+6q+1$

Case 5: when $r=4$

$a^2=6\left[6q^2+2q\cdot 4\right]+4^2$

$\Rightarrow a^2=6\left[6q^2+8q\right]+12+4$

$\Rightarrow a^2=6\left[6q^2+8q+2\right]+4$

$\Rightarrow a^2=6m+4$, where $m=6q^4+8q+2$

Case 6: When $r=5$ 

$a^2=6\left[6q^2+2q\cdot 5\right]+5^2$

$\Rightarrow a^2=6\left[6q^2+10q\right]+24+1$

$\Rightarrow 6\left[6q^2+10q+4\right]+1$

$\Rightarrow a^2=6m+1$, where $m=6q^2+10q+4$

Therefore, the form of $6m,(6m+1),(6m+3)$ and $(6m+4)$ are squares and $(6m+2)$ and $(6m+5)$ are not perfect square for some value of $m$.
Hence, the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer$m$.

5. Show that the square of any odd integer is of the form $4q+1$, for some integer $q$.

Ans: Using Euclid’s division lemma method, i.e., $a=bq+r,0⩽r<b$.

Let $b=4$ then $\text{a}=4\text{q}+\text{r}$, where $0⩽r<4$ i.e., $\text{r}=0,1,2,3\dots$ (i)

Case 1: If $r=0\Rightarrow a=4q$ 

$4q$is divisible by $2$. 

$4q$is even 

Case 2: If $r=1\Rightarrow a=4q+1$ 

$(4q+1)$is not divisible by $2$.

Case 3: If $r=2\Rightarrow a=4q+2$

 $2(2q+1)$is divisible by $2$ 

$2(2q+1)$is even.

Case 4: If $r=3\Rightarrow a=4q+3$ 

$(4q+3)$ is not divisible by  $2$.

$(4q+1)$ and $(4q+3)$ are odd integers, which are not divisible by $2$.

Now, squaring the odd number.

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

Since, $\left[{(a+b)}^2=a^2+2ab+b^2\right]$