Let a = 7\(\sqrt5\) be a rational number.
\(\Rightarrow \frac{a}{7}=\sqrt{5}\)
Now, \(\frac{a}{7}\) is a rational number since product of two rational number is a rational number.
The above will imply that \(\sqrt5\) is a rational number. But \(\sqrt5\)  is an irrational number.
his contradicts our assumption. Therefore we can conclude that 7\(\sqrt5\) is an irrational number and hence the result.

Let us assume, to the contrary, that 3\(\sqrt 2\) is rational.
That is, we can find coprime a and b (b ≠ 0) such that 3 \(\sqrt 2\) =\(\frac{a}{b}\).
Rearranging, we get \(\sqrt 2\)  = \(\frac{a}{3b}\).
Since 3, a and b are integers, \(\frac{a}{3b}\) is rational, and so \(\sqrt 2\) is rational.
But this contradicts the fact that \(\sqrt 2\) is irrational.
So, we conclude that 3\(\sqrt 2\) is irrational.

We have: 6 = 2 \(\times\) 3, 72 = 2³ \(\times\) 3², 120 = 2³ \(\times\) 3 \(\times\) 5
Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ \(\times\) 3¹ = 2 \(\times\) 3 = 6
2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So,    LCM (6, 72, 120) = 2³ \(\times\) 3² \(\times\) 5¹ = 360
Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

Smallest number exactly divisible by 520 and 468 = LCM. of 520 and 468 = 4680
\(\therefore\) Required number = 4680 - 17 = 4663

Given, rational number is \(\frac{13}{125}\)
\(\therefore \)  Prime factorisation of 125 = 5³
Now, \(\frac { 13 }{ 125 } =\frac { 13 }{ { 5 }^{ 3 } } =\frac { 13\times { 2 }^{ 3 } }{ { 5 }^{ 3 }\times { 2 }^{ 3 } } \)
[make the denominator in the power of 10]
\(=\frac { 104 }{ { 10 }^{ 3 } } =0.104\)
Thus, \(\frac{13}{125}\) has decimal expansion, which terminate.

On applying Euclid's division lemma for 960 and 432, we get
960 = (432 x 2) + 96
Here, remainder = 96 \(\neq \) 0,
so take new dividend as 432 and divisor as 96.
Then, we get 432 = (96 x 4) + 48
Here, remainder = 48 \(\neq \) 0,
so take new dividend as 96 and divisor as 48.
Then. we get 96 = (48 x 2) + 0
Here, the remainder is 0 (zero) and last divisor is 48.
Hence, HCF of 960 and 432 is 48.

Let us assume, to the contrary, that \(\sqrt 2\) is rational.
So, we can find integers r and s (≠ 0) such that \(\sqrt 2\) =\(\frac{r}{s}\) .
Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get \(\sqrt 2\) = \(\frac{a}{b}\) , where a and b are coprime.
So, b\(\sqrt 2\) = a.
Squaring on both sides and rearranging, we get 2b ² = a ² .Therefore, 2 divides a ² .
Now, by  it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c² , that is, b² = 2c² .
This means that 2 divides b² , and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that \(\sqrt 2\) is rational.
So, we conclude that \(\sqrt 2\) is irrational.

Given, \(a+\sqrt { b } =c+\sqrt { d } \)
If a = c, then \(\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \) b = d      [squaring both sides]
If \(a\neq c\), then there exists a positive rational number x such that a = c + x.
Now, \(a+\sqrt { b } =c+\sqrt { d } \)
\(\Rightarrow \quad c+x+\sqrt { b } =c+\sqrt { d } \quad \quad \left[ \because \quad a=c+x \right] \)
\(\Rightarrow \quad x+\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \quad \left( x+\sqrt { b } \right) ^{ 2 }=\left( \sqrt { d } \right) ^{ 2 }\)    [squaring both sides]
\(\Rightarrow \quad { x }^{ 2 }+b+2x\sqrt { b } =d\quad \Rightarrow { x }^{ 2 }+2x\sqrt { b } +b-d=0\)
\(\Rightarrow \quad 2x\sqrt { b } =d-b-{ x }^{ 2 }\Rightarrow \sqrt { b } =\frac { d-b-{ x }^{ 2 } }{ 2x } \)
Since, d, x and b are rational numbers and x > 0.
So, \(\frac { d-b-{ x }^{ 2 } }{ 2x } \) is a rational number.
\(\Rightarrow \) \(\sqrt { b } \) is a ratonal number.
\(\Rightarrow \)  b is the square of a rational number.
from Eq.(i)
\(\sqrt { d } =x+\sqrt { b } \quad \Rightarrow \quad \sqrt { d } \) is a rational number.
\(\Rightarrow \) d is the square of a rational number.
Hence, either a = c and b = d or b and d are the squares of rationals.

Given, HCF of 657 and 963 = 657 x + 963 (- 15)   .... (i) 
By using Euclid's division lemma, we get
963 = (657 x 1) + 306
Here, divisor is 657 and remainder is 306.
so, again applying Euclid's division lemma, we get
657 = (306 x 2) + 45
Here, divisor is 306 and remainder is 45.
So, again applying Euclid's division lemma, we get
306 = (45 x 6) + 36
Here, divisor is 45 and remainder is 36.
So, again applying Euclid's division lemma, we get
45 = (36 x 1)  + 9
Here, divisor is 36 and remainder is 9.
So, again applying Euclid's division lemma, we get
36 = (9 x 4) + 0
Here, remainder is zero and last divisor is 9.
\(\therefore \)  HCF of 657 and 963 = 9   ... (ii)
From Eqs.(i) and Eqs. (ii),
9 = 657 x  + 963 (- 15)
\(\Rightarrow \)  657x = 9 + 963 x 15 = 9 + 14445
\(\Rightarrow \)  657x = 14454  \(\Rightarrow \)  \(x=\frac { 14454 }{ 657 } =22\)

We have, (7 x 11 x 13) + 13 = 1001 + 13 = 1014
\(\Rightarrow \) 1014 = 2 x 3 x 13 x 13
We know that, a number is called composite number, if it has atleast one factor other than 1 and the number itself. Here, 1014 is the product of more than two prime numbers, i.e. 2, 3 and 13.
So, it is a composite number.
Now, (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 = 5045
\(\Rightarrow \) 5045 = 5 x 1009
Thus, it is the product of prime factors 5 and 1009.
Hence, it is also a composite number.

9072 = 2⁴ x 3⁴ x 7

8, 9 and 25
8 =  2³ x 1
9 = 3² x 1
25 = 5² x 1
From the above HCF (8, 9, 25) = 1 and LCM (8, 9, 25) = 1800

Given, HCF of 306 and 657 = 9
We know that
LCM \(\times\) HCF = Product of two numbers
\(\Rightarrow\) LCM \(\times\) 9 = 306 \(\times\) 657
\(\Rightarrow\)   LCM = \( \frac{306 \times 657}{9}\)
= 34 \(\times\) 657 = 22338
\(\therefore\) LCM of 306 and 657 = 22338

If the number 4ⁿ , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4ⁿ would contain the prime 5. This is not possible because 4ⁿ = (2)²ⁿ ; so the only prime in the factorisation of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4ⁿ . So, there is no natural number n for which 4ⁿ ends with the digit zero.
You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it. This method is also called the prime factorisation method. Let us recall this method through an example.

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.

(i) (c): Here, the simplest form of given options are
125/441 = 5³/(3² x 7²), 77/210 = 11/(2 x 3 x 5),
15/1600 = 3/(2⁶ x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal.
(ii) (b): 23/(2³ x 5²) = 23/200 = 0.115
(iii) (a): 441/(2² x 5⁷ x 7²) = 9/(2² x 5⁷), which is a terminating decimal.
(iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15.
(v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.

(i) (b): Here 80 = 2⁴ x 5, 85 = 17 x 5
and 90 = 2 x 3² x 5
L.C.M of 80, 85 and 90 = 2⁴ x 3 x 3 x 5 x 17 = 12240
Hence, the minimum distance each should walk when they at first time is 12240 cm.
(ii) (c): Here 594 = 2 x 3³ x 11 and 189 = 3³ x 7
HCF of 594 and 189 = 3³= 27
Hence, the maximum number of columns in which they can march is 27.
(iii) (c) : Here 768 = 2⁸ x 3 and 420 = 2² x 3 x 5 x 7
HCF of 768 and 420 = 2² x 3 = 12
So, the container which can measure fuel of either tanker exactly must be of 12litres.
(iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm
Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5
HCF = 5 x 5 x 3 = 75
Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm.
(v) (a): LCM of 8 and 12 is 24.
\(\therefore \)The least number of pack of pens = 24/8 = 3
\(\therefore \)The least number of pack of note pads = 24/12 = 2

(i) (d) : For a number to end in zero it must be divisible by 5, but 4ⁿ = 22ⁿ is never divisible by 5. So, 4ⁿ never ends in zero for any value of n.
(ii) (c) : We know that product of two rational numbers is also a rational number.
So, a² = a x a = rational number
a³ = a² x a = rational number
a⁴ = a³ x a = rational number
................................................
...............................................
aⁿ = a^n-1 x a = rational number.
(iii) (d): Let x = 2m + 1 and y = 2k + 1
Then x² + y² = (2m + 1)² + (2k + 1)²
= 4m² + 4m + 1 + 4k² + 4k + 1 = 4(m² + k² + m + k) + 2 So, it is even but not divisible by 4.
(iv) (a): Let three consecutive positive integers be n, n + 1 and n + 2.
We know that when a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
So, n = 3p or 3p + lor 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 Wi always divisible by 3.
(v) (d): Any odd number is of the form of (2k +1), where k is any integer.
So, n² - 1 = (2k + 1)2 -1 = 4k² + 4k
For k = 1, 4k² + 4k = 8, which is divisible by 8.
Similarly, for k = 2, 4k² + 4k = 24, which is divisible by 8.
And for k = 3, 4k² + 4k = 48, which is also divisible by 8.
So, 4k² + 4k is divisible by 8 for all integers k, i.e., n² - 1 is divisible by 8 for all odd values of n.