Let  \(3\sqrt { 2 } \)  be a rational number. Then, it will be of the form  \(\frac { p }{ q } \) , where p, q are coprime integers and  \(q\neq 0\).
Now, \(\frac { p }{ q } \) = \(3\sqrt { 2 } \)   \(\Rightarrow \quad \frac { p }{ 3q } =\sqrt { 2 } \)
Since, p is an integer and 3q is also an integer \(\left( 3q\neq 0 \right) \).
So, \(\frac { p }{ 3q } \) is a rational number.
\(\Rightarrow \quad \sqrt { 2 } \) is a rational number.
But this contradicts the fact that \(\sqrt { 2 } \) is an irrational number.
Hence, \(3\sqrt { 2 } \) is an irrational number.

Let \(3+\sqrt { 5 } \) is a rational number
\(3+\sqrt { 5 } =\frac { p }{ q } ,\quad q=0\)
\(3+\sqrt { 5 } =\frac { p }{ q } \)
\(\Rightarrow \sqrt { 5 } =\frac { p }{ q } -3\)
\(\Rightarrow \quad \sqrt { 5 } =\frac { p-3q }{ q } \)
\(\sqrt { 5 } \) is irrational and \(\frac { p-3q }{ q } \) is rational
But rational number cannot be equal to an irrational number.
\(\therefore \quad 3+\sqrt { 5 } \) is an irrational number.

A.T.Q. greatest number will divide 445 - 4, 572 - 5 and 699 - 6
\(\Rightarrow\) greatest number will be HCF of 441, 567 and 693
Now 447 = 32 x 72
567 =34 x 7
693 = 32 x 7 x 11
\(\therefore\)  HCF = 32 x 7 = 63
\(\therefore\) 63 is the required number.

Hint Given numbers are 404 and 96. Using prime factorisation method,
404 = 2 x 2 x 101
and 96 = 2 x 2 x 2 x 2 x 2 x 3
 HCF of (404. 96) = 2 x 2 = 4
and LCM (404, 96) = 2 x 2 x 2 x 2 x 2 x 3 x 101
 = 9696
Verification HCF x LCM= 4 x 9696= 38784        ... (i)
and product of t'te given numbers
= 404 x 96 = 38784          ... (ii)
 From Eqs. (i) and (ii), we get
HCF x LCM = Product of the given numbers.

Time taken by Sonia to drive one round of the field = 18 min
Time taken by Ravi  to drive one round of the field = 12 min
The LCM of 18 and 12 gives the exact number of minutes after which they will meet at the starting point again.
Now,  18 = 2 x 3 x 3 = 2 x 3²
and   12 = 2 x 2 x 3  = 2² x 3
\(\therefore \)  LCM of 18 and 12 = 2² x 3² = 2 x 2 x 3 x 3 = 36
Hence, Sonia and Ravi will meet again at the starting point after 36 min.

Suppose, \(\sqrt { 5 } \) is a rational number. Then, \(\sqrt { 5 } \) can be expressed in the form \(\frac{a}{b}\), where a and b are coprime integers and \(b\neq 0\).
\(\therefore \sqrt { 5 } =\frac { a }{ b } \)
On squaring both sides, we get
\(5=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \quad \Rightarrow { a }^{ 2 }=5{ b }^{ 2 }\)     ....(i)
\(\Rightarrow \)  5 divides a².
\(\Rightarrow \)  5 divides a. [by theorem 1]   ...(ii)
So, we can take a = 5m
\(\Rightarrow \)   a²  = 25m² [squaring both sides]
On putting the value of a² in Eq. (i), we get
\(\Rightarrow \)  5 divides b²
\(\Rightarrow \)  5 divides b. [by theorem 1]  ... (iii)
Thus, from Eq.(ii), 5 divides a and from Eq. (iii), 5 divides b. It means 5 is a common factor of a and b. This contradicts that there is no common factor of a and b.
This contradiction arises by assuming that \(\sqrt { 5 } \) is rational.
Hence, \(\sqrt { 5 } \) is irrational number.

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.

(i) We, have

Product of prime factors of 140
= 2 x 2 x 5 x 7= 2² x 5 x 7
(ii) 2² x 3 x 13
(iii) 3² x 5² x 17
(iv) 5 x 7 x 11 x 13
(v) 17 x 19 x 23

Given, HCF (420, 189) = 21

Let LCM of 420 and 189 be x.

We know that LCMx HCF = st number x IInd number

\(\begin{aligned} x \times 21 & =420 \times 189 \\ x & =\frac{420 \times 189}{21}=3780 \end{aligned}\)

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.

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

17, 23 and 29
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
From the above, HCF (17, 23, 29) = 1 and LCM (17, 23, 29) = 11339

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

(a) (ii)
96 = 2⁵ x 3
240 = 2⁴ x3  x5
(b) (iii)
Total number of books = 96 +240+336=672
Number of books in each stack = 48
\(\therefore\) Number of stacks formed -= \(\frac{672}{48}=14\) 
(c) (i)
Number of mathmatics books = 336
Number of stacks of mathematics books formed = \(\frac{336}{48}\) 
= 7
(d) (iv)
Number of books in each stack of english books = 48
Thickness of each english book = 3 cm
\(\therefore\)  Height of each stack of english books = (48X3) cm
= 144cm
(e) (iii)
Number of books in a stack of hindi books = 48
Weight of each hindi book = 1.5kg
\(\therefore\) The weight of books in a stack of hindi books
= (48X1.5)kg = 72kg

(d) Assertion is incorrect but Reason is correct.

Explanation:

We know that

HCF (4, b) x LCM (a, b) = a x b

5 x LCM (a, b) = 150

\(\operatorname{LCM}(a, b)=\frac{150}{5}=30\)

Assertion is incorrect but Reason is correct