The greatest number of cartons is the HCF of 144 and 90
\(144={ 2 }^{ 4 }\times { 3 }^{ 2 }\)
\(90=2\times { 3 }^{ 2 }\times 5\)
\(HCF=2\times { 3 }^{ 2 }=18\)
∴ The greatest number of cartons = 18

Let \(\sqrt { q } \) be a rational number and its simplest form is a/b. where q is a prime, a and b are coprime integers and \(b\neq 0\).
Now, \(\sqrt { q } =\frac { a }{ b } \)
On squaring both sides, we get
\(q=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \Rightarrow q{ b }^{ 2 }={ a }^{ 2 }\)              .... (i) (1)
Since, qb² is divisible by q, so a² is also divisible by q.
\(\Rightarrow \)  a is also divisible by q.    [by using theorem 1] ... (ii)
Now, a can be written as a = qc for some integer c.
On substituting a = qc in Eq.(i) , we get
qb² = (qc)² \(\Rightarrow \) qb² = q²c² \(\Rightarrow \)  qc²
\(\Rightarrow \) b² is divisible by q.
\(\Rightarrow \) b is divisible by q.     [ by using theorem 1] ... (iii) (1)
From Eqs.(ii) and (iii), q is a common factor of a and b. But this contradicts our assumption that a and b have no common factor. So, our assumption that \(\sqrt { q } \) is a rational number, is wrong. Hence, \(\sqrt { q } \) is an irrational number, if q is prime.

Let us assume to the contrary that \(3+2\sqrt { 5 } \)  is a rational number. Then, it can be expressed in the form \(\frac{a}{b}\), where a, b are coprime integers and \(b\neq 0\)
Now,  \(3+2\sqrt { 5 } \) = a/b, where a,b are integers and \(b\neq 0\)
On rearranging, we get
\(2\sqrt { 5 } =\frac { a }{ b } -3\quad or\quad \sqrt { 5 } =\frac { a }{ 2b } -\frac { 3 }{ 2 } \)
Since, a, b are integers and \(b\neq 0\) , therefore \(\frac{a}{2b}\) is rational number and so \(\frac{a}{2b}\) - \(\frac{3}{2}\) is a rational number.
[since, difference of two rational numbers is also a rational number]
\(\Rightarrow \sqrt { 5 } \) is a rational number. But \(\sqrt { 5 } \) is an irrational number.
This shows that our assumption is incorrect.
So, \(3+2\sqrt { 5 } \) is irrational.

Let the two number are a and b.
Given, HCF(a, b) = 2
and product (a x b) = 120
We know that, 
HCF (a, b) x LCM (a, b) = Product of a and b
\(\therefore \)   2 x LCM (a, b) = 120
\(\Rightarrow \)  LCM (a, b) = \(\frac{120}{2}\) = 6
Hence, the required LCM is 60.

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.

Let a=4q+r, 0\(\le \)r<4
⇒ a=4q,4q+1,4q+2 Or 4q+3
Case I: \({ a }^{ 2 }=({ 4q })^{ 2 }-16{ q }^{ 2 }=4({ 4q }^{ 2 })=4m\)
\(m=4{ q }^{ 2 }\)
Case II: \({ a }^{ 2 }=(4q+1)^{ 2 }=16{ q }^{ 2 }+8q+1\)
\(=4(4{ q }^{ 2 }+2q)+1\)
\(=4m+1\)
where \(m=4{ q }^{ 2 }+2q\)
\({ a }^{ 2 }=(4q+2)^{ 2 }\)
\(=16{ q }^{ 2 }+16q+4\)
\(\\ =4(4{ q }^{ 2 }+4{ q }^{ 2 }+1)=4m\)
where \(m=4{ q }^{ 2 }+4{ q }+1\)
Case IV: \({ }^{ }\)\({ a }^{ 2 }=(4q+3)^{ 2 }=16{ q }^{ 2 }+24q+9\)
\(=16{ q }^{ 2 }+24q+8+1\)
\(=4(4{ q }^{ 2 }+6q+2)+1\)
\(=4m+1\)
\(m=4{ q }^{ 2 }+6q+2\)
From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form 4m or 4m +1.

If 6" ends with 0, then it must have 5 as a factor. But we know that only prime factor of 6ⁿ are 2 and 3.
∴  6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
From the fundamental theorem of arithmetic, we know that the prime factorization of every composite numbers is unique.
∴  6ⁿ can never end with 0.

a is any positive integer

Denominator = 500
= 2² x 5³
Decimal expansion, \(\frac { 257 }{ 500 } =\frac { 257\times 2 }{ 2\times { 2 }^{ 2 }\times { 5 }^{ 3 } } =\frac { 514 }{ 10^{ 3 } } \)
= 0.514

\(\frac { 6 }{ 1250 } =\frac { 6 }{ 2\times { 5 }^{ 4 } } \)
 \(=\frac { 6\times { 2 }^{ 3 } }{ 2\times { 2 }^{ 3 }\times { 5 }^{ 4 } } \)
 \(=\frac { 6\times { 2 }^{ 3 } }{ { 2 }^{ 4 }\times { 5 }^{ 4 } } \)
\(=\frac { 6\times { 2 }^{ 3 } }{ (10)^{ 4 } } =0.0288\)
\(\frac { 6 }{ 1250 } \) will terminate after 4 decimal places.

Since ​​​​​​ ​\(\frac { 1 }{ 7 } +\frac { 7 }{ 100 } =\frac { 1 }{ 100 } =0.01\)
Thus smallest rational number is \(\\ \frac { 7 }{ 100 } \)

y = 5 x 13 = 65
and x = 3 x 195 = 585

We know that a x b = HCF (a, b) x LCM (a, b)
⇒ 1,800 = 12 x LCM (a, b)
⇒  LCM(a,b)=\(\frac { 1,800 }{ 12 } =150\)

The smallest prime number is 2 and the smallest composite number is 2² Hence, required HCF (2²,2) = 2.

The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number

Required number=LCM (1, 2, 3, 4, 5)
60

138

\(\because \) smallest odd composite number = 9
and smallest odd prime number = 3.
\(\therefore \) HCF of 9 and 3 = 3
and LCM of 9 and 3 = 9
Now, if an odd number p divides q², then p is one of the factors of q², i.e. q² = pm, for some integer m.    .... (i) 
Now, q³ = q² . q \(\Rightarrow \) q³ = pm . q   [from Eq.(i)]
\(\Rightarrow \)  q³ = p (mq)
\(\Rightarrow \)  p is a factor of q³ also \(\Rightarrow \)  p divides q³