a x 13 = 39
= 3

(i) 16  (ii) Unity and Discipline.

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.

Decimal expansion of \(\frac { 43 }{ { 2 }^{ 4 }\times { 5 }^{ 3 } } =\frac { 43\times 5 }{ { 2 }^{ 4 }\times { 5 }^{ 3 }\times 5 } =0.0215\) 4 places of decimal

If the factors of denominator of the given rational number is the form 2ⁿ 5^m , where n and m are non-negative integers, then the decimal expansion of the rational number is terminating otherwise non-terminating recurring.
Ans . Non-terminating recurring decimal expansion.

Given, HCF (150, 100) = 50
\(\therefore \ LCM(150,100)=\frac { 150\times 100 }{ HCF(150,100) } =\frac { 150\times 100 }{ 50 } =300\) 

Given, LCM (26, 91) = 182
\(\therefore \quad HCF(26,91)=\frac { 26\times 91 }{ LCM(26,91) } =\frac { 26\times 91 }{ 182 } =13\)

First, use Euclid's division algorithm for two larger numbers any of three and get the HCF of these two. After that take the third number and resulting HCF of two numbers and apply again Euclid's division algorithm and get the required HCF.
(i)  63  (ii) 15

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

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.