**Q.1: Prove that \(2 \sqrt{ 3 } – 1\) is an irrational number.**

**Sol: **

Let x = 2 \(\sqrt{ 3 }\)

x =2 \(\sqrt{ 3 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Since x is a rational number, \(x ^{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

But \(\sqrt{ 3 }\)

Hence, our assumption is wrong.

Thus, (\(2 \sqrt{ 3 }\)

**Q.2: Prove that \(\left ( 4 – 5 \sqrt{ 2 } \right )\) is an irrational number.**

**Sol:**

Let x = 4-\(5 \sqrt{ 2 }\)

x = 4 – \(5 \sqrt{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\frac{ 66 – x ^{ 2 } } { 40 }\)

Since x is a rational number, \(x ^{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\sqrt{ 2 }\)

But \(\sqrt{ 2 }\)

Hence, our assumption is wrong.

Thus, (4-\(5 \sqrt{ 2 }\)

**Q.3: Prove that \(5 \sqrt{ 2 }\) is irrational.**

**Sol:**

Let \( 5 \sqrt{ 2 }\)

Therefore, \(5 \sqrt{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Let p = 2m, where m is some integer.

Therefore, \(5 \sqrt{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong.

Thus, \(5\sqrt{ 2 }\)

**Q.4: Prove that \(\frac{ 2 }{ \sqrt{ 7 } }\) is irrational.**

**Sol:**

\(\frac{ 2 } { \sqrt{ 7 } }\)

Let \(\frac{ 2 }{ 7 } \sqrt{ 7 }\)

Therefore, \(\frac{ 2 }{ 7 } \sqrt{ 7 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Let q = 7m, where in is some integer.

\(2 \sqrt{ 7 } q\)

\([2 \sqrt{ 7 }( 7m ) ] ^{2} )\)

343(\(4 m ^{ 2 }\)

7 ( \(4 m ^{ 2 }\)

\(p ^{ 2 }\)

p is divisible by 7 Â Â â€¦â€¦. (3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong.

Thus, \(\frac{ 2 }{ \sqrt{ 7 } }\)

**Q.5: State Euclidâ€™s division lemma.**

**Sol: **

Euclid’s division lemma, states that for any two positive integers Â a and b, there exist unique whole numbers q and r, such that a=b x q + r Â where \(0\leq r< b\)

**Q.6: State fundamental theorem of arithmetic.**

**Sol:**

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

**Q.7: Express 360 as the product of its prime factors.**

**Sol:**

Prime factorization:

360 = 23 x 32 x 5

**Q.8: If a and b are two prime numbers then find HCF (a , b).**

**Sol:**

Prime factorization:

a = a

b = b

HCF = product of smallest power of each common prime factor in the numbers = 1

Thus, HCF (a, b) = 1

**Q.9: If a and b are two prime numbers then LCM (a, b).**

**Sol:**

Prime factorization:

a = a

b=b

LCM = product of greatest power of each prime factor involved in the numbers = a x b

Thus, LCM (a, b) = ab.

**Q.10: If the product of two numbers is 1050 and their HCF is 25, find their LCM. **

**Sol:**

HCF of two numbers = 25

Product of two numbers = 1050

Let their LCM be x.

Using the formula, Product of two numbers = HCF x LCM

We conclude that,

1050 = 25.X

X = \(\frac{ 1050 }{ 25 }\)

Hence, their LCM is 42.

**Q.11: What is a composite number?**

**Sol:**

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

**Q.12: If a and b are relatively prime then what is their HCF?**

**Sol:**

If two numbers are relatively prime then their greatest common factor will be 1.

Thus, HCF (a, b) = 1.

**Q.13: If the rational number \(\frac{ a }{ b }\) has a terminating decimal expansion, what is the condition to be satisfied by b?**

**Sol: **

Let x be a rational number whose decimal expansion terminates.

Then, we can express x in the form \(\frac{a}{b}\)

**Q.14: Simplify : \(\frac{ (2 \sqrt{ 45 } + 3 \sqrt{ 20 } ) }{ 2 \sqrt{ 5 } }\)**

**Sol: **

\(\frac{ 2\sqrt{ 45 }+ 3 \sqrt{ 20 } } { 2 \sqrt{ 5 } }\)

\(=\)

\(=\)

\(=\)

Thus, simplified form of Â Â \(\frac{ 2 \sqrt{ 45 } + 3 \sqrt{ 20 } }{ 2 \sqrt{ 5 } }\)

**Q.15: Write the decimal expansion of \(\frac{ 73 }{ ( 2 ^{ 4 } \times 5 ^{ 3 } ) }\) **

**Sol:**

Decimal expansion:

\(\frac{ 73 }{ 2 ^{ 4 } \times 5^{ 3 } }\)

= \(\frac{ 365 }{ ( 2 X 5 )^{ 4 } }\)

= \(\frac{ 365 }{ 10000 }\)

Thus, the decimal expansion of Â Â \(\frac{ 73 }{ 2^{ 4 } X 5^{ 3 } }\)

**Q.16: Show that there is no value of n for which \(\left ( 2 ^{ n } \times 5 ^{ n } \right )\) ends in 5.**

**Sol:**

We can write:

(2″ x 5″) = (2 x 5)^{n}

= 10^{n }

For any value of n, we get 0 in the end.

Thus, there is no value of n for which (2″ x 5″) ends in 5.

**Q.17: Is it possible to have two numbers whose HCF is 25 and LCM is 520?**

**Sol:**

No, it is not possible to have two numbers whose I-ICF is 25 and LCM is 520.

Since HCF must be a factor of LCM, but 25 is not a factor of 520.

**Q.18: Give an example of two irrationals whose sum is rational. **

**Sol:**

Let the two irrationals be 4 – \(\sqrt{ 5 }\)

(4 – \(\sqrt{ 5 }\)

Thus, sum (i.e., 8) is a rational number.

**Q.19: Give an example of two irrationals whose product is rational.**

**Sol:**

Let the two irrationals be 4 \(\sqrt{ 5 }\)

(4 \(\sqrt{ 5 }\)

Thus, sum (i.e., 60) is a rational number.

**Q.20: If a and b are relatively prime, what is their LCM?**

**Sol:**

If two numbers are relatively prime then their greatest common factor will be 1.

HCF (a, b) = 1

Using the formula, Product of two numbers = HCF x LCM we conclude that,

a x b = 1 x LCM

LCM = ab

Thus, LCM (a, b) is ab.

**Q.21: The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?**

**Sol: **

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500.

Since HCF must be a factor of LCM, but 500 is not a factor of 1200.

**Q.22: Express \(0.\bar{4}\) as a rational number in simplest form.**

**Sol:**

Let x be \(0.\bar{4}\)

x = \(0.\bar{4}\)

Multiplying both sides by 10, we get

10x = \(4.\bar{4}\)

Subtracting (1) from (2), we get

10x â€“ x = \(4.\bar{4}\)

\(\Rightarrow\)

\(\Rightarrow\)

Thus, simplest form of \(0.\bar{4}\)

**Q.23: Express \(0.\bar{ 23 }\) as a rational number in simplest form. **

**Sol:**

Let x be \(0.\bar{23}\)

x = \(0.\bar{4}\)

Multiplying both sides by 100, we get

100x = \(23.\bar{23}\)

Subtracting (1) from (2), we get

100x â€“ x = \(23.\bar{23}\)

\(\Rightarrow\)

\(\Rightarrow\)

Thus, simplest form of \(0.\bar{23}\)

**Q.24: Explain why 0.15015001500015 â€¦ is an irrational number.**

**Sol: **

Irrational numbers are non-terminating non-recurring decimals.

Thus, 0.15015001500015 … is an irrational number.

**Q.25: Show that \(\frac{ \sqrt{ 2 } }{ 3 }\) is irrational.**

**Sol:**

Let Â \(\frac{ \sqrt{ 2 } }{ 3 }\)

Therefore, \(\frac{ \sqrt{ 2 } }{ 3 }\)

where p and q are some integers and HCF(p, q) = 1 Â Â Â Â Â Â Â â€¦ ….(1)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Let p = 2m, where in is some integer.

Therefore, \(\sqrt{ 2 }\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).

Hence, our assumption is wrong.

Thus, \(\frac{ \sqrt{ 2 } }{ 3 }\)

**Q.26: Write a rational number between \(\sqrt{ 3 }\) and 2.**

**Sol:**

Since, \(\sqrt{ 3 }\)

So, we may take 1.8 as the required rational number between \(\sqrt{ 3 }\)

Thus, the required rational number is 1.8

**Q.27: Explain why \(3.\overline{1416}\) is a rational number.**

**Sol: **

Since, \(3.\bar{1}\bar{4}\bar{1}\bar{6}\)

Hence, is a rational number.