RS Aggarwal Solutions Class 10 Ex 1C

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

Sol:

Let x = 2 \(\sqrt{ 3 }\) -1 be a rational number.

x =2 \(\sqrt{ 3 }\)-1

\(\Rightarrow\) \(x ^{ 2 }\) = \(( 2 \sqrt{ 3 } – 1) ^{ 2 }\)

\(\Rightarrow\) \(x ^{ 2 }\)  = \(( 2\sqrt{ 3 } ) ^{ 2 }\) +\(1 ^{ 2 }\)\(2 ( 2\sqrt{ 3 } ) ( 1 )\)

\(\Rightarrow\) \(x ^{ 2 }\) = 12 + 1 – \(4 ( \sqrt{ 3 } )\)

\(\Rightarrow\) \(x ^{ 2 }\) – 13 =- \(4 ( \sqrt{ 3 } )\)

\(\Rightarrow\) \(\frac{ 13 – x ^{ 2 } } { 4 }\)= \(\sqrt{ 3 }\)

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

\(\Rightarrow\) 13 — \(x ^{ 2 }\)  is a rational number

\(\Rightarrow\) \(\frac{ 13 – x ^{ 2 } } { 4 }\) is a rational number

\(\Rightarrow\)  \(\sqrt{ 3 }\)  is a rational number

But \(\sqrt{ 3 }\)  is an irrational number, which is a contradiction.

Hence, our assumption is wrong.

Thus, (\(2 \sqrt{ 3 }\) – 1) is an irrational number.

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

Sol:

Let x = 4-\(5 \sqrt{ 2 }\) be a rational number.

x = 4 – \(5 \sqrt{ 2 }\)

\(\Rightarrow\) \(x ^{ 2 }\)= \(( 4 – 5 \sqrt{ 2 } ) ^{ 2 }\)

\(\Rightarrow\) \(x ^{ 2 }\) = \(4 ^{ 2 }\) + \(( 5 \sqrt{ 2 } )^{ 2 }\) – 2 ( 4 ) ( \(( 5 \sqrt{ 2 } )\))

\(\Rightarrow\) \(x ^{ 2 }\) = 16 + 50 – \(40 \sqrt{ 2 }\)

\(\Rightarrow\) \(x^{2}\) – 66 = – \(40 \sqrt{ 2 }\)

\(\frac{ 66 – x ^{ 2 } } { 40 }\) = \(\sqrt{ 2 }\)

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

\(\Rightarrow\)  66 — \(x ^{ 2 }\)  is a rational number

\(\Rightarrow\)  \(\frac{66 – x ^{ 2 } } { 40 }\)is a rational number

\(\sqrt{ 2 }\) is a rational number

But \(\sqrt{ 2 }\)  is an irrational number, which is a contradiction.

Hence, our assumption is wrong.

Thus, (4-\(5 \sqrt{ 2 }\)) is an irrational number.

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

Sol:

Let \( 5 \sqrt{ 2 }\)  is a rational number.

Therefore, \(5 \sqrt{ 2 }\) = \(\frac{ p } { q }\) ,  where p and q are some integers and HCF(p, q) = 1 ….(1)

\(\Rightarrow\)  \(5 \sqrt{ 2 }\) q = p

\(\Rightarrow\) \(( 5 \sqrt{ 2 } q ) ^{ 2 }\)= \(p ^{ 2 }\)

\(\Rightarrow\) \(2 ( 25 q ^{ 2 } )\) = \(p ^{ 2 }\)

\(\Rightarrow\) \(p ^{ 2 }\)  is divisible by 2

\(\Rightarrow\) p is divisible by 2      ………(2)

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

Therefore, \(5 \sqrt{ 2 }\)  = 2 m

\(\Rightarrow\) \(( 5 \sqrt{ 2 } q ) ^{ 2 }\)= \(2m ^{ 2 }\)

\(\Rightarrow\) \(2 ( 25 q ^{ 2 } )\) = \(4m ^{ 2 }\)

\(\Rightarrow\) 25 \(q ^{ 2 }\) = 2 \(m ^{ 2 }\)

\(\Rightarrow\) \(q ^{ 2 }\)  is divisible by 2

\(\Rightarrow\) q is divisible by 2   ………..(3)

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

Thus, \(5\sqrt{ 2 }\)  is irrational.

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

Sol:

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

Let \(\frac{ 2 }{ 7 } \sqrt{ 7 }\) is a rational number.

Therefore, \(\frac{ 2 }{ 7 } \sqrt{ 7 }\) = \(\frac{ p }{ q }\)  , where p and q are some integers and HCF(p, q) = 1 ….(1)

\(\Rightarrow\) \(2\sqrt{ 7 } q =7 p\)

\(\Rightarrow\) (\(2\sqrt{ 7 } q ) ^{ 2 } = 7 q ^{ 2 }\)

\(\Rightarrow\) \(7 ( 4 q ^{ 2 } )\) = \( 49 p ^{ 2 }\)

\(\Rightarrow\) \(4 q ^{ 2 }\) = \(7 p ^{2}\)

\(\Rightarrow\) \(q ^{2}\)  is divisible by 7

\(\Rightarrow\) q is divisible by 7   …… (2)

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

\(2 \sqrt{ 7 } q\) = 7p

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

343(\(4 m ^{ 2 }\) ) = 49 \(p ^{ 2 }\)

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

\(p ^{ 2 }\) is divisible by 7

p is divisible by 7   ……. (3)

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

Thus, \(\frac{ 2 }{ \sqrt{ 7 } }\) is irrational.

Q.5: State Euclid’s division lemma.

Sol:

Euclid’s division lemma, states that for any two positive integers  a and b, there exist unique whole numbers q and r, such that a=b x q + r  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 = \(\frac{ 1050 }{ 25 }\) = 42

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 \(\frac{a}{b}\), where a and b are co prime, and prime factorization of b is of the form (\(2^{m}\)  x\(5^{n}\) ), wherein and n are non negative integers.

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

Sol:

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

\(=\)  \(\frac{ 2 X 3 \sqrt{ 5 } + 3 X 2\sqrt{ 5} } {2 \sqrt{ 5 } }\)

\(=\) \(\frac{ 6 \sqrt{ 5 } + 6 \sqrt{ 5 } } { 2 \sqrt{ 5 } }\)

\(=\) \(\frac{ 12 \sqrt{ 5 } }{ 2 \sqrt{ 5 } }\) = 6

Thus, simplified form of   \(\frac{ 2 \sqrt{ 45 } + 3 \sqrt{ 20 } }{ 2 \sqrt{ 5 } }\) is 6.

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

Sol:

Decimal expansion:

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

= \(\frac{ 365 }{ ( 2 X 5 )^{ 4 } }\) = \(\frac{ 365 }{ ( 10 ) ^{ 4 } }\)

= \(\frac{ 365 }{ 10000 }\) = 0.0365

Thus, the decimal expansion of   \(\frac{ 73 }{ 2^{ 4 } X 5^{ 3 } }\) is 0.0365.

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

= 10n

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 – \(\sqrt{ 5 }\) and 4 + \(\sqrt{ 5 }\).

(4 – \(\sqrt{ 5 }\)) + (4 + \(\sqrt{ 5 }\)) = 8

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 \(\sqrt{ 5 }\) and 3 \(\sqrt{ 5 }\).

(4 \(\sqrt{ 5 }\)) X (3 \(\sqrt{ 5 }\) ) = 60

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 = \(0.\bar{4}\)  ………  (1)

Multiplying both sides by 10, we get

10x = \(4.\bar{4}\)       ………(2)

Subtracting (1) from (2), we get

10x – x = \(4.\bar{4}\)– 0.4

\(\Rightarrow\) 9x = 4

\(\Rightarrow\)x=\(\frac{ 4 }{ 9 }\)

Thus, simplest form of \(0.\bar{4}\) — as a rational number is \(\frac{4}{9}\).

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

Sol:

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

x = \(0.\bar{4}\).   ……. (1)

Multiplying both sides by 100, we get

100x = \(23.\bar{23}\)  ……   (2)

Subtracting (1) from (2), we get

100x – x = \(23.\bar{23}\)\(0.\bar{23}\).

\(\Rightarrow\) 99 x = 23

\(\Rightarrow\)x=\(\frac{23}{99}\)

Thus, simplest form of \(0.\bar{23}\) as a rational number is \(\frac{ 23 }{ 99 }\)

Q.24: Explain why 0.15015001500015 … 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  \(\frac{ \sqrt{ 2 } }{ 3 }\) is a rational number.

Therefore, \(\frac{ \sqrt{ 2 } }{ 3 }\) = \(\frac{ p } { q }\),

where p and q are some integers and HCF(p, q) = 1        … ….(1)

\(\Rightarrow\)  \(\sqrt{ 2 }\) q = 3p

\(\Rightarrow\) \(( \sqrt{ 2 } q ) ^{ 2 }\) = \(( 3 p ) ^{ 2 }\)

\(\Rightarrow\)  \(2 ( q ) ^{ 2 }\) = \(9 ( p ) ^{ 2 }\)

\(\Rightarrow\)  \(( p ) ^{ 2 }\) is divisible by 2

\(\Rightarrow\) p is divisible by 2         …….(2)

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

Therefore, \(\sqrt{ 2 }\) q = 3p

\(\Rightarrow\)  \(\sqrt{ 2 }\) q = 3 ( 2m )

\(\Rightarrow\)  \((\sqrt{ 2 } q ) ^ {2 }\) = \(( 3 ( 2 m ) ) ^{ 2 }\)

\(\Rightarrow\) \(2 q ^{ 2 } = 4 ( 9 p ^{ 2 } )\)

\(\Rightarrow\) \(q ^{ 2 } = 2 ( 9 p ^{ 2 } )\)

\(\Rightarrow\) \(q ^{ 2 }\)  is divisible by 2

\(\Rightarrow\) q is divisible by 2 …….(3)

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

Hence, our assumption is wrong.

Thus, \(\frac{ \sqrt{ 2 } }{ 3 }\)  is irrational.

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

Sol:

Since, \(\sqrt{ 3 }\)  = 1.732….

So, we may take 1.8 as the required rational number between \(\sqrt{ 3 }\)  and 2.

Thus, the required rational number is 1.8

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

Sol:

Since, \(3.\bar{1}\bar{4}\bar{1}\bar{6}\)  is a non-terminating repeating decimal.

Hence, is a rational number.


Practise This Question

Pick the odd pair out