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.

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