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(i) Classify the given number as rational or irrational with justification.\sqrt{196}.
(ii) Classify the given number as rational or irrational with justification 3\sqrt{18}
(iii) Classify the given number as rational or irrational with justification.\sqrt{\frac{9}{27}}
(iv) Classify the given number as rational or irrational with justification.\frac{\sqrt{28}}{\sqrt{343}}
(v) Classify the given number as rational or irrational with justification.-\sqrt{0\cdot 4}
(vi) Classify the given number as rational or irrational with justification.\frac{\sqrt{12}}{\sqrt{75}}
(vii) Classify the given number as rational or irrational with justification.0.5918
(viii) Classify the given number as rational or irrational with justification.\left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right )
(ix) Classify the given number as rational or irrational with justification.10.124124 ………..
(x) Classify the given number as rational or irrational with justification.1.010010001 ………….

 

 

 

 

 

 


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(i)Answer.   [Rational]
Solution.        
We have,
\sqrt{196} = 14 = \frac{14}{1} which follows rule of rational number.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.

So, \sqrt{196} is a rational number.

(ii)Answer.     [Irrational]
Solution.        
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,

3\sqrt{18}= 3\sqrt{9\times 2}
= 3\sqrt{9}\sqrt{2}
= 3\times 3\sqrt{2}= 9\sqrt{2}
So, it can be written in the form of \frac{p}{q} as \frac{9\sqrt{2}}{1} 

But we know that 9\sqrt{2} is irrational
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence, 3\sqrt{18} is an irrational number
(iii)Answer.          [Irrational]
Solution.        
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,

\sqrt{\frac{9}{27}}= \sqrt{\frac{3\times 3}{3\times 3\times 3}}
= \sqrt{\frac{1}{3}}= \frac{\sqrt{1}}{\sqrt{3}}= \frac{1}{\sqrt{3}}
So this can be written in the form of \frac{p}{q} as \frac{1}{\sqrt{3}} but we can see that \sqrt{3} (denominator) is irrational.
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence \sqrt{\frac{9}{27}} is irrational

(iv)Answer.          [Rational]
Solution.        
We have,

\frac{\sqrt{28}}{\sqrt{343}}= \frac{\sqrt{4\times 7}}{\sqrt{49\times 7}}
= \frac{2\times \sqrt{7}}{7\times \sqrt{7}}= \frac{2}{7}

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Hence \frac{\sqrt{28}}{\sqrt{343}} is a rational number.

(v)Answer.    [Irrational]
Solution.        
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
-\sqrt{0\cdot 4}= -\sqrt{\frac{4}{10}}
= -\sqrt{\frac{2}{5}}= -\frac{\sqrt{2}}{\sqrt{5}}
So, it can be written in the form of \frac{p}{q} as \frac{-\sqrt{2}}{\sqrt{5}}
But we know that both \sqrt{2},\sqrt{5} are irrational

(Irrational numbers are real numbers which cannot be represented as simple fractions.)

Hence, -\sqrt{0\cdot 4} is an irrational number


(vi)Answer.          [Rational]
Solution.        
We have,

\frac{\sqrt{12}}{\sqrt{75}}= \frac{\sqrt{4\times 3}}{\sqrt{25\times 3}}
= \frac{\sqrt{4}\sqrt{3}}{\sqrt{25}\sqrt{3}}= \frac{\sqrt{4}}{\sqrt{25}}
= \frac{2}{5}
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.

So, \frac{\sqrt{12}}{\sqrt{75}} is a rational number.
(vii)Answer.          [Rational]
Solution.        
We have,

0\cdot 5918= \frac{0\cdot 5918\times 10000}{1\times 10000}
= \frac{5918}{10000}= \frac{2959}{5000}

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.

Also we can see that 0.5918 is a terminating decimal number hence it must be rational.

So, 0.5918 is a rational number.

(viii)Answer.          [Rational]
Solution.        
We have,

\left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right )
= 1+\sqrt{5}-4-\sqrt{5}
= -3= \frac{-3}{1}
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, \left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right ) is a rational number.

(ix)Answer.       [Rational]
Solution.        

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating
itself after the decimal point. They are also called repeating decimals.
Examples: 1/3 = 0.33333…, 4/11 = 0.363636…
Now, 10.124124 ………. is a decimal expansion which is a non-terminating recurring.
So, it is a rational number.

(x)Answer.          [Irrational]
Solution.        

Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Non Recurring decimals are those decimals which do not have a particular pattern/sequence after the decimal point and it does not end.
All non-terminating non-recurring decimal numbers are irrational numbers.
1.010010001 ………. is non-terminating non-recurring decimal number, therefore it cannot be written in the form \frac{p}{q};q\neq 0,with p,q both as integers.

Thus, 1.010010001 ……….. is an irrational number.

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A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the formp/q? Give reason.

The given decimal number is 327.7881.

                      327\cdot 7081= \frac{3277081}{10000}= \frac{p}{q}
The q is 10000, therefore its prime factorization, can be written as:
q = 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5                      
  = 24 × 54 = (2 × 5)4

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Without actually performing the long division, find if 987/10500 will have terminating or non-terminating decimal expansion give reason for your answer.

We can also write this number as
:\frac{987}{10500}= \frac{3\times 7\times 47}{3\times 7\times 2^{2}\times 5^{3}}= \frac{47}{2^{2}\times 5^{3}}

= \frac{47}{2^{2}\times 5^{3}}\times \frac{2}{2}= \frac{47\times 2}{5^{3}\times 2^{3}}
= \frac{94}{\left ( 10 \right )^{3}}= \frac{94}{1000}= 0\cdot 094
Here, we have a finite number of digits after the decimal point.
 Hence it is terminating decimal expansion.
 

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Can two numbers have 18 as their HCF and 380 as then LCM? Give reasons.

As we know, LCM is a multiple of HCF
\frac{LCM}{HCF}has to be an integer.
If the given statement is true:
380 = 18n where n is any integer.
But there is no such type of n exist as we can say that 380 is not divisible by 18.
  Hence two numbers, with 18 as their HCF and 380 as their LCM, do not exist.

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Explain why 3 \times 5 \times 7 + 7 is a composite number

Composite number – the number which is divisible by a number other than one and itself.  Here, 3 x 5 x 7 + 7 = 112 and 112 is divisible by 2, 4, 7, 8, 14, 16, 28 and 56. Hence it is a composite number.

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The number 525 and 3000 are both divisible only by 3. 5, 15, 25 and 75 what is HCF (525, 3000)? Justify your answer.

The highest common divisor of any two numbers is called HCF.
Here 75 is the highest common divisor.
We can also find HCF:
Applying Euclid’s division algorithm on 525 and 3000 we get

Hence 75 is the HCF of numbers 525 and 3000

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A positive integer is of the form 3q + 1, being a natural number. Can you write its square in any form other than 3m + 1, i.e. 3m or 3m + 2 for some integer m? Justify your answer

Let us take the square of (3q + 1)                       

(3q + 1)2 = (3q) 2 + (1)2 + 2 × 3q × 1

 = 9q2 + 1 + 6q

= 9q2 + 6q + 1                                      

= 3(3q2 + 2q) + 1                     

= 3m + 1                                                  

{where m = 3q2 + 2q}

Hence, the square of 3q + 1 (where q is a positive integer) cannot be
written in any form other than 3m + 1
Hence the given statement is false.

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Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer

Any positive integer can be written in the form of 3q or 3q+1 or 3q + 2.
Since square of 3q = 9q² = 3(3q²) = 3m (where 3q² = m)
Square of (3q+1) = (3q+1) ² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1
(where, m = 3q² + 2q)
Square of (3q+2) = (3q+2) ² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (where, m = 3q²+2q)

Thus, there is not any square of a positive integer, which can be written in the form of 3m + 2
Hence the given statement is false.

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The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer

The given statement is true in all the conditions.
  Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
\thereforen = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.

\Rightarrow x (x + 1) (x + 2) is divisible by 3.
Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
\thereforen = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.
\Rightarrowx (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
   \therefore x (x + 1) (x + 2) is divisible by 6.

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The product of two consecutive positive integer is divisible by 2”. Is this statement true or false”? Justify your answer.

The above statement is true in all the conditions. Consider a number P = q (q + 1)
  If q is odd then q+ 1 will be even, hence P will be divisible by 2.
If q is even then q+ 1 will be odd, but P will be divisible by 2.
For example:
let us take numbers  2 and 3: 
product is 2 x 3 = 6 which is divisible by 2
Hence given statement is true
 

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