Chapter 1: Real Numbers

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Q
Real Numbers Solutions

Question:

Prove that the following are irrationals : (i) 1/โˆš2 (ii) 7โˆš5 (iii) 6 + โˆš2

Answer:

  1.     frac{1}{sqrt{2}}

To prove that  frac{1}{sqrt{2}}  is an irrational number. 

Let us assume that  frac{1}{sqrt{2}}   is rational. Then  frac{1}{sqrt{2}}   =   frac{a}{b}

Since a and b are co primes with only 1 common factor and b≠0.  

⇒ frac{1}{sqrt{2}}   =   frac{a}{b}  

frac{a}{b}  is a rational number, so  sqrt{2} โ€‹ should be a rational number.

But   sqrt{2}โ€‹ is an irrational number, so it is contradicted.

Therefore,  frac{1}{sqrt{2}}  is an irrational number.

 

(ii) 7sqrt{5}              

To prove that 7sqrt{5} is an irrational number 

Let us assume that  7sqrt{5}  is rational. Then 7sqrt{5}  =  frac{a}{b}

Since a and b are co primes with only 1 common factor and b≠0.  

⇒ 7sqrt{5}  =  frac{a}{b}  

Now divide by 7 into both sides,  we get

  sqrt{5}  =  frac{a}{7b}   

Here a and b are integer so frac{a}{b}  is a rational number, so sqrt{5}โ€‹ should be a rational number.

But  sqrt{5} โ€‹ is an irrational number, so it is contradicted.
Therefore, 7sqrt{5} is an irrational number.

 

 

(iii) 6 + sqrt{2}

To prove that 6 + sqrt{2} is an irrational number 

Let us assume that 6 + sqrt{2}  is rational. Then 6 + sqrt{2}  = frac{a}{b}  

Since a and b are co primes with only 1 common factor and b≠0.  

⇒ 6 + sqrt{2}  =    frac{a}{b}

  sqrt{2} = frac{a}{b}   – 6

  sqrt{2} =    frac{a-6b}{b}

Here a and b are integer so frac{a-6b}{b}   is a rational number, so  sqrt{2} โ€‹ should be a rational number.


But  sqrt{2} โ€‹ is an irrational number, so it is contradicted.

Therefore, 6 + 
sqrt{2}  is an irrational number.

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