Real Numbers : Exercise 1.3 (Mathematices NCERT Class 10th)



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Q.1     Prove that \sqrt 5 is irrational.
Sol.       Let us assume, to the contrary, that \sqrt 5 is rational.
             Now, let \sqrt 5 = {a \over b} , where a and b are coprime and b \ne 0. Squaring on both side, we get
             5 = {{{a^2}} \over {{b^2}}} \Rightarrow 5{b^2} = {a^2}                                                          ...(1)
             This shows that {a^2} is divisible by 5
             It follows that a is divisible by 5                                              ...(2)
              \Rightarrow a = 5m for some integer m.
             Substituting a = 5m in (1), we get
             5{b^2} = {\left( {5m} \right)^2} = 25{m^2}
             or {b^2} = 5{m^2}
              \Rightarrow {b^2} is divisible by 5
             and hence b is divisible by 5                                                   ...(3)
             From (2) and (3), we can conclude that 5 is a common factor of both a and b.
             But this contradicts our supposition that a and b are coprime.
             Hence, \sqrt 5 is irrational.


Q.2     Prove that 3 + 2\sqrt 5 is irrational.
Sol.       Let us assume, to the contrary, that 3 + 2\sqrt 5, is a rational number.
             Now, let  3 + 2\sqrt 5 = {a \over b}, where a and b are coprime and b \ne 0
              \Rightarrow 2\sqrt 5 = {a \over b} - 3 or \sqrt 5 = {a \over {2b}} - {3 \over 2}
             Since, a and b are integers.
             Therefore, {a \over {2b}} - {3 \over 2} is a rational number
              \Rightarrow \sqrt 5 is a rational number.
             But \sqrt 5 is an irrational number.
             This shows that our assumption is incorrect.
             So, 3 + 2\sqrt 5 is an irrational number .


Q.3     Prove that the following are irrationals :
(i) {1 \over {\sqrt 2 }}     (ii) 7\sqrt 5     (iii) 6 + \sqrt 2
Sol.       (i) Let us assume, to the contrary, that {1 \over {\sqrt 2 }} is rational. That is, we can find co-prime integers p and q( \ne 0) such that
              {1 \over {\sqrt 2 }} = {p \over q} \Rightarrow {{1 \times \sqrt 2 } \over {\sqrt 2 \times \sqrt 2 }} = {p \over q} \Rightarrow {{\sqrt 2 } \over { 2 }} = {p \over q}
               \Rightarrow \sqrt 2 = {{2p} \over q}
              Since p and q are integers, {{2p} \over q} is rational, and so \sqrt 2 is rational.
              But this contradicts the fact that \sqrt 2 is irrational.So, we conclude that {1 \over {\sqrt 2 }} is irrational.

             (ii) Let us assume, to the contrary, that 7\sqrt 5 is rational. 
             That is, we can find co-prime integers p and q( \ne 0) such that 7\sqrt 5 = {p \over q}
             Since p and q are integers, {p \over {7q}} is rational and so is \sqrt 5  
             But this contradicts the fact that \sqrt 5 is irrational . So, we conclude that 7\sqrt 5 is irrational.

             (iii) Let us assume, to the contrary, that \sqrt 2 is rational. That is, we can find integers p and q( \ne 0) such that
             6 + \sqrt 2 = {p \over q} \Rightarrow 6 - {p \over q} = \sqrt 2
              \Rightarrow \sqrt 2 = 6 - {p \over q}
             Since p and q are integers, we get 6 - {p \over q} is rational, and so \sqrt 2 is rational.
             But this contradicts the fact that \sqrt 2 is irrational.
             So, we conclude that 6 + \sqrt 2 is irrational

 

 



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