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Last Updated On September 19, 2022

Exercise 1.3 - Chapter 1 Real Numbers class 10 ncert solutions Maths - Free PDF Download 2022 - 2023

Question 1:

Prove that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_6af01362.gif is irrational.

Answer:

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_6af01362.gif is a rational number.

Therefore, we can find two integers ab (b ≠ 0) such that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_78e6dc15.gif

Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_m390f70ad.gif

Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_m1d93c50f.gifThis means that b2 is divisible by 5 and hence, b is divisible by 5.

This implies that a and b have 5 as a common factor.

And this is a contradiction to the fact that a and b are co-prime.

Hence,https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_6af01362.gifcannot be expressed as https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_65af025c.gifor it can be said thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1862/Chapter%201_html_6af01362.gif is irrational.

 

Question 2:

Prove that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_36598035.gif is irrational.

Answer:

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_36598035.gifis rational.

Therefore, we can find two integers ab (b ≠ 0) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_5959cefb.gif

Since a and b are integers, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_m18567cc4.gifwill also be rational and therefore,https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_6af01362.gifis rational.

This contradicts the fact thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_6af01362.gif is irrational. Hence, our assumption that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_36598035.gifis rational is false. Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1868/Chapter%201_html_36598035.gif is irrational.

 

Question 3:

Prove that the following are irrationals:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_38e28781.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_72800357.gif

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_4aa28d1a.gifis rational.

Therefore, we can find two integers ab (b ≠ 0) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_5b0410f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m6bef5b7c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m1e38d54f.gif is rational as a and b are integers.

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m74e5c629.gifis rational which contradicts to the fact that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m74e5c629.gif is irrational.

Hence, our assumption is false and https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_4aa28d1a.gifis irrational.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m50ec7feb.gif

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_596701c8.gif is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_1a858bcb.gif for some integers a and b

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m577b6f0f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_mffa8ec9.gifis rational as a and b are integers.

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_6af01362.gif should be rational.

This contradicts the fact thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_6af01362.gifis irrational. Therefore, our assumption that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_596701c8.gifis rational is false. Hence, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_596701c8.gif is irrational.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m2a1af84c.gif

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_aeac9b1.gif be rational.

Therefore, we can find two integers ab (b ≠ 0) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_273c8fef.gif

Since a and b are integers, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m65b553cc.gif is also rational and hence, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m74e5c629.gifshould be rational. This contradicts the fact thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_m74e5c629.gif is irrational. Therefore, our assumption is false and hence, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1873/Chapter%201_html_aeac9b1.gif is irrational.

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