Prove that following numbers are irrational
1) √2
II) √3
III) √7
Answers
Answer:
Let us assume 1/√2 is a rational number
Let us assume 1/√2 = r where r is a rational number
On further calculation we get
1/r = √2
Since r is a rational number, 1/r = √2 is also a rational number
But we know that √2 is an irrational number
So our supposition is wrong.
Hence, 1/√2 is an irrational number.
(ii) 7√5
Solution:
Let’s assume on the contrary that 7√5 is a rational number. Then, there exist positive integers a and b such that
7√5 = a/b where, a and b, are co-primes
⇒ √5 = a/7b
⇒ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]
This contradicts the fact that √5 is irrational. So, our assumption is incorrect.
Hence, 7√5 is an irrational number.
(iii) 6 + √2
Solution:
Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist co prime positive integers a and b such that
6 + √2 = a/b
⇒ √2 = a/b – 6
⇒ √2 = (a – 6b)/b
⇒ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number]
This contradicts the fact that √2 is irrational. So, our assumption is incorrect.
Hence, 6 + √2 is an irrational number.
Answer:
i hope it will help you good luck
Step-by-step explanation:
root 2
let root 2 is a rational number
root 2= a/b ( where a and b are coprime number )
squaring both side
2 = a2/b2
b2= a2/2
2 divide a2 Wem than also 2 divide a
2c put the value in equation 1,
b2 = 2c2/ 2
b2 = 2c2
b2 / 2=c2
2 divide b2 , than 2 divide b
thus 2 is a common factor of a and b but this contradic the fact that a and b have common factor other than 1.
This contradic are due to wrong assumition hence root 2is irrational number.
all you questions solve by this mathed
you can practice other 2 questions if you are unable to do it then you can send again your question then I will solve it
but now you can do I hope you can do it quickly.
it is very important question because it will always ask in board exam that's why tb you need to clear your conseptes.
if my answer is right so please give me thank.