Question

3)
Prous that root 3, root 5 and root7 are irrational number.​

Answer 1

Answer:

Let 3√5 -7 is rational no. we can find coprime a a and b ,(b isn't equal to zero) in such a way that, 3√5-7 =a/b.

3√5=a/b+7. [ HCF (a,b)=1,since a,b are coprime]

3√5=a+7b/b

√5=a+7b/3b

√5=integer + integer×integer/ integer × integer

√5= rational no.[ since a,b,3 and 7 are integers]

This contradicts the fact that √5 is irrational no.

our supposition is wrong that 3√5-7 is a rational no.

so we conclude that 3√5-7 is irrational no.

Step-by-step explanation:

Answer 2

Answer:

Answer. Let 3√5 -7 is rational no. we can find coprime a a and b ,(b isn't equal to zero) in such a way that, 3√5-7 =a/b. ... so we conclude that 3√5-7 is irrational no.