Question

prove that √5 is irrational​

Answer 1

Answer:

when u divide root of 5 you get a non terminating and non repeating decimal which proves that it is irrational. irrational numbers always have non terminating and non repeating decimal

Answer 2

Hii

here's your answer

It is known that if a prime number is a factor of a^2 then p should be a factor of a.

Also every fraction can be written in it's simplest form.

Suppose √5 is rational .

Them it can be written in the form a/b where a, b are integers with b is not equal to 0 and a,b are Co primes.

So √5 = a/b

=> 5 = a^2/b^2

=> 5b^2 = a^2…..(1)

=> 5 is a factor of a^2.

=> 5 is a factor of a.

So a = 5m for some integer m.

Using this in (1) we get

5b^2 = (5m)^2

=> 5b^2 = 25 m^2

=> b^2

= 5m^2

=> 5 is a factor of b^2

=> 5 is a factor of b

=> 5 is a common factor of a and b .

But this is a contradiction to our assumption that a and b are Co primes.

So our assumption that √5 is rational is wrong. So it should be irrational.

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