Question

Prove that √5 is irrational and hence show that 3+√5 is also irrational?? ​

Answer 1

Answer:

Hope this answer helps you

Answer 2

Step-by-step explanation:

let us assume on the contrary that √5

is rational

therefore √5 = a/b where a and b are coprime Integers

a=√5b

(squaring)

a^2 = 5b^2 .................... 1

therefore a^2 is divisible be 5

thus a is also divisible by 5.......................2

a = 5c. where can is some integer

(squaring)

a^2 = 25c^2

(from statement 1)

5b^2 = 25c^2

(divide by 5)

b^2 = 5c^2

therefore b^2 us divisible by 5

thus b is also divisible by 5..................... 3

from statement 2 and 3 , a and b are not coprime

this contradicts the fact that a and b are coprime

therefore our assumption that √5 is rational is wrong

thus √5 is irrational.

now let's assume on a contrary that 3+√5 is rational

therefore 3+√5= a/b where a and b are coprime Integers

therefore √5= a/b-3

√5 = a-3b

b

since a, -3b, b are the factors

√5 is rational

but this contradicts the fact that√5 is irrational

therefore our assumption that 3+√5 is rational was wrong

therefore 3+√5 is irrational

hence proved!