Squareroot 3 is an irrational number If a b is divisible by

Squareroot 3 is an irrational number If a - b is divisible by c, then a^n - b^n is divisible by c. If 2^n - 1 is a prime number then n is a prime number

Solution

1) Assume sqrt(3) = a/b where a, b are integers and b cannot equal to 0. (sqrt(3)= (a/b)^2 turns 3 = a^2 / b^2 and 3b^2 = a^2. we know that 3|a^2 iff 3|a. using this theorem we know that 3(b^2) = a^2 must be a multiple of 3|a. 3|a then becomes 3r = a where r is some integer. Then squaring both side 3r =a becomes 9r^2 = a^2.Since 3b^2 =a^2 that means that 3b^2 = 9r^2 which simplifies into b/r= sqrt(3).

since b/r is a rational number this is a contradiction. so sqrt(3) is irrational.

2) if a- b is divisible by c,then a^n - b^n is divisible by c or not?

let us consider some values for a, b, c,n .

let a= 5 , b=1 , c= 2 , n= 2

a-b =5 - 1 = 4

4 is divisible by 2

(5)^2 - (1)^2 = 25 - 1 = 24

24 is divisible by 2

therefore a^n - b^n is divisible by c.

3 ) if 2^n -1 is a prime number then n is a prime number

let us take n= 2

2^2 - 1 = 4 - 1 = 3

3 is a prime number but 2 is not prime.