Prove 5^ is irrational

We have to prove that √5 is an irrational number It can be proved using the contradiction method Assuming √5 as a rational number, i.e., can be written in the form a/b where a and b are integers with no common factors other than 1 and b is not equal to zero. √5/1 = a/b √5b = a By squaring on both sides 5b2 = a2 b2 = a2/5 .... (1) It means that 5 divides a2. It means that it also divides a a/5 = c a = 5c By squaring on both sides a2 = 25c2 Substituting the value of a2 in equation (1) 5b2 = 25c2 b2 = 5c2 Prove that √5 is irrational. b2/5 = c2 As b2 is divisible by 5, b is also divisible by 5 a and b have a common factor as 5 It contradicts the fact that a and b are coprime This has arisen due to the incorrect assumption as √5 is a rational number. Therefore, √5 is irrational. Prove that √5 is irrational. It is proved that √5 is irrational.

Let us assume that 5^ is an rational number. So we have to find co-prime in the form of p/q where q is not equals to zero. Such that 2^ = p/q So b 2^=a 2b^2 =a^ According according to 1.8 divisible by 2 squarring on both sides =2^b^2 =a^2. Substituting the values we get 2c=a Substituting for a we get 2b^2=4c^2 B^2=2c^2 So this means that 2 / b^2 square and also divide by b therefore A and B have at least common factor but this contradict ic the no common factors rather than one this contradiction has been that due to our wrong assumption. So we concluded that it is irrational