Question

prove that root 5 is irrational number by method of contradiction

Answer 1

Suppose we want to prove that a math statement is true. Simply put, we assume that the math statement is false and then show that this will lead to a contradiction.

If it leads to a contradiction, then the statement must be true

To show that √5 is an irrational number, we will assume that it is rational


Then, we need to find a contradiction when we make this assumption

If we are going to assume that √5 is rational, then we need to understand what it means for a number to be rational



Basically, if square root of 5 is rational, it can be written as the ratio of two numbers as shown below:





Square both sides of the equation above

5 = 

x2y2


Multiply both sides by y2

5 × y2 = 

x2y2

 × y2


We get 5 × y2 = x2


{

Another important concept before we finish our proof: Prime factorization 

Answer 2

Hey user here is your answer......

----->To prove
$\sqrt{5}$
is an irrational number


◆Let the root5 is rational number

So,

◆root5=a/b ( where a & b are co-prime number)

◆(root5)(a)=(b)

◆Square both side so,

◆5a^2=b^2------(1)
So,by Theorem if b^2 divides 5 then b also divides 5, so 5 is the factor of b

Let b=5c

◆Put in equation(1)

◆5a^2=(5c)^2

◆5a^2=25c^2

◆a^2=5c^2

By theorem if a^2 divides 5 then a divides 5 so a is factor of 5

◆Here a & b both are factors of 5 that is not possible. So,this contradiction is arises due to our wrong assumption.

So,

◆root5 is irrational


Thanks☺️