Prove 3-√5 is irrational

Question

Prove 3-√5 is irrational

 

Answer 1

We need the following supporting theorem:

 

   Theorem:   If x^2 is a multiple of 5, then X is a multiple of 5

 

   Proof:

     [proof by contradiction]

 

     x^2 is a multiple of 5 is the GIVEN hypothesis.

     Suppose X is NOT a multiple of 5. 

 

    Then X = 5N + 1 for some integer N

    So X^2 = (5N +1)^2 = 25N^2 + 10N + 1  by FOIL method

 

    25 and 10 are divisible by 5 but 1 is NOT.

   So X^2 is NOT Divisible by 5, which contradicts the GIVEN

   hypothesis.

 

   [end of proof]

 

--------------------------------------

 

Now we can prove the sqrt(5) is irrational

 

The Proof is done once again by contradiction.

 

 Suppose  sqrt(5) is rational.

 The sqrt(5) can be written as a fraction, specifically a quotient of integers

 

 THen there exists two integers A and B such that

 

  sqrt(5) = A/B    AND   A and B are relatively prime; 

 

 that is the fraction A/B is in lowest terms

 

 

Squaring both sides:  5 = A^2/B^2

 

5*B^2 = A^2     <--- please label this equation ALPHA

 

So A^2 is a multiple of 5.

 

By previous theorem, A is a multiple of 5.

So A = 5S for some integer S

 

Then  5 * B^2 = A^2 <--- by equation ALPHA

 

           = (5S)^2 = 25*S^2

 

 

Dividing both sides of 5*B^2 = 25*S^2:

 

        B^2 = 5 * S^2

 

 So B ^2 is a multiple of 5.

By previous theorem, B is also a multiple of 5.

So B = 5T for some integer T.

 

So A/B = 5S/5T = S/T which reduces the fraction

to lowest terms.

 

By transitive property, sqrt(5) = A/B = S/T.

 

The argument can be repeated indefinitely,

which means the fraction will NEVER get reduced

to lowest terms.

 

This contradicts the ASSUMPTION/SUPPOSITION

that sqrt(5) can be written as a fraction in lowest terms.

 

[end of proof]