Thread: Arithmetic proof

The problem is to proove that the square root of 5 is irrational via contradiction. I have gotten this far -

1.5^(1/2) is rational
2. 5^(1/2) = (a/b), a,b are coprime integers
3. 5 = (a^2/b^2)
4. 5b^2 = a^2
5. a^2 mod 5 = 0
6. a^2 = 5k, k is an integer

From here I don't know where to go. Honestly I'm not sure how I'm going to show a false statement. I understand the statements I have so far, but they are followed from a previous problem in my notes. The problem I have in my notes has this next step -

7. a^2 mod 5 = 0 -> a mod 5 = 0

Does this show a false statement? If so, I don't really see why. Can someone point me in the right direction? Thanks for any help.

Originally Posted by SterlingM

The problem is to proove that the square root of 5 is irrational via contradiction. I have gotten this far -

1.5^(1/2) is rational
2. 5^(1/2) = (a/b), a,b are coprime integers
3. 5 = (a^2/b^2)
4. 5b^2 = a^2
5. a^2 mod 5 = 0
6. a^2 = 5k, k is an integer

From here I don't know where to go. Honestly I'm not sure how I'm going to show a false statement. I understand the statements I have so far, but they are followed from a previous problem in my notes. The problem I have in my notes has this next step -

7. a^2 mod 5 = 0 -> a mod 5 = 0

Does this show a false statement? If so, I don't really see why. Can someone point me in the right direction? Thanks for any help.

a = 5m where m is an integer.

Therefore 5b^2 = 25m^2 => b^2 = 5m^2 => b mod 5 = 0 => b = 5n where n is an integer. Therefore a and b have a common factor 5 => contradiction.