- #1

- 97

- 5

## Main Question or Discussion Point

Proof by contradiction that cube root of 2 is irrational:

Assume cube root of 2 is equal to a/b where a, b are integers of an improper fraction in its lowest terns. So the can be even/odd, odd/even or odd/odd.

The only one that can make mathematical sense is even/odd. That is

##2=(2m)^3/(2n+1)^3 ##or

##2(8n^3+12n^2+6n+1)=8m^3##

Dividing both sides of the equation by two we see that the left side is always odd and the right side is always even for all m,n. That is the contradiction.

However when I try it for cube root of 3 which has to be odd/odd I get

##3=(2n+1)^3/(2m+1)^3##

or ##3(8m^3+12m^2+6m+1)=8n^3+12n^2+6n+1##

##24m^3+36m^2+18m+3=8n^3+12n^2+6n+1##

Subtracting one from both sides and dividing both sides by 2 I get

##12m^3+18m^2+9m+1=4n^3+6n^2+3n##

In this case you can't say both sides are always even or odd because of the 9m on the left side of the equation and the 3n on the right side. There is no proof by contradiction. Why doesn't it work?

Assume cube root of 2 is equal to a/b where a, b are integers of an improper fraction in its lowest terns. So the can be even/odd, odd/even or odd/odd.

The only one that can make mathematical sense is even/odd. That is

##2=(2m)^3/(2n+1)^3 ##or

##2(8n^3+12n^2+6n+1)=8m^3##

Dividing both sides of the equation by two we see that the left side is always odd and the right side is always even for all m,n. That is the contradiction.

However when I try it for cube root of 3 which has to be odd/odd I get

##3=(2n+1)^3/(2m+1)^3##

or ##3(8m^3+12m^2+6m+1)=8n^3+12n^2+6n+1##

##24m^3+36m^2+18m+3=8n^3+12n^2+6n+1##

Subtracting one from both sides and dividing both sides by 2 I get

##12m^3+18m^2+9m+1=4n^3+6n^2+3n##

In this case you can't say both sides are always even or odd because of the 9m on the left side of the equation and the 3n on the right side. There is no proof by contradiction. Why doesn't it work?