# Proof that cube roots of 2 and 3 are irrational

• I

## 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?

Mark44
Mentor
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.
Why improper? All you care about is that a/b is a rational number in its lowest terms; I.e., all common factors of a and b have been removed.
Thecla said:
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.
You claim that the odd/even and odd/odd cases don't make sense. If that's actually true, you should show why this is so. Actually, there's a better way to do this than a proof by exhaustion, which I give a hint on, below.
Thecla said:
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?
If $\frac a b$ is a cube root of 3, then it must be that $3 = \frac{a^3}{b^3} \Rightarrow 3b^3 = a^3$. From this, it must be that there is a factor of 3 on the right side of the equation. You need to come up with a good reason why this can't be true.

Delta2
Homework Helper
Gold Member
My approach will be something in between your work and what Mark suggests. You have find that
$3(4m^3+6m^2+3m)+1=3(2n^2+n)+4n^3$ (1) or equivalently that $4n^3-1=3(\lambda-k)$ is a multiple of 3. But $4n^3-1$ can be a multiple of 3 only when n=1 (a rigorous proof of this will need induction on n). After that you can plug n=1 in (1) and get an easy contradiction.

And just to say something else, the method you use which is based on an "even=odd " contradiction doesn't mean that it is necessarily has to work, sometimes it just doesn't work...

mfb
Mentor
But $4n^3-1$ can be a multiple of 3 only when n=1 (a rigorous proof of this will need induction on n). After that you can plug n=1 in (1) and get an easy contradiction.
4n3-1 is a multiple of 3 for every n=3k+1. As an example, 4*43-1 = 255 = 3*85.

Delta2
Homework Helper
Gold Member
4n3-1 is a multiple of 3 for every n=3k+1. As an example, 4*43-1 = 255 = 3*85.
Ok thanks for the correction (embarrassed :S :P), maybe then plugging n=3k+1 and doing the algebra could lead to a contradiction.

But I don't think so all I could get to was something like that 9 x odd1=odd2 which doesn't lead to a contradiction of the type odd=even...

Maybe you will find the following mini lemma useful :
For any natural number $n$, $n^3$ is a multiple of 3 implies that $n$ is a multiple 3.

<Moderator's edit: solution removed.>

Last edited by a moderator:
fresh_42
Mentor
The essential part of the standard proof is the definition of prime numbers:

A number $p$ is prime, if it isn't a unit (= not invertible) and $p\,|\,ab \Longrightarrow p\,|\,a \text{ or } p\,|\,b\,.$

Now use that both, $2$ and $3$ are prime.

Math_QED
Homework Helper
2019 Award
The essential part of the standard proof is the definition of prime numbers:

A number $p$ is prime, if it isn't a unit (= not invertible) and $p\,|\,ab \Longrightarrow p\,|\,a \text{ or } p\,|\,b\,.$

Now use that both, $2$ and $3$ are prime.
This might be the definition of prime numbers in ring theory, but I wouldn't call it standard. Rather, a prime number is a positive number that has exactly two divisors, namely 1 and itself.

fresh_42
Mentor
This might be the definition of prime numbers in ring theory, but I wouldn't call it standard. Rather, a prime number is a positive number that has exactly two divisors, namely 1 and itself.
Standard clearly referred to the proof, not the definition. Your definition describes irreducibilty. Since $\mathbb{Z}$ is a ring, the definition from ring theory is the correct one. The "school definition" is simply wrong. There is no "sometimes true" in mathematics. Prime is what makes the quotient ring an integral domain. The fact that in some rings irreducibilty and primarity is equivalent doesn't make the definitions interchangeable.

Math_QED
Homework Helper
2019 Award
This is irreducibilty. Since $\mathbb{Z}$ is a ring, the definition from ring theory is the correct one. The "school definition" is simply wrong. There is no "sometimes true" in mathematics. Prime is what makes the quotient ring an integral domain. The fact that in some rings irreducibilty and primarity is equivalent doesn't make the definitions interchangeable.
Not saying you were wrong in any way, just stating that there is a possibility that this is not the definition the OP is using, and becomes a theorem then.

fresh_42
Mentor
Not saying you were wrong in any way, just stating that there is a possibility that this is not the definition the OP is using, and becomes a theorem then.
Sorry, I'm fighting this irreducibility nonsense whenever I can. It is a theorem, not a definition and students should learn the difference as early as possible. It really drives me mad to teach students a wrong version of anything, only to admit a few years later, that the truth is a different one. This is typically for school math, at least here, and it assumes that kiddies are too stupid to cope with the exact version. Negative numbers in primary school are of similar questionable treatment.