# prove that 3^1/3 is irrational.

#### Habil

##### New member
I assume that it's rational to prove by contradiction.
3^1/3 = p/q where q and p are whole numbers. I get 3=p^3 /q^3. from there I say that 3q^3 = p^3 and if p is a whole number then p^3 is a perfect cube. and 3q^3 cant be a perfect cube and there I have shown that it's not rational. can you guys show me another way I don't really like the perfect cube argument.

#### pka

##### Elite Member
I assume that it's rational to prove by contradiction.
3^1/3 = p/q where q and p are whole numbers. I get 3=p^3 /q^3. from there I say that 3q^3 = p^3 and if p is a whole number then p^3 is a perfect cube. and 3q^3 cant be a perfect cube and there I have shown that it's not rational. can you guys show me another way I don't really like the perfect cube argument.
Why do you just say without proof that [imath]3q^3[/imath] cannot be a perfect cube?

• JeffM, topsquark and lev888

#### Jomo

##### Elite Member
Are you saying that 3q^3 is not a perfect cube since 3 is not a perfect cube? What is it that you are trying to prove??

#### Habil

##### New member
Sorry, i wasn't clear, since p is an integer <=>p^3 is a perfect cube and so is q with the same logic. And since 3q^3 is equal to p^3 which is a perfect cube, that means that 3q^3 is a perfect cube and så is q^3 which leads to this: if q^3 and 3q^3 are perfect cubes then 3^(1/3) is an integer. and we know that there is no integer between 1^(1/3) and 8^1/3 thas 3^1/3 is not an integer and its proof by contradiction

Again I am very messy i anywhere you see x^1/3 i mean x^(1/3)

#### JeffM

##### Elite Member
The problem with helping on proofs is that we have no idea what axioms and theorems are available to you.

#### Habil

##### New member
The problem with helping on proofs is that we have no idea what axioms and theorems are available to you.
That's an interesting thing to say, I had never thought about it like that. Any proof is welcome I am just trying to see other solutions to it.

I solved it in another way too where I used the fact that every rational number can be written as p/q where p and q have only 1 as their greatest common denominator. And from there we can go 3=(p/q)^3 >>> 3q^3=p^3 [corrected] and since q is an integer we know that p has a factor of 3 and can be written as 3q^3=(m*3)^3 where m is an integer. now we get q^3=(27m^3)/3 >>> q^3=9m^3=3*3m^3 with same logic as before we can see that if m is an integer and q is an integer then q has a factor of 3 in it and now we have shown that both q and p have a factor of 3 in them which is larger then the GCD 1 thas contradiction.

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#### Jomo

##### Elite Member
And from there we can go 3=(p/q)^3 >>> 3q^3=p................corrected

Shouldn't it be that 3q^3 = p^3. So p^3 is a multiple of 3==> p is a multiple of 3. ....

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#### Habil

##### New member
Yes, I missed putting the ^3 there you are right. And
3p^3 = p^3
should be 3q^3=p^3.