# Proof with the cube root of 2

1. Jun 29, 2009

### MLeszega

Hey guys. I forget where I found this problem but it goes as follows: Prove that $$\sqrt[3]{2}$$ cannot be represented in the form p+q$$\sqrt{r}$$ where p,q, and r are rational numbers.

It is easy to show that $$\sqrt[3]{2}$$ is irrational, so it cannot be put in the form m/n, where m and n are integers. However, I do not know where to go from here. I figured that since $$\sqrt[3]{2}$$ is irrational it cannot be put in any form using rational numbers, but I am not sure.

Any help is appreciated, thanks in advance.

2. Jun 29, 2009

### jgens

This seems like a homework question. If this is a homework question, you should post this in the homework forums:

https://www.physicsforums.com/forumdisplay.php?f=152

Anyway, try to prove it by assuming that 2^(1/3) can be represented in that form and derive a contradiction.

3. Jun 29, 2009

### MLeszega

It isn't a homework question, I don't go to school anymore. I did find it in a notebook that I used in school though. I just like working problems out for fun now.

Anyways, I actually did try proving it by contradiction. You get 2=(p+q$$\sqrt{r}$$)3, which I then expanded. It didn't seem to help though.

4. Jun 29, 2009

### jgens

First, assume that r1/2 is irrational (why is this a reasonable assumption?) Then, after expanding you get,

2 = p3 + 3(p2)(q)(r)1/2 + 3(p)(q)2(r) + (q)3(r)3/2

How might this expression help?

5. Jun 30, 2009

### MLeszega

I really don't think that assuming $$\sqrt{r}$$ is irrational is a reasonable assumption. There are plenty of rational numbers whose square roots are also rational numbers, i.e 4/9, 1/16 etc

6. Jun 30, 2009

### Moo Of Doom

In this situation, one usually says "without loss of generality, assume $$\sqrt{r}$$ is irrational." Here there are two cases: $$\sqrt{r}$$ is rational, and $$\sqrt{r}$$ is irrational, but one of these cases is trivial (since if $$\sqrt{r}$$ is rational, then $$p + q\sqrt{r}$$ is rational), so we don't bother with it and only look at the interesting case. This is a common phenomenon in proofs, so there is even an abbreviation for the phrase: "WLOG."

7. Jun 30, 2009

### HallsofIvy

But if $\sqrt{r}$ were rational here, that would imply $\sqrt[3]{2}= p+ q\sqrt{r}$ is rational which you said you knew not to be true.

8. Jun 30, 2009

### MLeszega

Ok, I think I understand what you guys are talking about now. So I will set $$\sqrt[3]{2}$$ = p + q$$\sqrt{r}$$, assume that $$\sqrt{r}$$ is irrational and show that this leads to a contradiction?

Would it be something like this: 2 = (p +q$$\sqrt{r}$$)3. Then you expand the RHS to get:

2 = p3 +3p2q$$\sqrt{r}$$ + 3pq2r + q3r3/2

Then I guess you look at the 2nd term 3p2q$$\sqrt{r}$$ because $$\sqrt{r}$$ is irrational. Would that $$\sqrt{r}$$ term make the whole RHS irrational and therefore not equal to 2? Or is it something different?

9. Jun 30, 2009

### g_edgar

Not quite. There are two irrational terms, you need to know their sum is irrational, which needs some more work.

10. Jun 30, 2009

### Petek

Here's a hint: Rewrite the last formula in your post in the form

2 = P + Q$$\sqrt{r}$$

where P and Q are rational. Observe that r > 0 (why?) and derive a contradiction.

11. Jun 30, 2009

### MLeszega

Ok, so i can take my equation 2 = p3 + 3p2q$$\sqrt{r}$$ +3pq2r + q3r3/2

and rewrite it as:
2 = p3 + (3p2q + 3pq2$$\sqrt{r}$$ + q3r3/4)$$\sqrt{r}$$

Set P = p3 and Q = 3p2q +3pq2$$\sqrt{r}$$ + q3r3/4

I now have the form 2 = P + Q$$\sqrt{r}$$

This is back to the original form of p + q$$\sqrt{r}$$ where we assumed that p,q and r are rational numbers. Now I can tell that Q is irrational (which would lead to a contradiction) because $$\sqrt{r}$$ and r3/4 are irrational, I just don't know how to show it mathematically.
Can you just say that if you have two irrational numbers a and b, that a + b is irrational? Like r1/2 and r3/4 are irrational, thus 3pq2r1/2 is irrational, as well as q3r3/4, then so is 3pq2r1/2 + q3r3/4 and then finally that Q is irrational?

12. Jul 1, 2009

### Moo Of Doom

You could rewrite it as $2 = \left( p^3 + 3pq^2r \right) + \left( 3p^2q + q^3r \right)\sqrt{r}$, since in this case P and Q are both actually rational.

You didn't quite do the algebra right, as you shouldn't get $r^{3/4}$. Remember that $r^{3/2} = \sqrt{r}^3 = \sqrt{r}\sqrt{r}\sqrt{r} = r\sqrt{r}$.

13. Jul 1, 2009

### jgens

Perhaps to help you finish the proof, using the equation as written by Moo Of Doom, you also know that r is rational.

The completed proof should look (roughly) something like this . . .

Suppose that 21/3 can be expressed as a number of the form, 21/3 = p + q(r)1/2 where q,p,r Є Q

Clearly, since 21/3 is irrational, (r)1/2 must also be irrational to ensure that p + q(r)1/2 is irrational.

Multiplying each side of the equation by itself three times yields the equality,

2 = p3 + 3(p)2(q)(r)1/2 + 3(p)(q)2(r) + (q)3(r)3/2

2 = [p3 + 3(p)(q)2(r))] + [3(p)2(q) + 3(r)(q)3](r)1/2

This equation is of the form 2 = P + Q(r)1/2 where P,Q,r Є Q. However, this contradicts the fact that 2 is a rational number since (r)1/2 is irrational; therefore, our initial assumption that 21/3 could be expressed in the form p + q(r)1/2 must have been incorrect. Q.E.D.

If you read the proof, it's pretty rough and the wording is awful (I guess that's what I get for trying to write it in 5 min.) but it should get the point across.

14. Jul 1, 2009

### Petek

The proof in the above spoiler is incomplete, as I'll explain in another spoiler:

If Q = 0, then you can't conclude that 2 is irrational. Therefore, assume that Q = 0 and derive a contradiction.

15. Jul 1, 2009

### jgens

Bah, I suppose that's true. Should be fairly obvious why we can't have Q = 0

16. Jul 1, 2009

### Petek

Did my last post offend you? If yes, I didn't mean to do so. In any event, the case Q = 0 needs to be considered, even though it's not all that difficult.

17. Jul 1, 2009

### jgens

Sorry, I didn't mean to convey that message. You're absolutely right, one should prove that case. I was just frustrated that I missed that piece of the proof.

18. Jul 1, 2009

No problem!