Proof with the cube root of 2

**MLeszega** · Jun29-09, 03:04 PM

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

It is easy to show that $LaTeX Code: \\sqrt[3]{2}$ 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 $LaTeX Code: \\sqrt[3]{2}$ is irrational it cannot be put in any form using rational numbers, but I am not sure.

Any help is appreciated, thanks in advance.

**jgens** · Jun29-09, 03:40 PM

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

http://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.

**MLeszega** · Jun29-09, 03:52 PM

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 $LaTeX Code: \\sqrt{r}$ )³, which I then expanded. It didn't seem to help though.

**jgens** · Jun29-09, 04:16 PM

First, assume that r^1/2 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?

**MLeszega** · Jun30-09, 02:00 AM

I really don't think that assuming $LaTeX Code: \\sqrt{r}$ 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

**Moo Of Doom** · Jun30-09, 03:02 AM

In this situation, one usually says "without loss of generality, assume $LaTeX Code: \\sqrt{r}$ is irrational." Here there are two cases: $LaTeX Code: \\sqrt{r}$ is rational, and $LaTeX Code: \\sqrt{r}$ is irrational, but one of these cases is trivial (since if $LaTeX Code: \\sqrt{r}$ is rational, then $LaTeX Code: p + q\\sqrt{r}$ 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."

**HallsofIvy** · Jun30-09, 06:43 AM

Originally Posted by MLeszega

I really don't think that assuming $LaTeX Code: \\sqrt{r}$ 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

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

**MLeszega** · Jun30-09, 04:20 PM

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

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

2 = p³ +3p²q $LaTeX Code: \\sqrt{r}$ + 3pq²r + q³r^3/2

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

**g_edgar** · Jun30-09, 05:12 PM

Originally Posted by MLeszega

2 = p³ +3p²q $LaTeX Code: \\sqrt{r}$ + 3pq²r + q³r^3/2

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

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

**Petek** · Jun30-09, 05:50 PM

Originally Posted by MLeszega

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

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

2 = p³ +3p²q $LaTeX Code: \\sqrt{r}$ + 3pq²r + q³r^3/2

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

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

2 = P + Q $LaTeX Code: \\sqrt{r}$

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

**MLeszega** · Jun30-09, 10:48 PM

Originally Posted by Petek

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

2 = P + Q $LaTeX Code: \\sqrt{r}$

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

Ok, so i can take my equation 2 = p³ + 3p²q $LaTeX Code: \\sqrt{r}$ +3pq²r + q³r^3/2

and rewrite it as:
2 = p³ + (3p²q + 3pq² $LaTeX Code: \\sqrt{r}$ + q³r^3/4) $LaTeX Code: \\sqrt{r}$

Set P = p³ and Q = 3p²q +3pq² $LaTeX Code: \\sqrt{r}$ + q³r^3/4

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

This is back to the original form of p + q $LaTeX Code: \\sqrt{r}$ 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 $LaTeX Code: \\sqrt{r}$ and r^3/4 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 r^1/2 and r^3/4 are irrational, thus 3pq²r^1/2 is irrational, as well as q³r^3/4, then so is 3pq²r^1/2 + q³r^3/4 and then finally that Q is irrational?

**Moo Of Doom** · Jul1-09, 02:10 AM

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

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

**jgens** · Jul1-09, 12:38 PM

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 . . .

Spoiler

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

Clearly, since 2^1/3 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 2^1/3 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.

**Petek** · Jul1-09, 02:53 PM

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

Spoiler

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

**jgens** · Jul1-09, 03:28 PM

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

**Petek** · Jul1-09, 06:44 PM

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.