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Proof with the cube root of 2

  1. Jun 29, 2009 #1
    Hey guys. I forget where I found this problem but it goes as follows: Prove that [tex]\sqrt[3]{2}[/tex] cannot be represented in the form p+q[tex]\sqrt{r}[/tex] where p,q, and r are rational numbers.

    It is easy to show that [tex]\sqrt[3]{2}[/tex] 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 [tex]\sqrt[3]{2}[/tex] 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. jcsd
  3. Jun 29, 2009 #2

    jgens

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    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.
     
  4. Jun 29, 2009 #3
    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[tex]\sqrt{r}[/tex])3, which I then expanded. It didn't seem to help though.
     
  5. Jun 29, 2009 #4

    jgens

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    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?
     
  6. Jun 30, 2009 #5
    I really don't think that assuming [tex]\sqrt{r}[/tex] 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
     
  7. Jun 30, 2009 #6
    In this situation, one usually says "without loss of generality, assume [tex]\sqrt{r}[/tex] is irrational." Here there are two cases: [tex]\sqrt{r}[/tex] is rational, and [tex]\sqrt{r}[/tex] is irrational, but one of these cases is trivial (since if [tex]\sqrt{r}[/tex] is rational, then [tex]p + q\sqrt{r}[/tex] 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."
     
  8. Jun 30, 2009 #7

    HallsofIvy

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    But if [itex]\sqrt{r}[/itex] were rational here, that would imply [itex]\sqrt[3]{2}= p+ q\sqrt{r}[/itex] is rational which you said you knew not to be true.
     
  9. Jun 30, 2009 #8
    Ok, I think I understand what you guys are talking about now. So I will set [tex]\sqrt[3]{2}[/tex] = p + q[tex]\sqrt{r}[/tex], assume that [tex]\sqrt{r}[/tex] is irrational and show that this leads to a contradiction?

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

    2 = p3 +3p2q[tex]\sqrt{r}[/tex] + 3pq2r + q3r3/2

    Then I guess you look at the 2nd term 3p2q[tex]\sqrt{r}[/tex] because [tex]\sqrt{r}[/tex] is irrational. Would that [tex]\sqrt{r}[/tex] term make the whole RHS irrational and therefore not equal to 2? Or is it something different?
     
  10. Jun 30, 2009 #9
    Not quite. There are two irrational terms, you need to know their sum is irrational, which needs some more work.
     
  11. Jun 30, 2009 #10
    Here's a hint: Rewrite the last formula in your post in the form

    2 = P + Q[tex]\sqrt{r}[/tex]

    where P and Q are rational. Observe that r > 0 (why?) and derive a contradiction.
     
  12. Jun 30, 2009 #11
    Ok, so i can take my equation 2 = p3 + 3p2q[tex]\sqrt{r}[/tex] +3pq2r + q3r3/2

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

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

    I now have the form 2 = P + Q[tex]\sqrt{r}[/tex]

    This is back to the original form of p + q[tex]\sqrt{r}[/tex] 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 [tex]\sqrt{r}[/tex] 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?
     
  13. Jul 1, 2009 #12
    You could rewrite it as [itex]2 = \left( p^3 + 3pq^2r \right) + \left( 3p^2q + q^3r \right)\sqrt{r}[/itex], since in this case P and Q are both actually rational.

    You didn't quite do the algebra right, as you shouldn't get [itex]r^{3/4}[/itex]. Remember that [itex]r^{3/2} = \sqrt{r}^3 = \sqrt{r}\sqrt{r}\sqrt{r} = r\sqrt{r}[/itex].
     
  14. Jul 1, 2009 #13

    jgens

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    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.
     
  15. Jul 1, 2009 #14
    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.
     
  16. Jul 1, 2009 #15

    jgens

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    Bah, I suppose that's true. Should be fairly obvious why we can't have Q = 0
     
  17. Jul 1, 2009 #16
    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.
     
  18. Jul 1, 2009 #17

    jgens

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    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.
     
  19. Jul 1, 2009 #18
    No problem!
     
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