1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof with the cube root of 2
  1. Calculating cube roots (Replies: 16)

Loading...