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Proof that sum of 3 roots of rationals is rational etc

  1. Apr 17, 2007 #1
    Excuse my typography - I'm new here...

    a, b, and c are rational numbers. I want to prove that

    * IF S = root(a) + root(b) + root(c) is rational THEN root(a), root(b) and root(c) are rational in themselves.

    Now I have done as follows: I reverse the problem and try to show that:

    * IF 0, 1 or 2 of root(a) etc are rational THEN the sum S must be irrational.

    If all three of the roots are rational then obviously the sum is too. This would conclude the proof.

    I can easily prove that if 1 or 2 of root(a) etc are rational, then the sum is, in fact, irrational. The trouble is that I can't prove the last case:

    * Show that IF none of root(a) root(b) and root(c) are rational, THEN their sum is irrational.

    I have tried the standard technique of assuming the sum is rational = m/n then rearranging terms and squaring successively to get rid of all root signs, then looking at parity but so far this seems to be a dead end - I can't draw ny conclusions from this.

    Could someone please help me with this - What insights do I need to attack this problem from some new angle? I need all help I can get here.
  2. jcsd
  3. Apr 17, 2007 #2


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    First, you are talking about the square root, are you not?

    Second when you say
    You mean "If 0, 1, or 2 of root(a) etc are irrational", right?

    Your statement "If all three of the roots are rational then obviously the sum is too." is irrelevant. This is not an "if and only if" theorem.

    How do you prove that?
  4. Apr 17, 2007 #3

    Nooo... I mean rational.

    If 2 of them, say root(a) and root(b) are rational then root(a) + root(b) = m/n and since the sum of a rational number and an irrational number is irrational this case is closed (I can prove it if you want, but this seems obvious).

    If only one of them is rational, say root(a) = p/q , I assume the sum is rational p/q + root(b) + root(s) = m/n. So r + s is rational, say v/w. So root(b) = v/w - root(c). Square this and isolate the remaining term containing root(c) to see that it must be rational. But this is in contradiction with my assumptions, i.e. the sum is not rational.

    It is correct that I added a bit that was irrelevant, but for me, at least it added a little clarity.

    Any news on the final case?
  5. Apr 18, 2007 #4


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    S^2-a-b-c is rational so we can denote it by D
    again D^2-4(Sa+bc) is rational so let's denote it by: E:
    E/8S=sqrt(abc) is rational, now if only one of the sqrts is irrational then obviously we get a contradiction cause the others are rational.
    if 2 of them are irrational then the third one can be both irrational or rational, if the third one is rational then the other are either a number and its reciprocal or the same number, if one of them is zero then this we obviously cannot have cause then we have an irrational plus a rational number which yields a rational number.
    now if the first case happens then the third number is E/8S and we have something like a rational number r=sqrt(s)+1/sqrt(s) now sqrt(s) is irrational, so we envatually get sqrt(s)=(s+1)/r which is a contradicition.
    so this case is eliminated, the same goes for the same number, cause then we have again the same contradiction.
    so we are left we the case that the third is irrational, i leave this as an excercise to prove that we get a contradiction.

    i hope i didnt type nonsense. (-:
  6. Apr 18, 2007 #5
    No nonsense there I think. Thank you. I had got to the point where I could show that if S is rational then so is the product = sqrt(abc) but I could not se how to continue from there so I abandoned that line of inquiry, as obviously, I shouln't have.

    I'm not sure I understand this though:

    I mean if you have for example sqrt(2)*sqrt(8) this product is rational while the roots themselves are not and 2 and 8 are neither the same or reciprocal. Am I missing something here?
  7. Apr 18, 2007 #6


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    yes iknew i missed something.
    in this case we have one number which is proportional to the other, i.e if a=2 b=8, then sqrtb=2sqrta, or in the genral case we have sqrt(b)=csqrt(a) where c is a rational number.
    i believe that this summarize all the cases, but i wont be amazed if there are other cases ive neglected.
    anyway, i dont see any other more elegant proof of this.
    perhaps halls know a better and elegant proof.
    but i think that any proof should get to what i got, that E/8S=sqrt(abc), cause usually in proof of irrationality you need fractions not summations.
  8. Apr 18, 2007 #7

    I'll try and get to grips over it. I realize now that there must be a relation between the numbers a and b if the product of their roots are to be rational. Perhaps reciprocity and proportionality can be combined to one "general" case... Anyway thanks a lot for the help!
  9. Apr 18, 2007 #8


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    anyway, if someone can give any other insight to this question which is better than what iv'e given, please do post it in.
  10. Apr 18, 2007 #9
    I'm still not convinced this gets me any further.

    Concider sqrt(2) x sqrt(3) x sqrt(6). This is a product which is rational though no factors are.

    So I can show that IF S is rational THEN P (for product) is rational,
    but P being rational does NOT lead to that all roots are rational which is what I originally want to prove. So I'm back to square one (excuse pun).

    On the other hand, this is no counter proof either for

    IF S is rational THEN P is rational

    does not lead to

    IF P is rational then S is rational (as is obvious from the numerical example).

    Unfortunately this seems to tell me nothing.

    I've tried to interpret the problem geometrically as well but it just translates to showing that if the sum of three distances is rational, so are the individual distances, and I'm even more clueless there.

    So please everyone - I still need help!
  11. Apr 18, 2007 #10
    abc when expressed as a product of primes, needs an even number (positive, negative, or zero), of all prime factors
  12. Apr 19, 2007 #11


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    this last case you gave is the case where two of them is irrational and the third one is also irrational.
    in this case sqrt(c)=E/8S(sqrtab)=r/sqrt(ab) where r is rational number.
    after rearranging, you get that g=sqrt(ab) where g is again rational, now if you put it back into the equation you get that:
    after rearranging you get that square root of a is rational although we assumed it to be irrational.
    anyway, as i said this isnt a complete solution and you rightfully should await to an elegant and shorter proof than that if possible.
  13. Apr 19, 2007 #12


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    Is this a possible solution? I'm working on loop's lead.

    Given [tex] \sqrt{a} + \sqrt{b} + \sqrt{c} = S [/tex]

    [tex] \sqrt{a} + \sqrt{b} = S - \sqrt{c} [/tex]

    Squaring both sides gives:
    [tex] a + b + 2\sqrt{ab} = S^2 - 2S\sqrt{c} + c [/tex]

    Isolating roots and denoting rational sum by D:
    [tex] \sqrt{ab} + S\sqrt{c} = \frac{S^2 + c - a - b}{2} = D [/tex]
    [tex] \sqrt{ab} = D - S\sqrt{c} [/tex]
    [tex] ab = D^2 - 2DS\sqrt{c} + S^2c [/tex]

    Isolating c:

    [tex] \sqrt{c} = \frac{D^2 - ab + S^2c }{ 2DS } [/tex]

    Which means that [tex]\sqrt{c}[/tex] is rational. Since a, b and c are arbitrarily chosen, then it follows from above that their roots are all rational.
    Last edited: Apr 19, 2007
  14. Apr 19, 2007 #13
    This looks too simple to be true! I went down that lane but didn't simplify enough by using S and D which is probably why I didn't see this. It seems robust enough - I'll mull over it for a while.

    As for NeoDevins suggestion I didn't get that to work. I thought it would yield something at first but All I could see was that if sqrt(abc) is rational then then either one or all three of a, b or c have an even number of prime factors. But these can be arranged so that all three roots still are irrational, as in sqrt(2)*sqrt(3)*sqrt(6) = 6.

    Btw - How do you do the nice typography?
  15. Apr 19, 2007 #14
    Hmmm... What about your final argument:

    Does this really hold? Why can't sqrt(c) be rational while the others aren't?
  16. Apr 19, 2007 #15
    This is what I was talking about when I posted that. You don't get a contradiction here if abc has only even numbers of prime factors.

    So that gives us that:

    [tex] S \in \mathbb{Q} \Rightarrow abc = \prod_p p^{n_p} [/tex]


    [itex] p [/itex] are prime and [itex] n_p [/itex] are even integers.

    But we don't have [itex] \Leftarrow [/itex] yet.
    Last edited: Apr 19, 2007
  17. Apr 19, 2007 #16


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    jonas after you showed that sqrtc is rational then you have reduced the problem to sqrt(a)+sqrt(b)=r which is rational, and again you have:
    a=b-2sqrt(b)r+r^2 and again you have the same proof as defender gave.

    i thought it should be much simpler than what i suggested... (-:
  18. Apr 23, 2007 #17


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    You know sqrt(c) is rational. Go back to the original problem. Do the exact same algebra, only to solve for sqrt(b). You'll get exactly the same thing, only some b's and c's will be switched. So sqrt(b) is also rational. Same thing for sqrt(a)
  19. Apr 27, 2007 #18
    If [itex]\sqrt{c}[/itex] is rational it's easy to see [itex]\sqrt{a}+\sqrt{b}[/itex] must be rational. Then [itex]\sqrt{a} = r - \sqrt{b}[/itex] where r is rational, ie. [itex]a = r + b - 2r\sqrt{b}[/itex], from which it follows that [itex]\sqrt{b}[/itex] is rational, and from that it follows that [itex]\sqrt{a}[/itex] must be rational as well.
  20. Apr 27, 2007 #19
    If [itex]\sqrt{c}[/itex] is rational it's easy to see [itex]\sqrt{a}+\sqrt{b}[/itex] must be rational. Then [itex]\sqrt{a} = r - \sqrt{b}[/itex] where r is rational, ie. [itex]a = r + b - 2r\sqrt{b}[/itex], from which it follows that [itex]\sqrt{b}[/itex] is rational (and the same argument shows [itex]\sqrt{a}[/itex] must be rational as well).
  21. Apr 27, 2007 #20
    Yes, from a scattered picture an image appears - I've now been able to piece it together - rather like Data suggested. I thank all of you who contributed to this. It is remarkable how stuck you can get when you dive in the wrong end at first. It now appears rather simple, though I assure you, it didn't appear so at first.

    Just from interest, does the same hold for 4 terms? 5? etc? Or does this stop at three and if so....WHY?
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