2

Prove that $5^{1/3}-3^{1/4}$ is irrational.

Let $x=5^{1/3}-3^{1/4}$. I started by:

\begin{align} (x+3^{1/4})^3&=5\\ x^3+3^{3/4}+3x\cdot3^{1/4}5^{1/3}&=5\\ 3^{3/4}+3x\cdot3^{1/4}5^{1/3}&=5-x^3 \tag{1} \end{align}

Now I wished to prove by contradiction that - assuming $x$ is rational - LHS of (1) is irrational whereas RHS is rational, so the equation cannot be satisfied. However, the LHS is a sum of irrational terms, and we know that the sum of irrational terms is not always irrational.

I obviously don't wish to apply power four on (1) . Is there any other shorter method?

Gaurang Tandon
  • 6,449
  • As indicated in another comment, I think there was a typo or oversight in the text you cited (in a comment). The method the problem is based on is discussed here. I'm pretty sure that the same method I used in this answer will work here, but my feeling is that this problem is a bit more involved than I suspect the author intended. – Dave L. Renfro Sep 04 '18 at 19:51

2 Answers2

2

~An answer involving elementary field theory

If it were rational then we would have $5^{1/3} = q + 3^{1/4}$ where $q$ is a rational number. This would then imply that the field $K = \mathbb{Q}(3^{1/4})$ would contain the field $\mathbb{Q}(5^{1/3})$ which obviously cannot be the case as $[\mathbb{Q}(3^{1/4}):\mathbb{Q}] = 4$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}] = 3$ and certainly $3$ does not divide $4$.

Dionel Jaime
  • 3,908
  • I understand that the solution must be easy if we know about fields, but we have not yet been taught fields yet :P I'll keep looking for a simpler solution. – Gaurang Tandon Sep 04 '18 at 19:08
1

The minimal polynomial of $5^{1/3} - 3^{1/4}$ is (according to Wolfram Alpha) $$P(x) = x^{12} - 20 x^9 - 9 x^8 + 150 x^6 - 720 x^5 + 27 x^4 - 500 x^3 - 2250 x^2 - 540 x + 598$$ According to the rational roots theorem every rational root of $P$ is a divisor of $598$. In particular every rational root of $P$ is an integer.

You can show that $0 < 5^{1/3} - 3^{1/4} < 1$ so it is not an integer, hence not rational.

Umberto P.
  • 52,165
  • We are not allowed to use calculators in the exam, and as I mentioned earlier, I wished to avoid taking powers of 12 by hand. Still though, thanks for including a solution that may be useful to those who're allowed calculators. – Gaurang Tandon Sep 04 '18 at 19:10
  • We are not allowed to use calculators in the exam --- I suspect many people here would be interested in knowing more about "the exam". Is this for a course in field theory? If so, then @Dionel Jaime gave you an answer. (A comment you posted while I was writing this comment says "NO".) Is this for a high school algebra class? If so, then there may be a typo or the teacher simply gave this as a challenge to work on. Is this for a math contest? Or is there some reason that details about "the exam" have to be kept secret? – Dave L. Renfro Sep 04 '18 at 19:18
  • @DaveL.Renfro This is an from undergraduate college level coursebook - Advanced Calculus by Wrede dan Murray - for a first-year course on Real Analysis (for computer science majors) – Gaurang Tandon Sep 04 '18 at 19:22
  • Sure enough, this is supplementary problem 1.4.1(a). Skimming through the chapter online, it doesn't look like the text has discussed any method other than the one used by Umberto to solve the problem. I can't imagine this problem's being given on an in-class exam. It would take all the time to do it. – saulspatz Sep 04 '18 at 19:30
  • @saulspatz Thanks for the information :) – Gaurang Tandon Sep 04 '18 at 19:32
  • The problem is 1.41(a) on p. 19, which I found online. I suspect this wasn't proof-read carefully enough (or it was not sufficiently thought out), since 1.41(b) is much easier and the two previous related examples in the text (1.8 on p. 10; 1.22 on p. 14) are nowhere near as computationally involved. I would forget this example. Can you do 1.4.1(b)? Can you do this for the sum of a cube root and a square root? That's enough, in my opinion. – Dave L. Renfro Sep 04 '18 at 19:37
  • @DaveL.Renfro Yes, I can do the other questions you mentioned. Thanks for the info :) – Gaurang Tandon Sep 04 '18 at 19:40
  • 1
    Sorry about the irritation my comment showed (e.g. "some reason ... kept secret"). In some cases, I sometimes wonder whether questioners, who say they saw something somewhere, are actually working on an assignment (maybe extra hard for extra credit) or they're just saying that in order to create a sense of legitimacy to something they were just curious about but didn't want to put forth any work trying to solve. Clearly, neither of these situations applies in your case. – Dave L. Renfro Sep 04 '18 at 20:01
  • @Dave that's alright :) – Gaurang Tandon Sep 05 '18 at 03:08