Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

[SOLVED] KingGeorge's Challenge of the Month! (this might be easier than previous challenges) Prove that \(\sqrt{n+\sqrt[3]{n+1}}\) is irrational for all \(n\in\mathbb{N}\).

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

Proof by contradiction?
If I wrote it formally, I believe I would use a proof by contradiction.
\[\Rightarrow\sqrt{n + \sqrt[3]{n + 1}} = {p \over q} \]p and q are co-prime. Square both sides.\[\Rightarrow n + \sqrt[3]{n + 1} = {p^2 \over q^2} \]\[\Rightarrow nq^2 + q^2\sqrt[3]{n + 1} = p\]I failed.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Oops. p^2 in the last reply*
I am not the best at classic proofs; this was my attempt(and I don't even know if it was correct).
Not the way I would do it.
*gives up*. Good luck, all of you geniuses who'd solve this problem. Cheers!
using @dominusscholae method of rational root test
\(x = \sqrt{n+\sqrt[3]{n+1}} \) \(x^2 = n+\sqrt[3]{n+1} \) \(x^2-n = \sqrt[3]{n+1} \) \((x^2-n)^3 = n+1 \) \(x^6-3x^4n + 3x^2n^2 - n^3 = n+1\) \(x^6-3nx^4 + 3n^2x^2 - (n^3+n+1)\)
this is a polynomial in x with integer coefficients
if this polynomial has any rational roots at all, they should be of form x = \(\pm(n^3+n+1/1), \pm(1/1)\) 1) put x =1, n = 1 in the polynomial 2) put n=1, x= 3 in the polynomial none of them satisfy the polynomial. and hence x is not rational
^Right idea, perhaps wrong explanation. You can say that \[\pm(n^3 + n + 1)\] has no factors, and then plug it into the polynomial constructed. I'll say this: This method is extremely algebra intensive.....
I'd probably say that mukushla's involves lesser steps.
I think there is a problem with the question ! Since \[\sqrt{n+\sqrt[3]{n+1}}\] is an integer for n= 0 and n=7
Lol. Perhaps a small definitional error. Still, if this is the case, the proof wouldn't matter. Maybe a rewording of the question saying all numbers of the form above are irrational :P.
WHOOPS! Did I say the proof wouldn't matter?!? XD. I think I meant to say that the fact that only n = 0 to n = 7 would make this valid wouldn't matter.
@Neemo....good Correction let \(\frac{a}{b}=\sqrt{n+\sqrt[3]{n+1}}\) \(a\) , \(b\) are positive integers and \(\gcd(a,b)=1\) . some algebra gives \(b^6 n^3-3a^2b^4n^2+(b^6+3b^2a^4)n+b^6-a^6=0\) easily u can see \(\gcd(b^6-a^6,b^6)=1\) since \(\gcd(a,b)=1\). if \(b \neq 1\) ....according to rational root theorem there is no natural root for \(n\). because \(\gcd(b^6-a^6,b^6)=1\)...this is a contradiction... so its better say if \(\sqrt{n+\sqrt[3]{n+1}}\) is not an integer then it will be irrational...
that's what I was thinking ! from the start @mukushla with same @mukushla's notations(and algebra too) \[\exists m \in \mathbb{N}-----bm=a^6\] then b divides a^6 But, gcd(a,b)=1=gcd(a^6,b) then b=1 so if \[\sqrt{n+\sqrt[3]{n+1}} \text{ is rational ,then It's an integer }\]
\(\text{or we can say if} \ \sqrt{n+\sqrt[3]{n+1}} \ \text{is not integer then it will be irrational...} \)
Thanks for pointing out 7 as a solution @neemo. I was not including 0 as part of the natural numbers which is why I wasn't worried about that. To make the rewording, show that 7 is the only rational number of the form \(\sqrt{n+\sqrt[3]{n+1}}\) where \(n\in\mathbb{N}\).
actually \(n=2196\) makes it integer wondering how we can find all solutions for which that expression becomes an integer
I guess I need to be more careful when writing these problems then. Good job for breaking the problem.
how about the proof...are u agree with our work...and the conclusion '' not integer then it will be irrational...''
I agree with your work and conclusion. Once again, good job.
Impressive Stuff @KingGeorge, @mukushla, @Neemo
@KingGeorge thank u.............:) and hero too
Now I need to come up with another problem of the month that is actually written correctly :/

Not the answer you are looking for?

Search for more explanations.

Ask your own question