A community for students.
Here's the question you clicked on:
 0 viewing
KingGeorge
 4 years ago
[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 \(n\in\mathbb{N}\).
KingGeorge
 4 years ago
[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 \(n\in\mathbb{N}\).

This Question is Closed

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Proof by contradiction?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0If I wrote it formally, I believe I would use a proof by contradiction.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[\Rightarrow\sqrt{n + \sqrt[3]{n + 1}} = {p \over q} \]p and q are coprime. 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.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Oops. p^2 in the last reply*

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I am not the best at classic proofs; this was my attempt(and I don't even know if it was correct).

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Not the way I would do it.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0*gives up*. Good luck, all of you geniuses who'd solve this problem. Cheers!

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1using @dominusscholae method of rational root test

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/users/kinggeorge#/updates/5019ff7ae4b02742c0b19282

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1\(x = \sqrt{n+\sqrt[3]{n+1}} \) \(x^2 = n+\sqrt[3]{n+1} \) \(x^2n = \sqrt[3]{n+1} \) \((x^2n)^3 = n+1 \) \(x^63x^4n + 3x^2n^2  n^3 = n+1\) \(x^63nx^4 + 3n^2x^2  (n^3+n+1)\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1this is a polynomial in x with integer coefficients

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1if 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

dominusscholae
 4 years ago
Best ResponseYou've already chosen the best response.0^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.....

dominusscholae
 4 years ago
Best ResponseYou've already chosen the best response.0I'd probably say that mukushla's involves lesser steps.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think there is a problem with the question ! Since \[\sqrt{n+\sqrt[3]{n+1}}\] is an integer for n= 0 and n=7

dominusscholae
 4 years ago
Best ResponseYou've already chosen the best response.0Lol. 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.

dominusscholae
 4 years ago
Best ResponseYou've already chosen the best response.0WHOOPS! 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Neemo....good catch...lol.... Correction let \(\frac{a}{b}=\sqrt{n+\sqrt[3]{n+1}}\) \(a\) , \(b\) are positive integers and \(\gcd(a,b)=1\) . some algebra gives \(b^6 n^33a^2b^4n^2+(b^6+3b^2a^4)n+b^6a^6=0\) easily u can see \(\gcd(b^6a^6,b^6)=1\) since \(\gcd(a,b)=1\). if \(b \neq 1\) ....according to rational root theorem there is no natural root for \(n\). because \(\gcd(b^6a^6,b^6)=1\)...this is a contradiction... so its better say if \(\sqrt{n+\sqrt[3]{n+1}}\) is not an integer then it will be irrational...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that'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 \[\sqrt{n+\sqrt[3]{n+1}} \text{ is rational ,then It's an integer }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\text{or we can say if} \ \sqrt{n+\sqrt[3]{n+1}} \ \text{is not integer then it will be irrational...} \)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks 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 \(n\in\mathbb{N}\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually \(n=2196\) makes it integer too...im wondering how we can find all solutions for which that expression becomes an integer

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0I guess I need to be more careful when writing these problems then. Good job for breaking the problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how about the proof...are u agree with our work...and the conclusion ''...is not integer then it will be irrational...''

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0I agree with your work and conclusion. Once again, good job.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Impressive Stuff @KingGeorge, @mukushla, @Neemo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge thank u.............:) and hero too

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Now I need to come up with another problem of the month that is actually written correctly :/
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.