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KingGeorge
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[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 \(n\in\mathbb{N}\).
 one year ago
 one year ago
KingGeorge Group Title
[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 \(n\in\mathbb{N}\).
 one year ago
 one year ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Proof by contradiction?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
If I wrote it formally, I believe I would use a proof by contradiction.
 one year ago

ParthKohli Group TitleBest 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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oops. p^2 in the last reply*
 one year ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Not the way I would do it.
 one year ago

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
using @dominusscholae method of rational root test
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
http://openstudy.com/users/kinggeorge#/updates/5019ff7ae4b02742c0b19282
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
\(x = \sqrt{n+\sqrt[3]{n+1}} \) \(x^2 = n+\sqrt[3]{n+1} \) \(x^2n = \sqrt[3]{n+1} \) \((x^2n)^3 = n+1 \) \(x^63x^4n + 3x^2n^2  n^3 = n+1\) \(x^63nx^4 + 3n^2x^2  (n^3+n+1)\)
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
this is a polynomial in x with integer coefficients
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

dominusscholae Group TitleBest 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.....
 one year ago

dominusscholae Group TitleBest ResponseYou've already chosen the best response.0
I'd probably say that mukushla's involves lesser steps.
 one year ago

Neemo Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

dominusscholae Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

dominusscholae Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.6
@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...
 one year ago

Neemo Group TitleBest ResponseYou've already chosen the best response.2
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 \[\sqrt{n+\sqrt[3]{n+1}} \text{ is rational ,then It's an integer }\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.6
\(\text{or we can say if} \ \sqrt{n+\sqrt[3]{n+1}} \ \text{is not integer then it will be irrational...} \)
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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 \(n\in\mathbb{N}\).
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.6
actually \(n=2196\) makes it integer too...im wondering how we can find all solutions for which that expression becomes an integer
 one year ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.6
how about the proof...are u agree with our work...and the conclusion ''...is not integer then it will be irrational...''
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I agree with your work and conclusion. Once again, good job.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Impressive Stuff @KingGeorge, @mukushla, @Neemo
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.6
@KingGeorge thank u.............:) and hero too
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Now I need to come up with another problem of the month that is actually written correctly :/
 one year ago
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