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KingGeorge
[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 \(n\in\mathbb{N}\).
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.
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
http://openstudy.com/users/kinggeorge#/updates/5019ff7ae4b02742c0b19282
\(x = \sqrt{n+\sqrt[3]{n+1}} \) \(x^2 = n+\sqrt[3]{n+1} \) \(x^2-n = \sqrt[3]{n+1} \) \((x^2-n)^3 = n+1 \) \(x^6-3x^4n + 3x^2n^2 - n^3 = n+1\) \(x^6-3nx^4 + 3n^2x^2 - (n^3+n+1)\)
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 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 \(b^6 n^3-3a^2b^4n^2+(b^6+3b^2a^4)n+b^6-a^6=0\) 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 too...im 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 ''...is 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 :/