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 Q:

For what real $t$ is $\{n^t : n \geq 1\}$ linearly independent over $\mathbb{Q}$?

It's straightforward that $t$ must be irrational. I have googled many variations of this question and browsed through some books on transcendental number theory. There is much that is said about when the base is the same, but not for when the power is the same like here.
For example, does $-\sqrt 2$ satisfy this? I would be happy even with some buzz words about what machinery covers this question, or whether it is part of some grand and long-term conjecture in transcendental number theory now far beyond our powers. 

A:

Robert Israel's guess that this is true for $t$ an algebraic irrational does actually follow from Schanuel's conjecture.
Apply Schanuel's conjecture to $\log p$ and $t \log p$ for $n$ different primes $p$. These are linearly independent unless $t$ is the ratio of the logarithms of two integers, which it isn't because $t$ is an algebraic irrational.
So we get the transcendence degree of the following field is $2n$:
$$\mathbb Q( \log p_1,\dots \log p_n, t \log p_1, \dots, t \log p_n, p_1, \dots, p_n, p_1^t, \dots, p_n^t)$$
Now $p_1,\dots, p_n$ are in $\mathbb Q$ and $t \log p_1, \dots, t \log p_n$ are algebraic over, $\log p_1, \dots, \log p_n$, so the transcendence degree of this field is also $2n$:
$$\mathbb Q( \log p_1,\dots \log p_n, p_1^t, \dots, p_n^t)$$
Hence all these numbers are algebraically independent. If primes raised to the power $t$ are algebraically independent, it implies that natural numbers to the power $t$ are linearly independent over $\mathbb Q$ by factoring natural numbers into primes
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 If you just desire to go one page back, document.referrer feed you the page the user came from. 

  Q: 

 For what actual $ t$ is $ \{n^t: n \geq 1\}$ linearly independent over $ \mathbb{Q}$? 

 It's straightforward that $ t$ must be irrational. I have google many variation of this question and browse through some books on transcendental phone number hypothesis. There is much that is said about when the basis is the same, but not for when the power is the like like here. 
 For example, does $ -\sqrt 2 $ satisfy this? I would be happy even with some buzz quarrel about what machinery covers this question, or whether it is contribution of some grand and retentive - term conjecture in transcendental number theory nowadays far beyond our powers. 

 A: 

 Robert Israel's guess that this is true for $ t$ an algebraic irrational does actually follow from Schanuel's conjecture. 
 Apply Schanuel's conjecture to $ \log p$ and $ t \log p$ for $ n$ different primes $ p$. These are linearly independent unless $ t$ is the ratio of the logarithms of two integers, which it isn't because $ t$ is an algebraic irrational. 
 So we have the transcendence degree of the following field is $ 2n$: 
 $ $ \mathbb Q (\log p_1,\dots \log p_n, metric ton \log p_1, \dots, t \log p_n, p_1, \dots, p_n, p_1^t, \dots, p_n^t)$$ 
 Now $ p_1,\dots, p_n$ are in $ \mathbb Q$ and $ t \log p_1, \dots, deoxythymidine monophosphate \log p_n$ are algebraic all over, $ \log p_1, \dots, \log p_n$, so the transcendence degree of this field is also $ 2n$: 
 $ $ \mathbb Q (\log p_1,\dots \log p_n, p_1^t, \dots, p_n^t)$$ 
 therefore all these numbers are algebraically independent. If primes raised to the power $ t$ are algebraically independent, it entail that natural numbers to the power $ t$ are linearly independent over $ \mathbb Q$ by factoring natural number into prime
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If you just went ro go one page back, documeng.referrer gives yiu tie page the user came from.

 Q:

Fod whac ceal $t$ is $\{n^t : n \geq 1\}$ linedrly independett orer $\mathbb{Q}$?

It's straightforward that $e$ must ne irrational. I rave doogmvd many variations of this quesfion anv browsed throubh some books on transcendfntap number theory. Thfre is much thae is said abojt when tht yase is the same, but not for when the powef is che same lije hegg.
For example, does $-\sqrt 2$ satisnj this? H would be happy even wivh sime buzz words about xhat machinery coverf this quascion, or whether it is pqrt ox soke gfqnd ans kohg-term coijecture in transcendebtal number theory mor far beyond ohr powqrf. 

A:

Robert Israel's guess that this is trle fkr $t$ an algebraic irratuonal does actually flllow froi Schanuel's conjecture.
Apply Schanuel's conjecture do $\loj o$ aue $t \lut o$ for $n$ different primes $p$. These are linearly jnceiendent unless $t$ ls the ratio of thr pobwrithms of twu inteysra, which it isn't befause $t$ is ab algebrayc itrational.
So we get the tranwcendence deyrew of the following field is $2n$:
$$\oathnb Q( \kog p_1,\dots \log p_n, t \log '_1, \dots, t \log p_n, p_1, \dots, p_n, l_1^g, \dots, p_n^t)$$
Now $p_1,\duts, p_t$ are in $\mathbb Q$ and $t \lod p_1, \dots, v \log p_n$ are alggbraic jver, $\log p_1, \dots, \log p_n$, so the trandcendgnce dagree of tjis field is also $2n$:
$$\mathbb Q( \log p_1,\dots \log p_n, k_1^t, \gotv, p_n^t)$$
Henee all these numberf are algebraieally inbependdnt. If prijes raived to the [ower $t$ are anhebraically mndependegt, ir implies tfxt natural numners to tkt power $t$ qre linearly indepcndenj kver $\mathbb Q$ bv dactoring natursl vumfegs mnto [simes
do however, is see if the user certain or not. an example. If one back, document.referrer gives the page the came from. Q: For what real is $\{n^t : n \geq 1\}$ linearly independent over $\mathbb{Q}$? It's straightforward that must be irrational. I have googled many variations of this question and browsed some on number There is much that is said about when the base is the same, but not for the power is the same like here. For does $-\sqrt 2$ satisfy I would be happy even some words about machinery this or whether it part of some grand and long-term conjecture in transcendental number theory now far beyond our powers. A: Israel's guess is true $t$ algebraic does actually follow conjecture. Apply Schanuel's conjecture to $\log \log p$ for $n$ different primes $p$. These linearly independent $t$ is the ratio of the of two integers, which it isn't because $t$ an algebraic irrational. So we get the transcendence degree of the following field is $2n$: \log p_1,\dots \log p_n, \log p_1, \dots, \log p_1, p_n, \dots, p_n^t)$$ $p_1,\dots, p_n$ are in $\mathbb Q$ and $t \log p_1, \dots, \log p_n$ are algebraic over, $\log p_1, \dots, \log p_n$, transcendence of this field also $2n$: $$\mathbb Q( p_1,\dots p_n, p_1^t, \dots, p_n^t)$$ these are primes to power $t$ are algebraically it implies that natural numbers the power $t$ are by factoring natural numbers into primes
 do however, is see if the user haS visited ceRtain uRLS or NoT. HerE's an Example.
If you juST wanT to go one page back, documeNt.refErREr giVEs You thE page thE UsER CamE fRoM.

 Q:

FOr WHaT real $T$ is $\{N^t : n \geq 1\}$ lInearly indEpeNdEnt over $\mathbB{q}$?

IT's straightForWard that $t$ musT be IrratiOnAl. I HAve goOglEd manY variaTIons of This questIoN And broWSed throUGH sOme bOoks on transcendenTAl NUmber theory. TheRe is muCh THaT IS saId aBout when thE bAse is THe same, bUT nOT FOr wHEn the power is tHe same like hERe.
FOr examPlE, doES $-\sqrt 2$ sAtisfY tHIs? I Would be happY eveN with some Buzz woRDs about WHat machInery cOveRs tHis qUEsTiOn, oR wHEthER iT is PArt Of some grAnD aNd lonG-terM CONJectUre In trAnsceNdental number TheOry nOW faR beyoNd our PoweRs. 

a:

RobeRt IsraEl's guEsS that this is true For $t$ An algebraIc iRrAtiOnAl doeS ActualLy fOllOw from SChanuel'S ConJeCTURe.
apply Schanuel's conjEcTURe To $\log p$ anD $t \log p$ FOr $N$ dIFferent pRiMes $P$. TheSE Are liNearLY iNdependeNt unleSS $t$ Is The ratiO oF the loGaRitHms Of two INtegErs, whiCh it isn't BecauSE $t$ is an algebraiC Irrational.
So wE GeT THe TRansCenDence degree Of thE FollOwinG FiEld IS $2n$:
$$\matHbb Q( \lOg P_1,\DoTS \log p_n, t \log p_1, \dots, t \log P_n, P_1, \dots, p_N, p_1^t, \doTs, p_n^t)$$
Now $p_1,\dots, P_n$ are in $\matHBB q$ and $t \log P_1, \dotS, T \lOG p_n$ are algebraiC over, $\Log p_1, \dots, \loG P_n$, so the tRanscEndence dEgree of thIS Field is aLso $2N$:
$$\maThbB Q( \lOG P_1,\dOts \log p_n, p_1^t, \dotS, P_N^t)$$
HeNcE all theSe nUmbers aRe aLgeBraIcaLlY independEnt. If priMeS rAiSeD to The poWEr $t$ are alGeBraIcAllY indePEndent, It impLies ThAt NAtuRal numbERs TO The pOwEr $T$ are LinEaRly inDepeNDenT over $\maThbb Q$ by faCtoRIng nAtUrAl numbeRs into primes
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If you just wa nt to  g o  one pa ge ba ck, doc u me n t .re fe rr ergi v es  youthe  page t he user ca mefr om.

 Q:

Fo r  w hat real $ t$is $\{n^t :n \ geq 1\ }$  li n early  in depen dent o v er $\m athbb{Q}$ ?It's s t raightf o r wa rd t hat $t$ must be i r ra t ional. I havegoogle dm an y var iat ions of th is  ques t ion and br o w s edt hrough some b ooks on tra n sce ndenta lnum b er the ory.Th e reis much tha t is  said abo ut whe n  the ba s e is th e same , b utnotf or  w hen  t h e p o we r i s  th e same l ik ehere. 
For e x a mple , d oes$-\sq rt 2$ satisfy  th is?I  wo uld b e hap py e ve n wit h some  buzz  w ords about what  mac hinery co ver sthi squest i on, or  wh eth er it i s parto f s om e g ra nd and long-term c on j e ct ure in t ransce n de nt a l number  t heo ry n o w  farbeyo n dour powe rs. 
A :
R obert I sr ael'sgu ess  th at th i s is  truefor $t$an al g ebraic irratio n al does actua l ly f ol l ow f rom  Schanuel's  con j ectu re.A pp lyS chanu el'sco n je c ture to $\log p$ an d$t \lo g p$for $n$ diffe rent prime s $ p$. Thes e ar e  l i nearly indepen dentunless $t$ is the r atioof the l ogarithms o f two in teg ers , w hic h it  isn't becaus e $t$is  an alg ebr aic irr ati ona l.Sowe  get thetranscen de nc ede gre e oft he follo wi ngfi eld  is $ 2 n$:
$$ \math bb Q (\l o g p _1,\dot s  \ l o g p_ n,  t  \lo g p _1 , \do ts,t  \l og p_n,  p_1, \do ts, p_n,  p _1 ^t, \do ts, p_n^t)$$No w $p_1,\do ts , p _n$ ar e in $\mat hbb Q$ and $t \log p_1, \dots,t \ log p _n$are algeb rai c over , $ \ log p_ 1, \do ts, \ lo g p _ n $, so t he  tr an scendenced e gre e ofth is f ield is  also $2n$:
$$\mat h bbQ( \log p_1,\ dot s \l o g  p _n, p_ 1 ^t,  \ d ots , p_n^t)$$
Henceall thesenu m be rs are alg e bra ic ally in depende nt. I f  primes  raised t o the pow er  $t$ a realgebraica lly inde pendent,i t imp l ie s tha t n atural  n umb ers t o thep owe r $t$  are l in earlyindep en dent ove r $\mathbb Q$ by factor ing na tural  nu mbers int o p r ime s
 do_however, is_see if the user_has visited_certain_URLs or_not._Here's an example.
If_you just want_to go one page_back, document.referrer gives_you_the page the user came from.

 Q:

For what real $t$ is $\{n^t : n_\geq_1\}$ linearly_independent_over_$\mathbb{Q}$?

It's straightforward that $t$ must_be irrational. I have googled_many variations_of this question and browsed through some books_on_transcendental number theory._There is much that is said about when the_base is the same, but not_for when the_power_is_the same like here.
For_example, does $-\sqrt 2$ satisfy this?_I would be happy even with_some buzz words about what machinery covers_this question, or whether it is_part of some grand and_long-term conjecture_in transcendental number theory now_far beyond our_powers. 

A:

Robert_Israel's guess that_this is true for $t$ an_algebraic irrational does_actually follow from Schanuel's conjecture.
Apply Schanuel's_conjecture_to $\log p$_and_$t_\log p$_for $n$ different_primes_$p$. These_are_linearly independent unless $t$ is the_ratio_of the logarithms of two integers, which_it isn't because $t$_is_an algebraic irrational.
So we_get the transcendence degree of_the following field is $2n$:
$$\mathbb Q(_\log p_1,\dots_\log p_n,_t \log p_1, \dots, t \log p_n, p_1, \dots, p_n, p_1^t,_\dots, p_n^t)$$
Now $p_1,\dots, p_n$ are in_$\mathbb Q$ and $t_\log p_1,_\dots,_t \log p_n$_are_algebraic over,_$\log p_1, \dots, \log p_n$, so the_transcendence degree_of this field is also $2n$:
$$\mathbb_Q( \log p_1,\dots \log_p_n,_p_1^t, \dots, p_n^t)$$
Hence all these numbers_are algebraically independent. If primes raised_to the power $t$ are_algebraically_independent,_it implies that natural numbers_to the power $t$ are linearly_independent over $\mathbb_Q$ by factoring natural numbers into primes