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| | It is very common to have a dental emergency -- a fractured tooth, an abscess, or severe pain when chewing. Over-the-counter pain medication is just masking the problem. Seeing an emergency dentist is critical to getting the source of the problem diagnosed and corrected as soon as possible.<br><br>Here are some common dental emergencies:<br>Toothache: The most common dental emergency. This generally means a badly decayed tooth. As the pain affects the tooth's nerve, treatment involves gently removing any debris lodged in the cavity being careful not to poke deep as this will cause severe pain if the nerve is touched. Next rinse vigorously with warm water. Then soak a small piece of cotton in oil of cloves and insert it in the cavity. This will give temporary relief until a dentist can be reached.<br><br>At times the pain may have a more obscure location such as decay under an old filling. As this can be only corrected by a dentist there are two things you can do to help the pain. Administer a pain pill (aspirin or some other analgesic) internally or dissolve a tablet in a half glass (4 oz) of warm water holding it in the mouth for several minutes before spitting it out. DO NOT PLACE A WHOLE TABLET OR ANY PART OF IT IN THE TOOTH OR AGAINST THE SOFT GUM TISSUE AS IT WILL RESULT IN A NASTY BURN.<br><br>Swollen Jaw: This may be caused by several conditions the most probable being an abscessed tooth. In any case the treatment should be to reduce pain and swelling. An ice pack held on the outside of the jaw, (ten minutes on and ten minutes off) will take care of both. If this does not control the pain, an analgesic tablet can be given every four hours.<br><br>Other Oral Injuries: Broken teeth, cut lips, bitten tongue or lips if severe means a trip to a dentist as soon as possible. In the mean time rinse the mouth with warm water and place cold compression the face opposite the injury. If there is a lot of bleeding, apply direct pressure to the bleeding area. If bleeding does not stop get patient to the emergency room of a hospital as stitches may be necessary.<br><br>Prolonged Bleeding Following Extraction: Place a gauze pad or better still a moistened tea bag over the socket and have the patient bite down gently on it for 30 to 45 minutes. The tannic acid in the tea seeps into the tissues and often helps stop the bleeding. If bleeding continues after two hours, call the dentist or take patient to the emergency room of the nearest hospital.<br><br>Broken Jaw: If you suspect the patient's jaw is broken, bring the upper and lower teeth together. Put a necktie, handkerchief or towel under the chin, tying it over the head to immobilize the jaw until you can get the patient to a dentist or the emergency room of a hospital.<br><br>Painful Erupting Tooth: In young children teething pain can come from a loose baby tooth or from an erupting permanent tooth. Some relief can be given by crushing a little ice and wrapping it in gauze or a clean piece of cloth and putting it directly on the tooth or gum tissue where it hurts. The numbing effect of the cold, along with an appropriate dose of aspirin, usually provides temporary relief.<br><br>In young adults, an erupting 3rd molar (Wisdom tooth), especially if it is impacted, can cause the jaw to swell and be quite painful. Often the gum around the tooth will show signs of infection. Temporary relief can be had by giving aspirin or some other painkiller and by dissolving an aspirin in half a glass of warm water and holding this solution in the mouth over the sore gum. AGAIN DO NOT PLACE A TABLET DIRECTLY OVER THE GUM OR CHEEK OR USE THE ASPIRIN SOLUTION ANY STRONGER THAN RECOMMENDED TO PREVENT BURNING THE TISSUE. The swelling of the jaw can be reduced by using an ice pack on the outside of the face at intervals of ten minutes on and ten minutes off.<br><br>If you loved this short article and you would such as to obtain more information pertaining to [http://www.youtube.com/watch?v=90z1mmiwNS8 Dentists in DC] kindly see our own page. |
| In [[mathematics]], the '''Lindemann–Weierstrass theorem''' is a result that is very useful in establishing the [[transcendental number|transcendence]] of numbers. It states that if α<sub>1</sub>, ..., α<sub>''n''</sub> are [[algebraic number]]s which are [[linearly independent]] over the [[rational number]]s '''Q''', then ''e''<sup>α<sub>1</sub></sup>, ..., ''e''<sup>α<sub>''n''</sub></sup> are [[algebraically independent]] over '''Q'''; in other words the [[extension field]] '''Q'''(''e''<sup>α<sub>1</sub></sup>, ..., ''e''<sup>α<sub>''n''</sub></sup>) has [[transcendence degree]] ''n'' over '''Q'''.
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| An equivalent formulation {{Harv|Baker|1975|loc=Chapter 1, Theorem 1.4}}, is the following: If α<sub>1</sub>, ..., α<sub>''n''</sub> are distinct algebraic numbers, then the exponentials ''e''<sup>α<sub>1</sub></sup>, ..., ''e''<sup>α<sub>''n''</sub></sup> are linearly independent over the algebraic numbers. This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over the '''Q''' by using the fact that a [[symmetric polynomial]] whose arguments are all [[algebraic conjugate|conjugates]] of one another gives a rational number.
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| The theorem is named for [[Ferdinand von Lindemann]] and [[Karl Weierstrass]]. Lindemann proved in 1882 that ''e''<sup>α</sup> is transcendental for every non-zero algebraic number α, thereby establishing that [[pi|π]] is transcendental (see below). Weierstrass proved the above more general statement in 1885.
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| The theorem, along with the [[Gelfond–Schneider theorem]], is extended by [[Baker's theorem]], and all of these are further generalized by [[Schanuel's conjecture]].
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| ==Naming convention==
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| The theorem is also known variously as the '''Hermite–Lindemann theorem''' and the '''Hermite–Lindemann–Weierstrass theorem'''. [[Charles Hermite]] first proved the simpler theorem where the α<sub>''i''</sub> exponents are required to be [[rational integer]]s and linear independence is only assured over the rational integers,<ref>''Sur la fonction exponentielle'', Comptes Rendus Acad. Sci. Paris, '''77''', pages 18–24, 1873.</ref> a result sometimes referred to as Hermite's theorem.<ref>A.O.Gelfond, ''Transcendental and Algebraic Numbers'', translated by Leo F. Boron, Dover Publications, 1960.</ref> Although apparently a rather special case of the above theorem, the general result can be reduced to this simpler case. Lindemann was the first to allow algebraic numbers into Hermite's work in 1882.<ref>''Über die Ludolph'sche Zahl'', Sitzungsber. Königl. Preuss. Akad. Wissensch. zu Berlin, '''2''', pages 679–682, 1882.</ref> Shortly afterwards Weierstrass obtained the full result,<ref>''Zu Hrn. Lindemanns Abhandlung: 'Über die Ludolph'sche Zahl' '', Sitzungber. Königl. Preuss. Akad. Wissensch. zu Berlin, '''2''', pages 1067–1086, 1885</ref> and further simplifications have been made by several mathematicians, most notably by [[David Hilbert]].
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| == Transcendence of ''e'' and π ==
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| The [[transcendental number|transcendence]] of [[e (mathematical constant)|''e'']] and π are direct corollaries of this theorem.
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| Suppose α is a nonzero algebraic number; then α is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {''e''<sup>α</sup>} is an algebraically independent set; or in other words ''e''<sup>α</sup> is transcendental. In particular, ''e''<sup>1</sup> = ''e'' is transcendental. (A more elementary proof that ''e'' is transcendental is outlined in the article on [[transcendental number]]s.)
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| Alternatively, by the second formulation of the theorem, if α is a nonzero algebraic number, then {0, α} is a set of distinct algebraic numbers, and so the set {''e''<sup>0</sup>, ''e''<sup>α</sup>} = {1, ''e''<sup>α</sup>} is linearly independent over the algebraic numbers and in particular ''e''<sup>α</sup> cannot be algebraic and so it is transcendental.
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| The proof that π is transcendental is [[proof by contradiction|by contradiction]]. If π were algebraic, π''i'' would be algebraic as well, and then by the Lindemann–Weierstrass theorem ''e''<sup>π''i''</sup> = −1 (see [[Euler's identity]]) would be transcendental, a contradiction.
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| A slight variant on the same proof will show that if α is a nonzero algebraic number then sin(α), cos(α), tan(α) and their [[hyperbolic function|hyperbolic]] counterparts are also transcendental.
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| == ''p''-adic conjecture ==
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| <blockquote>'''''p''-adic Lindemann–Weierstrass Conjecture.''' Suppose ''p'' is some [[prime number]] and α<sub>1</sub>, ..., α<sub>''n''</sub> are [[p-adic numbers|''p''-adic numbers]] which are algebraic and linearly independent over '''Q''', such that |α<sub>''i''</sub>|<sub>''p''</sub> < 1/''p'' for all ''i''; then the [[p-adic exponential function|''p''-adic exponential]]s exp<sub>''p''</sub>(α<sub>1</sub>), ..., exp<sub>''p''</sub>(α<sub>''n''</sub>) are ''p''-adic numbers that are algebraically independent over '''Q'''.</blockquote>
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| ==Modular conjecture==
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| An analogue of the theorem involving the [[modular function]] [[j-invariant|''j'']] was conjectured by Daniel Bertrand in 1997, and remains an open problem.<ref>Daniel Bertrand, ''Theta functions and transcendence'', The Ramanujan Journal '''1''', pages 339–350, 1997.</ref> Writing ''q'' = ''e''<sup>2π''i''τ</sup> for the [[Nome (mathematics)|nome]] and ''j''(τ) = ''J''(''q''), the conjecture is as follows. Let ''q''<sub>1</sub>, ..., ''q''<sub>''n''</sub> be non-zero algebraic numbers in the complex [[unit disc]] such that the 3''n'' numbers
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| :<math>\left \{ J(q_1), J'(q_1), J''(q_1), \ldots, J(q_n), J'(q_n), J''(q_n) \right \}</math>
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| are algebraically dependent over '''Q'''. Then there exist two indices 1 ≤ ''i'' < ''j'' ≤ ''n'' such that ''q<sub>i</sub>'' and ''q''<sub>''j''</sub> are multiplicatively dependent.
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| ==Lindemann–Weierstrass Theorem ==
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| <blockquote>'''Lindemann–Weierstrass Theorem (Baker's Reformulation).''' If ''a''<sub>1</sub>, ..., ''a''<sub>''n''</sub> are non-zero algebraic numbers, and α<sub>1</sub>, ..., α<sub>''n''</sub> are distinct algebraic numbers, then<ref>{{fr icon}}[http://nombrejador.free.fr/article/lindemann-weierstrass_ttj.htm Proof's Lindemann-Weierstrass (HTML)]</ref>
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| :<math>a_1 e^{\alpha_1} +\cdots + a_n e^{\alpha_n}\ne 0.</math></blockquote>
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| ==Proof==
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| ===Preliminary Lemmas===
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| <blockquote>'''Lemma A.''' Let ''c''(1), ..., ''c''(''r'') be non-zero integers and, for every ''k'' between 1 and ''r'', let {γ(''k'')<sub>''i''</sub>} (''i'' = 1, ..., ''m''(''k'')) be the roots of a polynomial with integer [[coefficient]]s ''T''<sub>''k''</sub>(''x'') = ''v''(''k'')''x''<sup>''m''(''k'')</sup> + ... +''u''(''k'') (with ''v''(''k''), ''u''(''k'') ≠ 0). If γ(''k'')<sub>''i''</sub>≠γ(''u'')<sub>''v''</sub> whenever (''k'',''i'')≠(''u'',''v''), then
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| : <math>c(1)\left (e^{\gamma(1)_1}+\cdots+ e^{\gamma(1)_{m(1)}} \right ) + \cdots + c(r) \left (e^{\gamma(r)_1}+\cdots+ e^{\gamma(r)_{m(r)}} \right) \ne 0.</math></blockquote> | |
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| '''Proof of Lemma A.''' To simplify the notation, let us put <math>n_0=0</math>, <math>n_i=\sum_{k=1}^i m(k)</math> (for <math>i=1,\dots,r</math>) and <math>n=n_r</math>. Let <math>\alpha_{n_i+j}=\gamma(i+1)_j</math> (for <math>0\le i<n_r</math> and <math>1\le j\le m(i)</math>). Let us also put <math>\beta_{n_i+j}=c(i+1)</math>.
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| The thesis becomes that <math>\sum_{k=1}^n \beta_k e^{\alpha_k}\neq 0</math>.
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| Let ''p'' be a [[prime number]] and define the following polynomials:
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| : <math>f_i(x) = \frac {l^{np} (x-\alpha_1)^p \cdots (x-\alpha_n)^p}{(x-\alpha_i)},</math>
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| where ''l'' is a non-zero integer such that <math>l\alpha_1,\dots,l\alpha_n</math> are all algebraic integers, and the integrals:
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| : <math>I_i(s) = \int^s_0 e^{s-x} f_i(x) \, dx.</math>
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| (Up to a factor, this is the same integral appearing in [[Transcendental number#Sketch of a proof that ''e'' is transcendental|the proof that ''e'' is a transcendental number]], where β<sub>1</sub>, ..., β<sub>''m''</sub> {{math|1==}} 1, ..., ''m''. The rest of the proof of the Lemma is analog to that proof.)
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| It can be shown by [[integration by parts]] that
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| : <math>I_i(s) = e^s \sum_{j=0}^{np-1} f_i^{(j)}(0) - \sum_{j=0}^{np-1} f_i^{(j)}(s),</math>
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| (<math>np-1</math> is the [[Degree of a polynomial|degree]] of <math>f_i</math>, and <math>f_i^{(j)}</math> is the ''j''th derivative of <math>f_i</math>). This also holds for ''s'' complex (in this case the integral has to be intended as a contour integral, for example along the straight segment from 0 to ''s'') because <math>-e^{s-x} \sum_{j=0}^{np-1} f_i^{(j)}(x)</math> is a primitive of <math>e^{s-x} f_i(x)</math>.
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| Let us consider the following sum:
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| : <math>J_i=\sum_{k=1}^n\beta_k I_i(\alpha_k)=\sum_{k=1}^n\left(\beta_k e^{\alpha_k}\sum_{j=0}^{np-1}f_i^{(j)}(0)\right)-\sum_{k=1}^n\left(\beta_k\sum_{j=0}^{np-1}f_i^{(j)}(\alpha_k)\right)=\left(\sum_{j=0}^{np-1}f_i^{(j)}(0)\right)\left(\sum_{k=1}^n \beta_k e^{\alpha_k}\right)-\sum_{k=1}^n\left(\beta_k\sum_{j=0}^{np-1}f_i^{(j)}(\alpha_k)\right)</math>
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| Suppose now that <math>\sum_{k=1}^n \beta_k e^{\alpha_k}=0</math>: we will reach a contradiction by estimating <math>|J_1\cdots J_n|</math> in two different ways.
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| We obtain <math>J_i=-\sum_{j=0}^{np-1}\sum_{k=1}^n\beta_k f_i^{(j)}(\alpha_k)</math>. Now <math>f_i^{(j)}(\alpha_k)</math> is an algebraic integer which is divisible by ''p''! for <math>j\ge p</math> and vanishes for <math>j<p</math> unless ''j''=''p''-1 and ''k''=''i'', in which case it equals <math>l^{np}(p-1)!\prod_{k\neq i}(\alpha_i-\alpha_k)^p</math>.
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| This is not divisible by ''p'' (if ''p'' is large enough) because otherwise, putting <math>\delta_i=\prod_{k\neq i}(l\alpha_i-l\alpha_k)</math> (which is an algebraic integer) and calling <math>d_i</math> the product of its conjugates, we would get that ''p'' divides <math>l^p(p-1)!d_i^p</math> (and <math>d_i</math> is a non-zero integer), so by Fermat's little theorem ''p'' would divide <math>l(p-1)!d_i</math>, which is false.
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| So <math>J_i</math> is a non-zero algebraic integer divisible by (''p''-1)!. Now
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| :<math>J_i=-\sum_{j=0}^{np-1}\sum_{t=0}^{r-1}c(t+1)\left(f_i^{(j)}(\alpha_{n_t+1})+\dots+f_i^{(j)}(\alpha_{n_{t+1}})\right).</math> | |
| Since each <math>f_i(x)</math> is obtained by dividing a fixed polynomial with integer coefficients by <math>(x-\alpha_i)</math>, it is of the form <math>f_i(x)=\sum_{m=0}^{np-1}g_m(\alpha_i)x^m</math>, where <math>g_m</math> is a polynomial (with integer coefficients) independent of ''i''. The same holds for the derivatives <math>f_i^{(j)}(x)</math>.
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| Hence, by the fundamental theorem of symmetric polynomials, <math>f_i^{(j)}(\alpha_{n_t+1})+\dots+f_i^{(j)}(\alpha_{n_{t+1}})</math> is a fixed polynomial with integer coefficients evaluated in <math>\alpha_i</math> (this is seen by grouping the same powers of <math>\alpha_{n_t+1},\dots,\alpha_{n_{t+1}}</math> appearing in the expansion and using the fact that <math>\alpha_{n_t+1},\dots,\alpha_{n_{t+1}}</math> are a complete set of conjugates). So the same is true of <math>J_i</math>, i.e. it equals <math>G(\alpha_i)</math>, where ''G'' is a polynomial with integer coefficients which is independent of ''i''.
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| Finally <math>J_1\dots J_n=G(\alpha_1)\dots G(\alpha_n)</math> is an integer number (again by the fundamental theorem of symmetric polynomials), it is non-zero (since the <math>J_i</math>'s are) and it is divisible by <math>(p-1)!^n</math>.
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| So <math>|J_1\dots J_n|\ge(p-1)!^n</math>, but clearly <math>|I_i(\alpha_k)|\le|\alpha_k|e^{|\alpha_k|}F_i(|\alpha_k|)</math>, where ''F''<sub>''i''</sub> is the polynomial whose coefficients are the absolute values of those of ''f''<sub>''i''</sub> (this follows directly from the definition of <math>I_i(s)</math>).
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| Thus <math>|J_i|\le\sum_{k=1}^n|\beta_k\alpha_k|e^{|\alpha_k|}F_i(|\alpha_k|)</math> and so by the construction of the <math>f_i</math>'s we have <math>|J_1\dots J_n|\le C^p</math> for a sufficiently large ''C'' independent of ''p'', which contradicts the previous inequality. This proves Lemma A.
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| <blockquote>'''Lemma B.''' If ''b''(1), ..., ''b''(''n'') are non-zero integers and γ(1), ..., γ(''n''), are distinct [[algebraic number]]s, then
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| : <math>b(1)e^{\gamma(1)}+\cdots+ b(n)e^{\gamma(n)}\ne 0.</math></blockquote>
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| '''Proof of Lemma B:''' Assuming
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| :<math>b(1)e^{\gamma(1)}+\cdots+ b(n)e^{\gamma(n)}= 0,</math>
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| we will derive a contradition, thus proving Lemma B.
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| Let us choose a polynomial with integer coefficients which vanishes on all the <math>\gamma(k)</math>'s and let <math>\gamma(1),\dots,\gamma(n),\gamma(n+1),\dots,\gamma(N)</math> be all its distinct roots. Let ''b''(''n''+1)=...=''b''(''N'')=0.
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| Let us consider the product <math>\prod_{\sigma\in S_N}(b(1) e^{\gamma(\sigma(1))}+\dots+b(N) e^{\gamma(\sigma(N))})</math>. This vanishes by assumption, but by expanding it we obtain a sum of terms of the form <math>e^{h_1\gamma(1)+\dots+h_N\gamma(N)}</math> multiplied by integer coefficients.
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| Since the product is symmetric, we have that, for any <math>\tau\in S_n</math>, <math>e^{h_1\gamma(\tau(1))+\dots+h_N\gamma(\tau(N))}</math> has the same coefficient as <math>e^{h_1\gamma(1)+\dots+h_N\gamma(N)}</math>.
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| Thus (after having grouped the terms with the same exponent) we see that the set of the exponents form a complete set of conjugates and, if two terms have conjugate exponents, they are multiplied by the same coefficient.
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| So we are in the situation of Lemma A. To reach a contradiction it suffices to see that at least one of the coefficients is non-zero.
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| This is seen by equipping <math>\mathbb{C}</math> with the lexicographic order and by choosing for each factor in the product the term with non-zero coefficient which has maximum exponent according to this ordering: the product of these terms has non-zero coefficient in the expansion and does not get simplified by any other term. This proves Lemma B.
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| ===Final step===
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| We turn now to prove the theorem: Let ''a''(1), ..., ''a''(''n'') be non-zero [[algebraic number]]s, and α(1), ..., α(''n'') distinct algebraic numbers. Then let us assume that:
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| : <math>a(1)e^{\alpha(1)}+\cdots + a(n)e^{\alpha(n)} = 0.</math>
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| We will show that this leads to contradiction and thus prove the theorem.
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| The proof is very similar to that of Lemma B, except that this time the choices are made over the ''a''(''i'')'s: | |
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| For every ''i'' ∈ {1, ..., ''n''}, ''a''(''i'') is algebraic, so it is a root of a polynomial with integer coefficients, we denote its degree by ''d''(''i''). Let us denote the roots of this polynomial ''a''(''i'')<sub>1</sub>, ..., ''a''(''i'')<sub>''d''(''i'')</sub>, with ''a''(''i'')<sub>1</sub> = ''a''(''i'').
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| Let σ be a function which chooses one element from each of the sequences (1, ..., ''d''(1)), (1, ..., ''d''(2)), ..., (1, ..., ''d''(''n'')), such that for every 1 ≤ ''i'' ≤ ''n'', σ(''i'') is an integer between 1 and ''d''(''i''). Then according to our assumption:
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| : <math>\prod\nolimits_{\{\sigma\}}\left(a(1)_{\sigma(1)}e^{\alpha(1)}+\cdots+ a(n)_{\sigma(n)} e^{\alpha(n)}\right) = 0</math>
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| where the product is over all possible choices. The product vanishes because one of the choices is just σ(''i'') = 1 for all ''i'', for which the term vanishes according to our assumption above.
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| By expanding this product we get a sum of the form:
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| : <math>b(1)e^{\beta(1)}+ b(2)e^{\beta(2)}+ \cdots + b(N)e^{\beta(N)}= 0.</math>
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| for some non-zero integer ''N'', some distinct algebraic β(1), ..., β(''N'') (these are indeed algebraic because each is a sum of α's which are algebraic themselves), and ''b''(1), ..., ''b''(''N'') are polynomial in ''a''(''i'')<sub>''j''</sub> (''i'' in 1, ..., ''n'' and ''j'' in 1, ..., ''d''(''i'')) with integer coefficients.
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| Since the product is over all possible choices, each of ''b''(1), ..., ''b''(''N'') is symmetric in ''a''(''i'')<sub>1</sub>, ..., ''a''(''i'')<sub>''d''(''i'')</sub> for every ''i''; therefore each of ''b''(1), ..., ''b''(''N'') is a polynomial with integer coefficients in elementary symmetric polynomials of the sets {''a''(''i'')<sub>1</sub>, ..., ''a''(''i'')<sub>''d''(''i'')</sub>} for every ''i''. Each of the latter is a rational number (as in the proof of Lemma B).
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| Thus ''b''(1), ..., ''b''(''N'') ∈ '''Q''', and by multiplying the equation with an appropriate integer factor, we get an identical equation except that now ''b''(1), ..., ''b''(''N'') are all integers.
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| Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof.
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| Note that Lemma A is sufficient to prove that π is irrational, since otherwise we may write π = ''k''/''n'' (''k'', ''n'', integers) and then ±''i''π are the roots of ''x''<sup>2</sup> + ''k''<sup>2</sup>/''n''<sup>2</sup>; thus 2 + ''e''<sup>''i''π</sup> + ''e''<sup>−''i''π</sup> ≠ 0; but this is false.
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| Similarly, Lemma B is sufficient to prove that π is transcendental, since otherwise we would have 1 + ''e''<sup>''i''π</sup> ≠ 0.
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| == References ==
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| {{Reflist|2}}
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| == Further reading ==
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| *{{Citation|authorlink=Alan Baker (mathematician)|last=Baker|first=Alan|title=Transcendental Number Theory|publisher=Cambridge University Press|year=1975|isbn=0-521-39791-X}}
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| {{DEFAULTSORT:Lindemann-Weierstrass theorem}}
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| [[Category:Exponentials]]
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| [[Category:Number theory]]
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| [[Category:Pi]]
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| [[Category:Transcendental numbers]]
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems in number theory]]
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| [[Category:E (mathematical constant)]]
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