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| In [[algebra]], the '''rational root theorem''' (or '''rational root test''') states a constraint on rational solutions (or [[root of a function|roots]]) of the [[polynomial]] equation
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| :<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!</math>
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| with [[integer]] coefficients.
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| If ''a''<sub>0</sub> and ''a''<sub>''n''</sub> are nonzero,
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| then each [[rational number|rational]] solution ''x'',
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| when written as a fraction ''x'' = ''p''/''q'' in lowest terms (i.e., the [[greatest common divisor]] of ''p'' and ''q'' is 1), satisfies
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| * ''p'' is an integer [[divisor|factor]] of the [[constant term]] ''a''<sub>0</sub>, and
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| * ''q'' is an integer factor of the leading [[coefficient]] ''a''<sub>''n''</sub>.
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| The rational root theorem is a special case (for a single linear factor) of [[Gauss's lemma (polynomial)|Gauss's lemma]] on the factorization of polynomials. The '''integral root theorem''' is a special case of the rational root theorem if the leading coefficient ''a''<sub>''n''</sub> = 1.
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| == Proofs ==
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| ===A proof===
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| Let ''P''(''x'') = ''a''<sub>''n''</sup>''x''<sup>''n''</sup> + ''a''<sub>''n''−1</sup>''x''<sup>''n''−1</sup> + ... + ''a''<sub>1</sup>''x'' + ''a''<sub>0</sup> for some ''a''<sub>0</sub>, ..., ''a''<sub>''n''</sub> ∈ '''Z''', and suppose ''P''(''p''/''q'') = 0 for some [[coprime]] ''p'', ''q'' ∈ '''Z''':
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| :<math>P\left(\tfrac{p}{q}\right) = a_n\left(\tfrac{p}{q}\right)^n + a_{n-1}\left(\tfrac{p}{q}\right)^{n-1} + \cdots + a_1\left(\tfrac{p}{q}\right) + a_0 = 0.</math>
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| If we shift the constant term to the right hand side, factor a ''p'' and multiply by ''q''<sup>''n''</sup>, we get
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| :<math>\qquad p(a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1}) = -a_0q^n.</math>
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| We see that ''p'' times the integer quantity in parentheses equals −''a''<sub>0</sub>''q''<sup>''n''</sup>, so ''p'' divides ''a''<sub>0</sub>''q''<sup>''n''</sup>. But ''p'' is coprime to ''q'' and therefore to ''q''<sup>''n''</sup>, so by (the generalized form of) [[Euclid's lemma]] it must divide the remaining factor ''a''<sub>0</sub> of the product.
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| If we instead shift the leading term to the right hand side and multiply by ''q''<sup>''n''</sup>, we get
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| :<math>\qquad q(a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1}) = -a_np^n.</math>
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| And for similar reasons, we can conclude that ''q'' divides ''a''<sub>''n''</sub>.<ref>{{cite book|author=D. Arnold, G. Arnold|title=Four unit mathematics|publisher=Edward Arnold|year=1993|isbn=0-340-54335-3|pages=120–121}}</ref>
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| === Proof using Gauss's lemma ===
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| Should there be a nontrivial factor dividing all the coefficients of the polynomial, then one can divide by the [[greatest common divisor]] of the coefficients so as to obtain a primitive polynomial in the sense of [[Gauss's lemma (polynomial)|Gauss's lemma]]; this does not alter the set of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in {{math|ℚ[''X'']}}, then it also factors in {{math|ℤ[''X'']}} as a product of primitive polynomials. Now any rational root {{math|''p''/''q''}} corresponds to a factor of degree 1 in {{math|ℚ[''X'']}} of the polynomial, and its primitive representative is then {{math|''qx'' − p}}, assuming that ''p'' and ''q'' are coprime. But any multiple in {{math|ℤ[''X'']}} of {{math|''qx'' − p}} has leading term divisible by ''q'' and constant term divisible by ''p'', which proves the statement. This argument shows that more generally, any irreducible factor of ''P'' can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of ''P''.
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| == Example ==
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| For example, every rational solution of the equation
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| :<math>3x^3 - 5x^2 + 5x - 2 = 0\,\!</math>
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| must be among the numbers symbolically indicated by
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| :± <math>\tfrac{1,2}{1,3}\,,</math>
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| which gives the list of 8 possible answers:
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| :<math>1, -1, 2, -2, \frac{1}{3}, -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\,.</math>
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| These root candidates can be tested using the [[Horner's method]] (for instance). In this particular case there is exactly one rational root. If a root candidate does not satisfy the equation, it can be used to shorten the list of remaining candidates. For example, ''x'' = 1 does not satisfy the equation as the left hand side equals 1. This means that substituting ''x'' = 1 + ''t'' yields a polynomial in ''t'' with constant term 1, while the coefficient of ''t''<sup>3</sup> remains the same as the coefficient of ''x''<sup>3</sup>. Applying the rational root theorem thus yields the following possible roots for ''t'':
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| :<math>t=\pm\tfrac{1}{1,3}</math>
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| Therefore,
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| :<math>x = 1+t = 2, 0, \frac{4}{3}, \frac{2}{3}</math>
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| Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just ''x'' = 2 and ''x'' = 2/3.
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| If a root ''r''<sub>1</sub>
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| is found, Horner's method will also yield a polynomial of degree ''n'' − 1 whose roots, together with ''r''<sub>1</sub>, are exactly the roots of the original polynomial. It may also be the case that none of the candidates is a solution; in this case the equation has no rational solution. If the equation lacks a constant term ''a''<sub>0</sub>, then 0 is one of the rational roots of the equation.
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| ==See also==
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| *[[Descartes' rule of signs]]
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| *[[Gauss–Lucas theorem]]
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| *[[Properties of polynomial roots]]
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| *[[Content (algebra)]]
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| == Notes ==
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| <references/>
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| == References ==
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| *Charles D. Miller, Margaret L. Lial, David I. Schneider: ''Fundamentals of College Algebra''. Scott & Foresman/Little & Brown Higher Education, 3rd edition 1990, ISBN 0-673-38638-4, pp. 216–221
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| *Phillip S. Jones, Jack D. Bedient: ''The historical roots of elementary mathematics''. Dover Courier Publications 1998, ISBN 0-486-25563-8, pp. 116–117 ({{Google books|7xArILpcndYC|online copy|page=116}})
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| *Ron Larson: ''Calculus: An Applied Approach''. Cengage Learning 2007, ISBN 978-0-618-95825-2, pp. 23–24 ({{Google books|bDG7V0OV34C|online copy|page=23}})
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| ==External links==
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| *{{MathWorld|urlname=RationalZeroTheorem|title=Rational Zero Theorem}}
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| *[http://planetmath.org/encyclopedia/RationalRootTheorem.html ''RationalRootTheorem''] at [[PlanetMath]]
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| * [http://www.cut-the-knot.org/Generalization/RationalRootTheorem.shtml Another proof that n<sup>th</sup> roots of integers are irrational, except for perfect nth powers] by Scott E. Brodie
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| *[http://www.purplemath.com/modules/rtnlroot.htm ''The Rational Roots Test''] at purplemath.com
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| [[Category:Polynomials]]
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| [[Category:Theorems in algebra]]
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| [[Category:Root-finding algorithms]]
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Apart buy yoga dvd for pregnant women from the religious objectives, the physical postures of yoga are used to alleviate health issues, cut back stress and make the backbone supple in modern occasions. Yoga can be used as an entire exercise program and bodily remedy routine. forty nine History edit I went to a couple periods at the 2014 Symposium on Yoga Remedy and Analysis. Here's what I learned about yoga for ache, most cancers, coronary heart illness and more. Can Yoga Assist You Lose Weight? We're excited to announce that the E-GUIDE model of our Yoga for the Special Little one guide is on the market for Kindle and Nook. It is a revised edition. Yoga to Strengthen the Legs and Ft Yoga to Mild Your Inner Fire Practices designed by skilled yoga instructors. Holistic Yoga Stream Immersion Asanas for Most cancers Patients
These instances may seem exceedingly rare, but surveys by the Client Product Security Commission showed that the variety of emergency-room admissions related to yoga, after years of gradual increases, was rising rapidly. They went from thirteen in 2000 eurotrip to 20 in 2001. Then they free sports betting tips odds more than doubled to 46 in 2002. These surveys depend on sampling somewhat than exhaustive reporting buy yoga dvds toronto — they reveal tendencies quite than totals — but the spike was nonetheless statistically important. Solely a fraction of the injured go to hospital emergency rooms. A lot of buy yoga dvds those suffering from much less severe yoga injuries go to household docs, chiropractors and various sorts of therapists.
", as in liberating it from the thralldom of the senses. Says that once we can do this well we shall actually possess character and have made a protracted step in direction of freedom; earlier than then, we are "machines". "Again bends are all about unlocking the power of the spine and nervous system," Bielkus says. Mendacity on your again, bend the knees and place your ft flat on the ground with arms by your sides. Carry the hips up high and interlace the hands collectively or leave the arms on the sides of the physique. Breathe deeply for five breaths and repeat several occasions.
Pre-philosophical speculations of yoga start to emerge in the texts of c. 500–200 BCE. Between 200 BCE–500 CE philosophical faculties of Hinduism, Buddhism and Jainism have been taking kind and a coherent philosophical system of yoga started to emerge. 52 The Middle Ages saw the event of many satellite tv for pc traditions of yoga. Yoga came to the attention of an informed western public in the mid nineteenth century together with other matters of Indian philosophy. Origins (earlier than 500 BCE) edit