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18.781 Problem Set 3
18.781 Problem Set 3

Solution - Stony Brook Mathematics
Solution - Stony Brook Mathematics

... than x (since x/2 < x by the multiplicative law for inequalities), a contradiction. Therefore, there does not exist a smallest positive real number. Problem 4. Use induction to prove Bernouilli’s inequality : (1 + x)n ≥ 1 + nx for all non-negative integers n and real numbers x > −1. Sol. Fix some re ...
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The ring of evenly weighted points on the projective line

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A Case of Depth-3 Identity Testing, Sparse Factorization and Duality

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Existence and uniqueness of Haar integrals

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Chapter 1 – ALGEBRA REVIEW

... 4.6 Factor a polynomial using only linear factors. 4.6 Determine the equation of a polynomial given its zeros or graph. 5.1 Define and apply rational exponents. 5.1 Simplify expressions containing radicals or rational exponents. 5.2 Graph and identify transformations of exponential functions. 5.3 U ...
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Solving Quadratic Equations by Finding Square Roots

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31 Semisimple Modules and the radical

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2 is irreducible in Q[ √ 2]

... to 7 so it cannot be isomorphic to Q[ 7]. • Exercises 4.1: 9, 14(a)(c). I went over these ones or very similar ones in class so I’m not going to do them here. • Exercises 3.3: 2(a)(c)(e), 4(a). 2(a) Is x3 + 4x2 − 3x + 5 irreducible in Q[x]? Its irreducible if and only if it has no rational roots whi ...
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Fundamental theorem of algebra

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with an imaginary part equal to zero.Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n roots. The equivalence of the two statements can be proven through the use of successive polynomial division.In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; its name was given at a time when the study of algebra was mainly concerned with the solutions of polynomial equations with real or complex coefficients.
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