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Get Full Access to Mathematics: A Discrete Introduction - 3 Edition - Chapter 1 - Problem 11
Get Full Access to Mathematics: A Discrete Introduction - 3 Edition - Chapter 1 - Problem 11

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# Suppose we are asked to prove the following identity: x.x C y 1/ y.x C 1/ D x.x 1/ y ISBN: 9780840049421 447

## Solution for problem 11 Chapter 1

Mathematics: A Discrete Introduction | 3rd Edition

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Problem 11

Suppose we are asked to prove the following identity: x.x C y 1/ y.x C 1/ D x.x 1/ y: The identity is true (i.e., the equation is valid for all real numbers x and y). The following proof is incorrect. Explain why. Proof. We begin with x.x C y 1/ y.x C 1/ D x.x 1/ y and expand the terms (using the distributive property) x 2 C xy x yx y D x 2 x y: We cancel the terms x 2 , x, and y from both sides to give xy yx D 0; and finally xy and yx cancel to give 0 D 0; which is correct.

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Step 1 of 3

Constant Coe¢ cient Homogeneeous DE. Case 1: Starting simple: Finding two Linearly independent solutions of Linear second order Constant Homogeneous DE s. Solutions to ▯ 2 ▯ 00 0 Ly = aD + bD + c y = ay + by + cy = 0: (1) Lets look at the simpler case of ▯ 2 ▯ 00 0 Ly = D + a D + 1 2 y = y + a y1+ a y 2 0: (2) 00 where the coe¢ cient of y is 1: To solve the homogeneous equation 00 0 y + a y1+ a y 2 0: (3) Every solution of this homogeneous equation is of the form e . When substituted into the homogeneous equation, we obtain an equation, called the characteristic (also called auxiliary) equation, given by r + a 1 + a =20 (4) Solving this equation leads to the characteristic roots values of r and r given as 1 2 p p ▯a a ▯ 4a ▯a a ▯ 4a r1= 1+ 1 2 ; r 2 1 ▯ 1 2

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##### ISBN: 9780840049421

This textbook survival guide was created for the textbook: Mathematics: A Discrete Introduction, edition: 3. The answer to “Suppose we are asked to prove the following identity: x.x C y 1/ y.x C 1/ D x.x 1/ y: The identity is true (i.e., the equation is valid for all real numbers x and y). The following proof is incorrect. Explain why. Proof. We begin with x.x C y 1/ y.x C 1/ D x.x 1/ y and expand the terms (using the distributive property) x 2 C xy x yx y D x 2 x y: We cancel the terms x 2 , x, and y from both sides to give xy yx D 0; and finally xy and yx cancel to give 0 D 0; which is correct.” is broken down into a number of easy to follow steps, and 111 words. Since the solution to 11 from 1 chapter was answered, more than 231 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 69 chapters, and 1110 solutions. The full step-by-step solution to problem: 11 from chapter: 1 was answered by , our top Math solution expert on 03/15/18, 06:06PM. Mathematics: A Discrete Introduction was written by and is associated to the ISBN: 9780840049421.

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Calculus: Early Transcendental Functions : Differentiation and Integration of Vector-Valued Functions
?Differentiation of Vector-Valued Functions In Exercises 1-6, find r’(t), $$\mathrm{r}\left(t_{0}\right)$$, and $$\mathrm{r’}\left(t_{0}\right)$$ for t

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