Show that x + y is a factor of x2n - 1+ y2n - 1 for all natural numbers n.Let P(n) denote the statement that x + y is a factor of x2n- 1+y2n – 1.P(1) is the statement that x + y is a factor of x+ ywhich is true.Assume that P(k) is true. Thus, our induction hypothesis is x + y is a factor of2k -2k -X+ yWe want to use this to show that P(k + 1) is true. Now,x2(k+1) – 1 + y2(k+1) – 1 =2k +2k ++ y2k +2k +x2k – 1y2 + x2k - ly2 + y°2k -2k -= X2k -(x² – y²) ++ yWe can see that x + y is a factor of this last step, since x - y = (x + y)(x - ), and since x + y is a factor ofby the induction hypothesis. Thus, P(k + 1)follows from P(k), and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that P(n) is true for all naturalnumbers n.

Name: Show that x + y is a factor of x2n - 1+ y2n - 1 for all natural numbe
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: Show that x + y is a factor of x2n - 1+ y2n - 1 for all natural numbers n.Let P(n) denote the statement that x + y is a factor of x2n- 1+y2n – 1.P(1) is the statement that x + y is a factor of x+ ywhich is true.Assume that P(k) is true. Thus, our in

Answered: Image /qna-images/answer/316f29b0-dcc5-46e2-9c4b-2f13441c1c29.jpg

Math

Calculus

Show that x + y is a factor of x2n - 1 + y2n - 1 for all natural numbers n. Let P(n) denote the statement that x + v is a factor of x2n - 1 + y2n – 1. P(1) is the statement that x + y is a factor of which is true. Assume that P(k) is true. Thus, our induction hypothesis is x + y is a factor of 2k 2k - + y We want to use this to show that P(k + 1) is true. Now, x2(k+1) – 1 + y2(k+1) – 1 = 2k + 2k + + y 2k + 2k + - x2k - 1y2 + x2k - 1y2 + y 2k - = X 2k - +y (x2 - y2) + 2k - We can see that x + y is a factor of this last step, since x2 - y2 = (x + y)(x - y), and since x + y is a factor of by the induction hypothesis. Thus, P(k + 1) follows from P(k), and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that P(n) is true for all natural numbers n.

Show that x + y is a factor of x2n - 1 + y2n - 1 for all natural numbers n. Let P(n) denote the statement that x + v is a factor of x2n - 1 + y2n – 1. P(1) is the statement that x + y is a factor of which is true. Assume that P(k) is true. Thus, our induction hypothesis is x + y is a factor of 2k 2k - + y We want to use this to show that P(k + 1) is true. Now, x2(k+1) – 1 + y2(k+1) – 1 = 2k + 2k + + y 2k + 2k + - x2k - 1y2 + x2k - 1y2 + y 2k - = X 2k - +y (x2 - y2) + 2k - We can see that x + y is a factor of this last step, since x2 - y2 = (x + y)(x - y), and since x + y is a factor of by the induction hypothesis. Thus, P(k + 1) follows from P(k), and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that P(n) is true for all natural numbers n.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter2: The Integers

Section2.3: Divisibility

Problem 49E: 49. a. The binomial coefficients are defined in Exercise of Section. Use induction on to prove...

See similar textbooks

Similar questions

31. Prove statement of Theorem : for all integers and .
Use generalized induction and Exercise 43 to prove that n22n for all integers n5. (In connection with this result, see the discussion of counterexamples in the Appendix.) 1+2n2n for all integers n3
In Exercise , use generalized induction to prove the given statement. for all integers
49. a. The binomial coefficients are defined in Exercise of Section. Use induction on to prove that if is a prime integer, then is a factor of for . (From Exercise of Section, it is known that is an integer.) b. Use induction on to prove that if is a prime integer, then is a factor of .
Assume the statement from Exercise 30 in section 2.1 that for all and in . Use this assumption and mathematical induction to prove that for all positive integers and arbitrary integers .
Let x and y be integers, and let m and n be positive integers. Use mathematical induction to prove the statements in Exercises 1823. ( The definitions of xn and nx are given before Theorem 2.5 in Section 2.1 ) (m+n)x=mx+nx
Let and be integers, and let and be positive integers. Use mathematical induction to prove the statements in Exercises. The definitions of and are given before Theorem in Section
Prove by induction that 1+2n3n for n1.