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# Solved: Assign a grade of A (correct), C (partially correct), or F (failure) to each

ISBN: 9780495562023 335

## Solution for problem 7 Chapter 1.6

A Transition to Advanced Mathematics | 7th Edition

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A Transition to Advanced Mathematics | 7th Edition

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Problem 7

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justifyassignments of grades other than A. (a) Claim. Every polynomial of degree 3 with real coefficients has a real zero.Proof. The polynomial has degree 3, real coeffi-cients, and a real zero Thus the statement Every polynomial ofdegree 3 with real coefficients does not have a real zero is false, andhence its denial, Every polynomial of degree 3 with real coefficientshas a real zero, is true. (b) Claim. There is a unique polynomial whose first derivative isand which has a zero atProof. The antiderivative of is If we letthen and So is thedesired polynomial.p(c) Claim. Every prime number greater than 2 is odd.Proof. The prime numbers greater than 2 are 3, 5, 7, 11, 13, 17,None of these are even, so all of them are odd. (d) Claim. There exists an irrational number r such that is rational.Proof. If is rational, then is the desired example.Otherwise, is irrational and which isrational. Therefore either or is an irrational number r such thatis rational.(e) Claim. For every real number x,Proof. We proceed by three cases: andCase 1. Choose, for example, Then ThusCase 2. Then Thus,Case 3. Choose, for example, Then Thus(f) Claim. If x is prime, then is composite.Proof. Let x be a prime number. If then which iscomposite. If then x is odd, so is even and greater than 2. Inthis case, too, is composite. Therefore, if x is prime, then iscomposite.(g) Claim. For all irrational numbers t, is irrational.Proof. Suppose there exists an irrational number t such that isrational. Then where p and q are integers and Thenwith and q integers and This is acontradiction because t is irrational. Therefore, for all irrational numberst, is irrational.(h) Claim. For real numbers x and y, if then orProof.Case 1. If thenCase 2. If thenIn either case, (i) Claim. For every real number there is a natural number K suchthat for all real numbers 14xx > K, < . > 0,xy = 0.y = 0, xy = x0 = 0.x = 0, xy = 0y = 0.xy = 0 x = 0 y = 0.t 8t = p + 8q q = 0. pq + 8 = p + 8qq ,t 8 = q = 0. pq,t 8t 8x + 7 x + 7x = 2, x + 7x = 2, x + 7 = 9,x + 7|x| 0.x < 0. x = 5. |5| = 5.x = 0. |0| = 0. |x| 0.|x| 0.x > 0. x = 4. |4| = 4.x > 0, x = 0, x < 0.|x| 0.r232 3(32)2 = (3)2 = 3, 32r = 3 32r219, .Proof. Let be a real number. Let K be Assume x is a real 12 > 0 .number and Then so Therefore, so(j) Claim. For every natural number n,Proof. Let n be a natural number. Since n is a natural number,Since n is positive, Therefore, for all naturalnumbers n.n

Step-by-Step Solution:
Step 1 of 3

R = PMT FV = R [(1 + i) - 1] / i n FV = PMT [(1 + i) -1] / I P ( 1+ I) = PMT [(1 + i) - 1] / i -n P = PMT [ 1 - (1 + i) / i PV = PMT [ 1 - (1 + i) / i -n

Step 2 of 3

Step 3 of 3

##### ISBN: 9780495562023

Since the solution to 7 from 1.6 chapter was answered, more than 228 students have viewed the full step-by-step answer. A Transition to Advanced Mathematics was written by and is associated to the ISBN: 9780495562023. This textbook survival guide was created for the textbook: A Transition to Advanced Mathematics, edition: 7. This full solution covers the following key subjects: . This expansive textbook survival guide covers 39 chapters, and 619 solutions. The full step-by-step solution to problem: 7 from chapter: 1.6 was answered by , our top Math solution expert on 03/05/18, 08:54PM. The answer to “Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justifyassignments of grades other than A. (a) Claim. Every polynomial of degree 3 with real coefficients has a real zero.Proof. The polynomial has degree 3, real coeffi-cients, and a real zero Thus the statement Every polynomial ofdegree 3 with real coefficients does not have a real zero is false, andhence its denial, Every polynomial of degree 3 with real coefficientshas a real zero, is true. (b) Claim. There is a unique polynomial whose first derivative isand which has a zero atProof. The antiderivative of is If we letthen and So is thedesired polynomial.p(c) Claim. Every prime number greater than 2 is odd.Proof. The prime numbers greater than 2 are 3, 5, 7, 11, 13, 17,None of these are even, so all of them are odd. (d) Claim. There exists an irrational number r such that is rational.Proof. If is rational, then is the desired example.Otherwise, is irrational and which isrational. Therefore either or is an irrational number r such thatis rational.(e) Claim. For every real number x,Proof. We proceed by three cases: andCase 1. Choose, for example, Then ThusCase 2. Then Thus,Case 3. Choose, for example, Then Thus(f) Claim. If x is prime, then is composite.Proof. Let x be a prime number. If then which iscomposite. If then x is odd, so is even and greater than 2. Inthis case, too, is composite. Therefore, if x is prime, then iscomposite.(g) Claim. For all irrational numbers t, is irrational.Proof. Suppose there exists an irrational number t such that isrational. Then where p and q are integers and Thenwith and q integers and This is acontradiction because t is irrational. Therefore, for all irrational numberst, is irrational.(h) Claim. For real numbers x and y, if then orProof.Case 1. If thenCase 2. If thenIn either case, (i) Claim. For every real number there is a natural number K suchthat for all real numbers 14xx > K, < . > 0,xy = 0.y = 0, xy = x0 = 0.x = 0, xy = 0y = 0.xy = 0 x = 0 y = 0.t 8t = p + 8q q = 0. pq + 8 = p + 8qq ,t 8 = q = 0. pq,t 8t 8x + 7 x + 7x = 2, x + 7x = 2, x + 7 = 9,x + 7|x| 0.x < 0. x = 5. |5| = 5.x = 0. |0| = 0. |x| 0.|x| 0.x > 0. x = 4. |4| = 4.x > 0, x = 0, x < 0.|x| 0.r232 3(32)2 = (3)2 = 3, 32r = 3 32r219, .Proof. Let be a real number. Let K be Assume x is a real 12 > 0 .number and Then so Therefore, so(j) Claim. For every natural number n,Proof. Let n be a natural number. Since n is a natural number,Since n is positive, Therefore, for all naturalnumbers n.n” is broken down into a number of easy to follow steps, and 490 words.

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Solved: Assign a grade of A (correct), C (partially correct), or F (failure) to each