Math.412.001& Math.612.001 Modern Algebra
Extra Problems

Proofs to Grade:  A number of the extra problems are designated Proofs to Grade (ptg).  In each one, you will be given a statement and a purported proof.  It is your job to judge the validity of the proof.  Assign a grade of

A (correct) if the statement and proof are correct.  Give this grade even if the given proof is not the simplest possible, or is not the one you would have given.

C (partially correct) if the statement is correct and the proof is largely valid, but contains a few statements that are incorrect or insufficiently justified.

F (incorrect) if the statement is incorrect, or the main idea of the proof is incorrect or has faulty logic, or if most of the statements in the proof are incorrect.

Be sure to justify grades of C or F.

Most of these problems, and the concept for this type of problem, are taken from A Transition to Advanced Mathematics by Smith, Eggen, and St. Andre, 6th edition, Brooks/Cole.

Section 1.1

x1.1.1  If it is 2AM now, what time will it be 3736 hours from now?  

x1.1.2   (ptg) Statement: There is a unique set of three consecutive odd integers that are all prime.
Proposed Proof:  The consecutive odd integers 3, 5, and 7 are all prime.  Suppose that x, y, and z, are a different set of three consecutive odd primes. Then x is not 3.  Since x is prime, when x is divided by 3 it must leave a remainder of 1 or 2.  In case the remainder is 1, we can write x = 3k + 1 for some positive integer k.  Then y = x + 2 = 3k + 3 = 3(k+1), implying that y is not prime.  Similarly, in case the remainder is 2, we can write x = 3k + 2 for some positive integer k.  Then z = x + 4 = 3k + 2 + 4 = 3(k+2), implying that z is not prime.  In either case, we reach the contradiction that y or z is not prime.  This rules out the possibility of three consecutive odd primes x, y, z, with x different from 3.  Therefore, 3, 5, 7, is the only set of three consecutive odd primes.

Section 1.2

x1.2.1 (ptg) Statement:  Suppose a, b, and c are integers.  If a | b and a | c, then a | (b+c).
Proposed Proof:  Since a | b we know that b = aq for some integer q.  Similarly, since a | c we know that c = aq for some integer q.   Then b + c = aq + aq = 2aq = a(2q).  This shows that a | (b + c).

x1.2.2  (ptg) Statement: For all natural numbers n, (n , n+1) = 1.
Proposed Proof:  1 | n  and 1 | n  + 1, so 1 is certainly a common divisor of  n  and n +1.  Now suppose that c is any other common divisor, so c | n and c | n +1.  Then 1 | c.  Therefore by Corollary 1.4  (n, n +1) = 1.