Math.412.001& Math.612.001 Modern Algebra Extra Problems
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
(correct) if the
statement and proof are correct. Give this grade even if the
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
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
if most of the statements in the proof are incorrect.
Be sure to
justify grades of C
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,
x1.1.1If it is now, what time will it be
hours from now?
There is a unique set of three consecutive odd integers that are all
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
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
Statement: Suppose a, b,
and c are integers. If a | b and a
then a | (b+c).
Proposed Proof: Since a | b we know that b
= aq for some integer q.
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).
Statement: For all natural numbers n, (n
, n+1) = 1.
Proposed Proof:1 | nand 1 | n+ 1, so 1 is certainly a common divisor ofn 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.