Extra Problems
Proofs to Grade: A number of the extra problems are to 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.
A.1 (ptg) Statement: If m is an integer, and m2 is odd, then m is odd.
Proposed Proof: Assume m is odd. Then m = 2k + 1 for some integer k. Therefore,
m2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2+2k) + 1, which is odd. Therefore, if m2 is odd, then
m is odd.
A.2 (ptg) Statement: For any positive real number x, the sum of x and its reciprocal is
greater than or equal to 2. That is,
x + 1/x > 2
Proposed Proof: Multiplying by x we get x2 + 1 > 2x. By algebra, x2 - 2x + 1 > 0. Thus
(x - 1)2 > 0. This is true for any real x, so x + 1/x > 2 is true.
A.3 (ptg) Statement: Every polynomial of degree 3 with real coefficients has a real zero.
Proposed Proof: The polynomial x3 - 8 has degree 3, real coefficients, and a real zero at x
= 2. Thus the statement "every polynomial of degree 3 with real coefficients does not
have a real zero" is false, and hence its denial, "every polynomial of degree 3 with real
coefficients has a real zero" is true.
A.4 (ptg) Statement: if x is any real number then either  - x or  + x is irrational.
Proposed Proof: It is known that is irrational, so  cannot be written in the form a/b
for integers a and b. This shows that 2 is also irrational, for otherwise, 2 = a/b for
integers a and b and that means  = a/(2b), which is impossible. Now consider x = 
Then  - x = 0 is rational, and  +x = 2 is irrational. So it is true that  - x or  + x is
irrational.
B.1 (ptg) Statement: If A, B, and C are sets and if A  B and B  C then A  C .
Proposed Proof: If x  C then since B  C we can conclude that x  B . Similarly, now
that we know that x  B , since A  B we can conclude that x  A . Thus, x  C implies
x  A . This shows A  C .
B.2 (ptg) Statement: If A, B, and C are sets then A  B if and only if A  B  A .
Proposed proof: Assume that A  B . Suppose that x  A  B . Then x  A
and x  B so x  A . This shows that A  B  A .
Now assume that A  B  A . Suppose x  A . Then x  A  B since A  B  A ,
and therefore x  B . This shows that x  A implies x  B , so A  B .
1.1.1 If it were 2AM now, what time would it be 3736 hours from now?
1.1.2 (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.
1.2.1 (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).
1.2.2 (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 (n, n +1) = 1.