Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.
(a) Claim. There is a unique three-digit number whose digits have sum 8 and product 10.
“Proof.” Let x, y, and z be the digits. Then
x
+
y
+
z
=
8
and
x
y
z
=
1
0
. The only factors of 10 are 1, 2, 5, and 10, but since 10 is not a digit, the digits must be 1, 2, and 5. The sum of these digits is 8. Therefore, 125 is the only three-digit number whose digits have sum 8 and product 10.
(b) Claim. There is a unique set of three consecutive odd numbers that are all prime.
“Proof.” The consecutive odd numbers 3, 5, and 7 are all prime. Suppose that x, y, and z are consecutive odd numbers, all prime, and
x
≠
3
. Then
y
=
x
+
2
and
z
=
x
+
4
. Since x is prime, when x is divided by 3, the remainder is 1 or 2. In case the remainder is 1, then
x
=
3
k
+
1
for some integer
k
≥
1
. But then
y
=
x
+
2
=
3
k
+
3
=
3
(
k
+
1
)
, so y is not prime. In case the remainder is 2, then
x
=
3
k
+
2
for some
integer
k
≥
1
. But then
z
=
x
+
4
=
3
k
+
2
+
4
=
3
(
k
+
2
)
, so z is not prime. In either case we reach the contradiction that y or z is not prime. Thus
x
=
3
and so
y
=
5
and
z
=
7
. Therefore, the only three consecutive odd primes are 3, 5, and 7.
(c) Claim. If x is any real number, then either
π
−
x
is irrational or
π
+
x
is irrational.
“Proof.” It is known that p is an irrational number; that is, p cannot be written in the form
a
b
for integers a and b. Consider
x
=
π
. Then
π
−
x
=
0
, which is rational, but
π
+
x
=
2
π
. If 2p were rational, then
2
π
=
a
b
for some integers a and b. Then
π
=
a
2
b
, so p is rational. This is impossible, so 2p is irrational. Therefore, either
π
−
x
or
π
+
x
is irrational.
(d) Claim. If x is any real number, then either
π
−
x
is irrational or
π
+
x
is irrational.
“Proof.” It is known that p is an irrational number; that is, p cannot be written in the form
a
b
for integers a and b. Let x be any real number. Suppose both
π
−
x
and
π
+
x
are rational. Then, since the sum of two rational numbers is always rational,
(
π
−
x
)
+
(
π
+
x
)
=
2
π
is rational. Then
2
π
=
a
b
for some integers a and b. Then
π
=
a
2
b
, so p is rational. This is impossible. Therefore, at least one of
π
−
x
or
π
+
x
is irrational.
(e) Claim. For all real numbers x and y,
x
2
−
3
x
=
y
2
−
3
y
2
if and only if
x
=
y
or
x
+
y
=
3
.
“Proof.” Suppose that
x
2
−
3
x
=
y
2
−
3
y
. Then
x
2
+
x
y
−
3
x
−
x
y
−
y
2
+
3
y
=
0
, so
(
x
+
y
−
3
)
(
x
−
y
)
=
0
. Therefore,
x
=
y
or
x
+
y
=
3
.
(f) Claim. For all real numbers x and y, the equality
x
y
=
1
2
(
x
+
y
)
2
holds if and only if
x
=
y
=
0
.
“Proof.”
Part (i) Suppose that
x
=
y
=
0
. Then
x
y
=
0
=
1
2
(
x
+
y
)
2
, so the equality holds.
Part (ii) Suppose that x and y are real numbers and
x
y
=
1
2
(
x
+
y
)
2
. Then
2
x
y
=
x
2
+
2
x
y
+
y
2
, so
x
2
+
y
2
=
0
. Since the square of a real number is never negative,
x
2
=
y
2
=
0
, so
x
=
y
=
0
.
(g) Claim. If n is prime and
n
+
5
or
n
+
12
is prime, then
n
=
2
.
“Proof.” Assume that n is a prime number. Then
n
≥
2
, so if
n
+
5
is prime, then
n
+
5
must be odd. Therefore, n must be even. Since 2 is the only even prime,
n
=
2
.
(h) Claim. Let a, b, and c be real numbers with
a
≠
0
. If
a
x
2
+
b
x
+
c
=
0
has no rational roots, then
c
x
2
+
b
x
+
a
=
0
has no rational roots.
“Proof.” Suppose that
c
x
2
+
b
x
+
a
=
0
has a rational root p/q. Then
c
(
p
/
q
)
2
+
b
(
p
/
q
)
+
a
=
0
. Then
c
+
b
(
q
/
p
)
+
a
(
q
/
p
)
2
=
0
, so q/p is a rational root of the equation
a
x
2
+
b
x
+
c
=
0
.