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MATH 2390 Exam 3 Solutions
October 23, 2014
S. F. Ellermeyer
Name
Instructions. Remember that writing and correct use of notation are very important. Write in
complete sentences.
1. Prove that if a and b are integers and a | b, then a 2 | b 2 .
Proof: Suppose that a | b. Then there exists an integer k such that b  ka. This implies that
b 2  k 2 a 2 . Since k 2 is an integer, we see that a 2 | b 2 .
2. If a, b, c and d are integers, n is an integer with n ≥ 2 and a ≡ b (mod n) and c ≡ d (mod
n), then we know (it has been proved in class) that
a  c ≡ b  d (mod n)
and
ac ≡ bd (mod n).
3.
4.
5.
6.
Prove one of these facts (whichever one you want to). Both of these facts can be used
throughout this exam if needed.
Proofs: Given in class.
Prove that if a, b and c are integers such that a 2  b 2  c 2 , then either 3 | a or 3 | b.
Proof: We first note that if x is any integer, then either x ≡ 0 (mod 3) or x ≡ 1 (mod 3) or
x ≡ 2 (mod 3).
If x ≡ 0 (mod 3), then x 2 ≡ 0 (mod 3).
If x ≡ 1 (mod 3), then x 2 ≡ 1 (mod 3).
If x ≡ 2 (mod 3), then x 2 ≡ 4 ≡ 1 (mod 3).
Hence the square of any integer is congruent either to 0 or to 1 (mod 3). It is not possible,
for any integer x, that x 2 ≡ 2 (mod 3). We will use this fact in our proof. Also, we will
prove the contrapositive of the given statement. That is, we will prove that if a, b and c are
any integers such that neither a nor b is divisible by 3, then a 2  b 2 ≠ c 2 .
Let a, b and c be integers such that neither a nor b is divisible by 3. This means that
neither a nor b is congruent to 0 (mod 3). Hence a 2 ≡ 1 and b 2 ≡ 1 (mod 3) which means
that a 2  b 2 ≡ 2 (mod 3). Since c is an integer, then either c 2 ≡ 0 or c 2 ≡ 1 (mod 3). We
conclude that a 2  b 2 ≠ c 2 .
Let m and n be integers with m ≥ 2 and n ≥ 2 and suppose that m | n. Prove that if a and b
are any integers such that a ≡ b (mod n), then a ≡ b (mod m).
Proof: We are given that m | n. This means that there exists an integer, k, such that n  km.
Now suppose that a and b are any integers such that a ≡ b (mod n). This means that there
exists an integer r such that b  a  rn. Hence
b  a  rkm  a  rkm.
Since rk is an integer, we conclude that a ≡ b (mod m).
Disprove the statement: If n ∈ 0, 1, 2, 3, 4, then 2 n  3 n  nn − 1n − 2 is prime.
Answer: The given statement if false because 2 4  3 4  44 − 14 − 2  121 is not
prime. (121  11 2 ).
Disprove the statement: If a and b are integers such that ab and a  b 2 are of opposite
parity, then a 2 b 2 and a  ab  b are of opposite parity.
Answer: The given statement is false. If a  3 and b  5, we have
ab  15 (odd)
a  b 2  64 (even)
a 2 b 2  225 (odd)
a  ab  b  23 (odd).
We see that ab and a  b 2 are of opposite parity, but a 2 b 2 and a  ab  b are of the same
parity.
7. Prove or disprove the statement: If x and y are irrational numbers, then x  y is an
irrational number.
Answer: The statement is false. As a counterexample, take x  2 and y  − 2 .
8. Prove that the number 10 is irrational.
Proof (by contradiction): Suppose that 10 is rational. There there exist integers p and q
such that
p
10  q
and p and q have no factors in common. From this, we obtain p 2  10q 2 , which tells us
that p 2 must be even. Hence p must be even. This tells us that there exists an integer k
such that p  2k. We now have 2k 2  10q 2 which can be written as 4k 2  10q 2 or
2k 2  5q 2 . Since 2k 2 is even, then 5q 2 must be even. However, this is possible only if q 2
is even, which is possible only if q is even. We have now arrived at the conclusion that p
and q must both be even, but this contradicts our assumption that p and q have no factors
in common. In conclusion, 10 is irrational.
9. Prove that if b and c are integers such that b  c is divisible by 3, then 10b  c is divisible
by 3.
Proof: Suppose that b  c is divisible by 3. Then there exists an integer k such that
b  c  3k. Now note that
10b  c  9b  b  c  9b  3k  33b  k.
Since 3b  k is an integer, then 10b  c is divisible by 3.
(Do only one of problems 10–12. I will grade only one of them. If you work on more
than one of them, then you must choose which one you want me to grade.)
10. Prove that there do not exist real numbers x and y with x  0 and y  0 such that
xy  x  y.
Proof: Suppose that there exist real numbers x and y with x  0 and y  0 such that
x  y  x  y . Then, by squaring both sides of this equation, we obtain
x  y  x  2 xy  y.
This implies that xy  0 and hence that xy  0. However, since x  0 and y  0, this is
a contradiction. We conclude that there do not exist real numbers x and y with x  0 and
y  0 such that x  y  x  y .
11. Prove that if a, b and c are real numbers such that ax 2  bx  c  0 for every real number
x, then a  0, b  0 and c  0.
Proof: Suppose that a, b and c are real numbers such that ax 2  bx  c  0 for every real
number x. Then, since 0 is a real number, we have
a0 2  b0  c  0
which implies that c  0. Thus ax 2  bx  0 for every real number x.
Since −1 is real number, then
a−1 2  b−1  0
which implies that a  b. We thus have ax 2  ax  0 for every real number x. This can be
written as ax 2  x  0 for every real number x.
Since 1 is a real number, then a1 2  1  0 which implies that 2a  0 which implies that
a  0. Also, since b  a, then b  0.
This completes the proof.
12. Prove that if a and b are any real numbers with a  0 and b  0, then
ab ≤ a  b .
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Proof: Let a and b be real numbers with a  0 and b  0. Then a − b 2 ≥ 0. This
implies that
0 ≤ a 2 − 2ab  b 2
which implies that
4ab ≤ a 2  2ab  b 2
or
4ab ≤ a  b 2 .
Taking the square root of both sides of the above inequality gives
2 ab ≤ a  b.
Thus
ab ≤ a  b .
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