Math 201 – Exam 2 – solutions
Problem 1 Professor D. U. Believitch teaches set theory. He claims that
for all sets A, B, C
(A − B) ∪ C = (A ∪ C) − (B ∪ C)..
Prove or disprove.
Solution. Make a Venn diagram for both given sets to see that they are
not equal. This also shows how to construct actual examples of sets A, B, C
where the given equation is wrong, eg
A = {1},
Problem 2 Prove that α =
square it!
Solution. First,
B = ∅,
√
2+
C = {2}.
√
3 is not a rational number. Hint –
√
√
α2 = 2 + 2 6 + 3 = 5 + 2 6.
Suppose that α ∈ Q. So α2 ∈ Q and
√
6=
α2 − 5
∈ Q.
2
Therefore there exist integers a, b with no proper common divisor such that
√
a
6= .
b
Multiply by b and square both sides, so 6b2 = a2 . Therefore 2|a, and we can
write a = 2c with c ∈ Z. Substitute 2c for a to get
6b2 = 4c2
therefore 3b2 = 2c2 , hence b2 is even and b must be even, too. This contradicts
the choice of a, b with no common divisor (a similar argument could have been
made with 3|a).
Problem 3 Let n be any positive integer.
a) Let p be a prime number. Prove np ≡ n (mod p). Hint – use induction
over n, the binomial formula, and the fact that all binomial coefficients kp
are congruent to zero modulo p.
b) If p does not divide np − n, prove that p is not a prime number.
Solution. a) Base Step: for n = 1, this is obviously true since every integer
divides 1p − 1 = 0.
Induction Step: assume np ≡ n (mod p). We then expand (n + 1)p using the
binomial formula, and then congruences modulo p.
!
p−1 X
p k
(n + 1)p = np +
n + 1p
k
k=1
≡ np + 1 (mod p)
≡ n + 1 (mod p)
because the binomial coeffs in the sum ≡ 0
because of the induction hypothesis.
b) If p is a prime number, p divides np − n – this has been shown in part a).
The contrapositive of that statement is the claim in part b). Note that this
can be made into a test whether p is prime, which does not require finding
factors of p – most of the modern primality tests for large numbers are based
on similar principles.
Problem 4 Let x be any integer. Prove that x ≡ 7 (mod 20) if and only if
x ≡ 3 (mod 4) and x ≡ 2 (mod 5).
Solution. Proof of ⇒: Let x ≡ 7 (mod 20). So x − 7 = 20k = 4(5k), and
x − 3 = 4(5k) + 4 = 4(5k + 1)
proving x ≡ 3 (mod 4). Similarly, x − 7 = 5(4k) and x − 2 = 5(4k + 1),
showing x ≡ 2 (mod 5).
Proof of ⇐: Suppose x ∈ Z with x ≡ 3 (mod 4) and x ≡ 2 (mod 5). So
x − 7 = (x − 3) − 4 = 4k
and
x − 7 = (x − 2) − 5 = 5m.
These are equal, 4k = 5m, in particular 5 divides 4k, hence 5 divides k since
5 is a prime. So k = 5n and x − 7 = 20n, proving x ≡ 7 (mod 20).
Problem 5 Consider a 2-by-n rectangle. Let Gn be the number of ways
to tile this rectangle by copies of a 2-by-1 rectangle (cover without gaps or
overlapping). Prove that Gn = Fn+1 , where Fn is the n-th Fibonacci number,
defined by F1 = 1, F2 = 1, and Fn+1 = Fn + Fn−1 for all n ≥ 2.
Solution. Base Step: for n=1, there is only one way to tile the rectangle,
obviously, and we check G1 = F2 = 1.
Induction Step: we use strong induction, in particular
Gn = Fn+1 ,
Gn−1 = Fn .
There are two types of tilings of a 2-by-(n + 1) rectangle. Either we start
with one vertical tile, or with two horizontal tiles. In the first case, we can
then complete the tiling in exactly Gn ways, because a 2-by-n rectangle is
left over. In the second, we can complete the tiling in exactly Gn−1 ways,
since a 2-by-(n − 1) rectangle is left over. The two types of tilings count each
possible tiling of the 2-by-(n + 1) rectangle exactly once, because each tiling
must fall into one of the two types, and the two types do not overlap. We
have shown
Gn+1 = Gn + Gn−1 .
Using the induction hypothesis, we can then replace the G’s by Fibonacci
numbers,
Gn+1 = Fn+1 + Fn = Fn+2 .