A slightly random selection of review problems
Math 1120b –
April 24, 2015
1. Functions and equivalence relations
(1.1) Remember the greatest integer function: for x ∈ R, bxc is the greatest integer less than
or equal to x. This is a function
b·c : R → Z.
(a) Compute b1.5c and b−1.5c.
Solution: b1.5c = 1 and b−1.5c = −2.
(b) Prove that b·c : R → Z is onto.
Solution: Since bnc = n for each integer n, the image of b·c is all of Z. That is, the function
is onto.
(c) Describe the fibre b·c
−1
({n}) for each integer n ∈ Z.
Solution: By definition, for any n ∈ Z,
b·c−1 ({n})
=
{x ∈ R : bxc = n}
=
{x ∈ R : n − 1 < x ≤ n.}
(1.2) Let S = x ∈ R : x2 ∈ Q . Show that Q ⊆ S, Q 6= S, but S is a countable set.
Solution: We know that the square of every rational number is rational,√so Q ⊆ S. On the other
hand, not every rational number is the square of a rational (remember 2) so Q 6= S.
To show that S is countable, recall
countable. Let f : N → Q≥0
npQ≥0 is p
o be a bijection, so
p
p
Q≥0 = {f (1), f (2), · · · }. Then S =
f (1), − f (1), f (2), − f (2), · · · , so S is countable
as well.
(1.3) Let S = T = [10] and define a function f : S → T by letting f (n) = bn/2c. Compute
f −1 ({n}) for each n ∈ T . Decide if f is 1-1 and/or onto.
Solution: Neither. To see f is not 1-1, it is enough to notice that f (1) = f (0) = 0 but (of course)
1 6= 0. To see f is not onto, we notice that (for example) f (n) = 10 implies n ≥ 20, so 10 6∈ im(f ).
(1.4) Let R be the relation defined on the set of all women by saying xRy if and only if either
x = y or x and y are sisters. Is R an equivalence relation?
Solution: Clearly R has the symmetric and reflexive properties. Transitivity depends on our
definition of sisters. If x and y are to be called sisters if and only if they have the same mother
and father, than R is transitive (so, an equivalence relation). But if we are willing to call x and y
sisters if they only share one parent, then R is not transitive: e.g., suppose Azalea and Rebecca
have the same father (but not mother) but Rebecca and Hazel have the same mother (but not
father) – then Azalea and Hazel need not have any parent in common.
(1.5) Define a relation on C by setting x ∼ y iff |x − y| = 1. Is ∼ an equivalence relation?
Prove or disprove.
Solution: No, because ∼ is not transitive. For a counterexample, check that 0 ∼ 1 because
|−1| = 1; also 1 ∼ 2 because |1 − 2| = 1; however, 0 6∼ 2 because |0 − 2| 6= 1.
(1.6) Define a relation on R × R by setting (x, y) ∼ (x0 , y 0 ) iff x + y = x0 + y 0 .
(a) Prove that ∼ is an equivalence relation.
Solution: Since x + y = x + y, we have (x, y) ∼ (x, y) for all x, y) ∈ R × R, so ∼ is reflexive.
Since x + y = x0 + y 0 if and only if x0 + y 0 = x + y, the relation ∼ is also symmetric.
1
Finally, if (x, y) ∼ (x0 , y 0 ) and (x0 , y 0 ) ∼ (x00 , y 00 ), then x + y = x0 + y 0 and x0 + y 0 = x00 + y 00 .
Substituting, we see x + y = x00 + y 00 , which (by definition) means (x, y) ∼ (x00 , y 00 ). That
is, ∼ is also transitive.
Since we showed ∼ is reflexive, symmetric, and transitive, we have proven ∼ is an equivalence
relation.
(b) Describe the equivalence classes of ∼.
Solution: If (x, y) ∈ R × R, let c = x + y. Then (x, y) ∼ (x0 , y 0 ) if and only if x0 + y 0 = c
as well. So to points in R2 are equivalent if and only if the sum of their coordinates is the
same. Another way to describe all pairs (x, y) for which x + y = c is as the points on a line.
That is, the equivalence classes are the (sets of points on the) lines of slope −1.
(c) Give a subset S ⊆ R × R so that each equivalence class contains exactly one element
of S.
Solution: There is a line of slope −1 given by the equation x + y = c for each choice of
c ∈ R. So let S = {(c, 0) : c ∈ R}. Each element of S is on some line of slope −1, but no
such line contains two different elements of S, because no line of slope −1 joins two points
on the x-axis.
2. Complex numbers, polynomials, inequalities. . .
(2.7) Find all z ∈ C for which z 5 = 2i.
Solution: Let’s write 2i in polar form, by taking r = |2i| = 2, and noting the angle is π/2, so
2i = 2eπ/2i . Then the fifth roots are given by
√
5
z = 2e(π/2+2πk)i/5 ,
where k is an integer. To get five distinct roots, take k = 0, 1, 2, 3, 4. That is, the five solutions
are given by
√
5
z = 2eθi ,
where θ ∈ {π/10, π/2, 9π/10, 13π/10, 17π/10}.
(2.8) Suppose, for some a, b ∈ R, we know z 2 + az + b = 0 when z = 1 + 2i. Find a, b.
Solution: We know that if z is a root of a polynomial with real coefficients, so is z. So the two
roots are 1 + 2i and 1 − 2i, and
z 2 + az + b
=
(z − (1 + 2i))(z − (1 − 2i))
=
z 2 − 2z + 5,
that is, a = −2 and b = 5.
(2.9) If z ∈ C and |z| = 1, say z has order n if n is the smallest positive integer for which
z n = 1. Prove that, if a complex number z has order n, then −z either has order n or 2n.
Describe those n for which −n has order 2n.
Solution: If z n = 1, then z = e2πik/n for some integer k. It is easy to check that if gcd(k, n) 6= 1,
then the order of z is strictly less than n, so we may assume that k and n are relatively prime.
Then
−z
=
eπi e2πik/n
=
e
=
e
2k+n
πi
n
2k+n
2πi
2n
.
Now check that gcd(2k + n, 2n) is either 1 or 2, depending on whether or not n is odd or even.
With that, we see −z has order 2n when n is odd, and order n when n is even.
(2.10) Evaluate (1 −
√
3i)100 .
Solution: Use polar form! 1 −
(1 −
√
√
3i = 2e2π/3i so
3i)100
=
2100 e200π/3i
=
2100 e2π/3i
√
3
100 1
2 ( −
i).
2
2
=
10
(2.11) Which is bigger, 10(10
)
, or (1010 )10 ?
10 10
Solution: Note (10 ) = 10100 . Since 1010 > 100 and 10 > 1, exponentiating both sides
10
preserves the inequality, and 1010 > 10100 .
(2.12) Describe all x for which
Solution: By definition,
x+3 x+3
8
8
= −1.
= −1 if and only if
−2 <
x+3
≤ −1.
8
Rearranging the left-hand inequality gives −19 < x. The right-hand inequality is equivalent to
x ≤ −11, so our equation holds exactly for those x in the range −19 < x ≤ −11.
3. Induction
(3.13) Use induction to show that 52n − 3n is divisible by 11 for all n ≥ 0.
Solution: First check the base case: 50 − 30 = 0, which is divisible by 11. Now suppose 11 divides
52n − 3n for some n ≥ 0. This means 52n − 3n ≡ 0 mod 11.
We prove that 11 also divides 52(n+1) − 3n+1 , by the following calculation:
52(n+1) − 3n+1
=
52n+2 − 3n+1
=
25 · 52n − 3 · 3n
≡
3 · 52n − 3 · 3n
≡
3(52n − 3n )
≡
3·0
≡
0
mod 11 because 25 ≡ 3
mod 11;
mod 11
mod 11 by the induction hypothesis
mod 11,
which is equivalent to saying 11 divides 52(n+1) − 3n+1 . By induction, the claim is true for all
n ≥ 0.
(3.14) Use induction to prove that, if n lines are drawn in the plane so that no two are parallel
and no three pass through the same point, then 2n of the regions are unbounded.
Solution: It’s not clear that this makes sense if n = 0, so the author probably should have clarified
that n ≥ 1. In that case, take n = 1 as the base case: one line splits the plane into 2 = 2(1)
unbounded regions.
Now suppose the claim is true for all arrangements of n lines, where n ≥ 1 is some integer.
Given an arrangement of n + 1 lines, choose any line L. By removing L, we get an arrangement
of n lines, so that arrangement has 2n unbounded regions, by induction. The line L intersects
the other n lines somewhere, since no two lines are allowed to be parallel.
That means that L intersects at least two regions of the smaller arrangement. [Draw a picture.]
Each region that intersects L is a union of two regions in the larger arrangement. Clearly if L
intersects a bounded region, the two subregions are bounded. If L intersects an unbounded region,
only one subregion is unbounded, except for the two regions containing the two ends of L. So
adding in the line L creates two new unbounded regions, which gives a total of 2n + 2 = 2(n + 1)
unbounded regions, as required.
[Looks like this result is still true if we drop the condition that no three lines meet at a point,
since we never used that fact in the argument. Making the argument completely rigorous would
call for a little more geometry, but that’s beyond the scope of this course.]
(3.15) Use induction to prove the following: for any arrangement of n lines in the plane, it is
possible to colour the regions red and blue so that no two neighbouring regions have the
same colour.
Solution: A similar argument applies. If n = 1, there are two regions: colour one red and one
blue. If n > 1, remove a line L. The smaller arrangement has such a colouring, by induction.
Now add the line L back in. Unfortunately, some neighbouring regions now have the same colour:
that is, the ones that intersect L. To fix this, choose a side of L arbitrarily. Call it “left”. Change
each red region on the left side of L to a blue, and vice-versa. Leave the colours unchanged on
the other side of L. Now if two neighbouring regions had different colours before our switch,
they still do. If they had the same colour, they intersected the line L, so exactly one of them
got changed, and now they have different colours. So we’ve coloured all the regions in the way
required. [Draw a picture!]
4. Logic
(4.16) Decide which of the following propositions are true and which are false. Prove accordingly.
(a) ∀x ∈ R ∃y ∈ R : y 2 = x
Solution: False. Pick x = −1. Then 6 ∃y ∈ R : y 2 = −1, since the square of every real
number is nonnegative.
(b) ∀x ∈ R ∃y ∈ R : x2 = y
Solution: True. For any x ∈ R, x2 is a real number too, so let y = x2 .
(c) ∀x ∈ R ∃n ∈ Z : |x − n| < 1.
Solution: True: if x is a real number, we can write x = n + r where n is an integer and r a
the decimal part satisfying |r| < 1. Some rearranging gives |x − n| = |r| < 1.
(d) ∀n ∈ N ∀S, T ⊆ [n] : |S| + |T | > n ⇒ S ∩ T 6= ∅.
Solution: True. Since S ∪T ⊂ [n], we know |S ∪ T | ≤ n. By the inclusion-exclusion formula,
n
≥
|S ∪ T |
=
|S| + |T | − |S ∩ T |
> n − |S ∩ T |,
since |S| + |T | > n.
Subtracting n − |S ∩ T | from both sides gives |S ∩ T | > 0, which means S ∩ T 6= ∅.