1.
Which of the following statements is true or false. (Give one or two sentences justifying
each answer, but you don’t need to give a full proof.)
1. (∀x ∈ Z)(∃y ∈ Z)(x2 = y)
True:
for any integer x, x2 is also an integer.
2. (∀y ∈ Z)(∃x ∈ Z)(x2 = y)
False.
work.
For example, when y = 2, there is no choice of an integer x which could
3. (∀x ∈ Z)(∀y ∈ Z)(∃z ∈ Z)(x2 + z 2 = y 2 )
False. For example, if x = 1 and y = 2, there is no z so that 1 + z 2 = 4,
because such a z would have to satisfy z 2 = 3, and no integer is a square root
of 3.
2.
Consider the statement:
Suppose x is a real number less than or equal to every square number. Then x is
less than every positive number.
In symbols, this says
(∀x ∈ R) (∀y ∈ R)(x ≤ y 2 ) ⇒ (∀z ∈ R)(0 < z ⇒ x < z) .
Prove this statement. (You don’t have to go all the way down to field axioms, but do give a
rigorous proof where every step would be obvious to any math 202 student. Make sure to be
careful and explicit about the steps you use for the connectives and quantifiers.)
(Of course, there are many possible ways to write this proof. This is one example.
In terms of how detailed a proof should be, this is about where to aim for; it’s
a little on the verbose side, but not much, and you’re better off aiming to be a
little verbose on an exam.)
Let x be given. We prove the the contrapositive: suppose ¬(∀z ∈ R)(0 < z ⇒ x <
z), so ∃z(0 < z ∧ z ≤ x). We must show that (∃y ∈ R)(y 2 < x).
We take a z so that 0 < z and z ≤ x.
Since 0 < z ≤ x, also 0 < z/2 < z ≤ x. Since z/2 is positive, z/2 has a real
square root: there is a y so that y 2 = z/2. Therefore y 2 < x.
3.
Prove that, for every natural number n,
−1 + 4 − 9 + 16 − · · · + (−1)n n2 = (−1)n
n(n + 1)
.
2
We prove this by induction on n.
Base case: when n = 1, −1 = (−1)1 1(2)
as needed.
2
Inductive case: assume
−1 + 4 − 9 + · · · + (−1)n n2 = (−1)n
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n(n + 1)
.
2
We want to show that
−1 + 4 − 9 + · · · + (−1)n n2 + (−1)n+1 (n + 1)2 = (−1)n+1
(n + 1)(n + 2)
.
2
We have
n(n + 1)
+ (−1)n+1 (n + 1)2
2
n(n + 1)
2
n+1
(n + 1) −
= (−1)
2
2
n+1 2(n + 1) − n(n + 1)
= (−1)
2
(2(n
+
1)
−
n)(n + 1)
= (−1)n+1
2
(2n
+
2
−
n)(n + 1)
= (−1)n+1
2
(n
+
2)(n
+ 1)
.
= (−1)n+1
2
−1 + 4 − 9 + · · · + (−1)n n2 + (−1)n+1 (n + 1)2 = (−1)n
This completes the inductive case.
4.
Here is a definition:
A tuple is a list with k entries for some positive integer k.
1. Give two examples of tuples of real numbers.
There are an unlimited number of examples here.
(7, 2, 15, 12, 35).
Some might be (3.1, π, 5), (7, 2),
2. Give two non-examples of tuples of real numbers.
Some non-examples would include tuples of non-real numbers like (3.1, π, i), or
sets {3.1, π, 5} (which are unordered).
3. Give a surprising example of a tuple of real numbers: a tuple which some might not realize
is a tuple.
This is pretty subjective, and any reasonable answer would give credit, but
it should have some surprising feature.
Some plausible answers are (π) (people might forget that a tuple of length 1
is still a tuple), (1, 2, 3) (perhaps someone might think ‘‘that’s a tuple of integers’’
and not realize that it is also a tuple of real numbers)
4. Give a surprising non-example of a tuple of real numbers: a non-tuple which some people
might mistake for a tuple.
Again, any reasonable answer will do. π (a real number by itself, rather than
the tuple (π) of length) is one. So is () (the ‘‘tuple’’ of length 0, which
was ruled out by the definition).
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5.
Either prove or give a counterexample to the statement:
1.
If f : A → B, g : A → B, h : B → C, h is injective, and h ◦ f = h ◦ g then f = g.
To show that f = g, are equal, we must show that for every x, f (x) = g(x).
Let x be given. Then (h◦f )(x) = (h◦g)(x) by assumption.
Therefore f (x) = g(x) because h is injective.
2.
So h(f (x)) = h(g(x)).
If f : A → B, g : A → B, h : B → C, h is surjective, and h ◦ f = h ◦ g then
f = g.
This is false. Suppose A = {a}, B = {b, c}, and C = {d}, and f (a) = b, g(a) =
c, and h(b) = h(c) = d. Then h ◦ f = h ◦ g, because (h ◦ f )(a) = d and (h ◦ g)(a) = d
and a is the only element of a. But f 6= g.
6.
Prove by induction:
Whenever S and T are sets with |S| = |T | = n (i.e. each set has n elements), there
are n! possible bijections f : S → T .
Note that this has the form
(∀n ∈ N)(∀S ∈ Sets)(∀T ∈ Sets) [(|S| = n ∧ |T | = n) → there exist n! possible bijections from S to T }] .
By induction on n.
Base case: when n = 1, so S = {s} and T = {t}, the only bijection from S to T
is f (s) = t. This is the 1 = 1! possible bijection.
Inductive case: Suppose that for any set S and any set T with |S| = |T | = n, there
are n! possible bijections from S to T .
Let S and T be given with |S| = |T | = n + 1. Pick some s ∈ S. For each t ∈ T ,
the sets S \ {s} and T \ {t} both have size n, so there are n! possible bijections
from S \ {s} to T \ {t}, each of which extends to exactly one bijection from S to
T (namely the one mapping s to t). So for each choice of t, there are n! bijections
mapping s to t, and since there are n+1 choices for t, there are (n+1)(n!) = (n+
1)! possible bijections from S to T .
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