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MAT 425: Introduction to Analysis I
Solution #1
Set 1 → Quantifiers.
a) (Strichartz, §1.1.3, Exercise #2.) For each of the following mathematical
statements, rewrite the statement making all the quantifiers explicit. Then form
the negation of the statement.
a. Every positive integer has a unique prime factorization.
Original statement: ∀n ∈ Z>0 , ∃!p1 , p2 , ..., pk , s.t. pi is prime for i = 1, 2, ..., k and
n = p1 p2 ...pk .
Negation: ∃n ∈ Z>0 s.t. ∀p1 , p2 , ..., pk , where pi is prime for i = 1, 2, ..., i, n 6= p1 p2 ...pk .
c. Multiplication of integers is associative.
Original statement: ∀m, n, k ∈ Z, m(nk) = (mn)k.
Negation: ∃m, n, k ∈ Z s.t. m(nk) 6= (mn)k.
d. Two points in the plane determine a line.
Original statement: ∀(x1 , y1 ), (x2 , y2 ) ∈ R2 , ∃ line joining both points.
Negation: ∃(x1 , y1 ), (x2 , y2 ) ∈ R2 , s.t. no line joins both points.
Set 2 → Cardinality.
a) (Strichartz, §1.2.3, Exercise #5.) Let A1 , A2 , A3 , ... be countable sets, and
let their Cartesian product A1 × A2 × A3 × ... be defined to be the set of all
sequences (a1 , a2 , a3 , ...), where ak is an element of Ak for some k ≥ 1. Prove that
the Cartesian product is uncountable. Show that the same conclusion holds if
each of the sets A1 , A2 , A3 , ... has at least two elements.
The trick here is to follow what we did in class when we showed that the open interval
(0, 1) is uncountable, i.e., we assume it is countable and prove (by contradiction) that if all
of its elements were to be enumerated, and therefore in 1-1 correspondence with N, that one
could construct a sequence that is not on that list.
Here, the elements of the list are not real numbers between 0 and 1, but sequences of
elements in the Cartesian product space A1 × A2 × A3 × .... Here is the list:
(a11 , a12 , a13 , a14 , a15 , ...)
(a21 , a22 , a23 , a24 , a25 , ...)
–2–
(a31 , a32 , a33 , a34 , a35 , ...)
(a41 , a42 , a43 , a44 , a45 , ...)
(a51 , a52 , a53 , a54 , a55 , ...)
...
By using Cantor’s diagonalization argument, we can construct the sequence (a11 , a22 , a33 , a44 , ...),
where the “bar” means any element from Ak which differs from akk (∀k ∈ N).
From this argumentation, one can see that no assumption on the cardinality (finite vs
infinite) of the sets A1 , A2 , A3 , ... was made. And in fact, they need not be infinite. As long
as each of the sets has at least two elements, we’re good. Why two? Just to make sure that
there is (at least one) option available to us when looking for the akk 6= akk .
b) Show that the open interval (0, 1) has the same cardinality as R.
One can show that f : (0, 1) → R, where f (x) = tan(πx − π/2) is 1-1 and onto. Another
function that can be used is g(x) = x1−2x
2 −x . There are probably many other functions that
work. Just remember to show that it is both 1-1 and onto.
Set 3 → Proofs. Prove the following statements.
a) The equality (1 + 2 + 3 + ... + n)2 = 13 + 23 + 33 + ... + n3 holds for any integer
n > 0.
Straigth-forward induction proof, using the fact that 1 + 2 + ... + n = n(n + 1)/2, which
was proven in class.
√
b) If p is a prime number, then p is an irrational number.
√
A generalization of the proof that 2 is irrational, which was proven in class. The main
√
√
difference is that for 2, we used the fact that ∀n ∈ N, 2|n2 ⇒ 2|n, whereas for p, we need
the fact that p|n2 ⇒ p|n, which can be proven, for example, by contrapositive and using the
unique prime factorization.
c) For every integer n > 0, 5n + (2)3n−1 + 1 is divisible by 8.
This is done by induction.
1) Let P (n) : 8|5n + (2)3n−1 + 1.
2) Show that P (1) is true: 51 + (2)31−1 + 1 = 5 + 2 + 1 = 8. Since 8|8, P (1) is true.
3) Assume that P (k) is true. Show that P (k + 1) is also true. (Note: In an induction
proof, the induction hypothesis (P (k)) is the most important tool in the proof, so we want
–3–
to make sure we use it.) We want to show that 5k+1 + (2)3k + 1 is divisible by 8. We can
say that
5k+1 + (2)3k + 1 =
(5)5k + (3)(2)3k−1 + 1 =
(4 + 1)5k + (2 + 1)(2)3k−1 + 1 =
k
k−1
k
k−1
5 + (2)3
+ 1 + (4)5 + (2)(2)3
.
Since we assumed that P (k) was true, we know that the left term is divisible by 8. So we
need to show that the right term is also divisible by 8. This right term can be rewritten
4(5k + 3k−1 ), which is obviously divisible by 4, so that we need to show that 5k + 3k−1 is
even. Well 5k is always odd (can easily be shown by induction) and the same is true about
3k−1 . And since the sum of two odd integers is even, we are done.