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MATH 201
Solutions: TEST 3-A
(in class)
(revised)
God created infinity, and man, unable to understand infinity, had to invent finite sets.
- Gian Carlo Rota
Part I [5 pts each]
1.
Let X be a set. Define P(X), the power set of X.
P(X) is the set of all subsets of X.
2. Let F: X →Y be a mapping. Define: F is injective.
F: X →Y is injective if:
∀𝑎, 𝑏 ∈ 𝑋 𝐹(𝑎) = 𝐹(𝑏) ⇒ 𝑎 = 𝑏
3. Let G: X →Y be a mapping. Define: G is surjective.
G: X →Y is surjective if:
4.
∀𝑐 ∈ 𝑌 ∃𝑎 ∈ 𝑋 𝐺(𝑎) = 𝑐
n
Give the combinatorial definition of   --- not the means of calculating this number or any other interpretation
r
 
using factorials.
n
  is the number of ways that an (unordered) subset can be chosen from a set of n items.
r 
5. Let S be a set. Define: S is countably infinite.
S is countably infinite if there exists a bijection from the set of natural numbers to S.
6. Let R and S be sets. What does it mean to assert that R and S are of the same cardinality?
Two sets, R and S, are of the same cardinality if there exists a bijection from R to S
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7.
If |A| = 9876, then |P(A)| =
If |A| = 9876, then |P(A)| = 29876
8. Let X and Y be sets and let G: X→Y be a bijection. Define G-1: Y → X
Let 𝑦 ∈ 𝑌. 𝑆𝑖𝑛𝑐𝑒 𝐺: 𝑋 → 𝑌 is a bijection, there exists one and only one element x ∈ 𝑋 for which G(x) = y.
We define G(y) = x.
Part II [10 pts each] Answer any 11 of the following 14. You may solve more for extra credit.
1.
Here is a partial proof of Cantor’s theorem. Your job is to complete it.
Theorem: For any set X, P(X) and X are not of the same cardinality.
Proof: Let X be a set and P(X) be the power set of X.
Strategy: proof by contradiction.
Assume that  bijection F: X →P(X).
Define 𝐷∗ = {𝑞 ∈ 𝑋 |𝑞 ∉ 𝐹(𝑞)}
𝑵𝒐𝒘, 𝒔𝒊𝒏𝒄𝒆 𝑭 𝒊𝒔 𝒔𝒖𝒓𝒋𝒆𝒄𝒕𝒊𝒗𝒆, ∃ 𝒙∗ ∈ 𝑿 for which F(𝒙∗ ) = 𝑫∗ .
Next, suppose that 𝒙∗ ∈ 𝑫∗ . Then, by definition of 𝑫∗ , 𝒙∗ ∉ 𝑭(𝒙∗ ) = 𝑫∗ .
This is impossible.
Thus 𝒙∗ ∉ 𝑫∗ . But, 𝑫∗ = 𝑭(𝒙∗ ); 𝒔𝒐 𝒙∗ ∉ 𝑭(𝒙∗ ).
𝑻𝒉𝒊𝒔 𝒎𝒆𝒂𝒏𝒔 𝒕𝒉𝒂𝒕 𝒙∗ ∈ 𝑫∗ . 𝑩𝒖𝒕 𝒕𝒉𝒊𝒔 𝒕𝒐𝒐 𝒊𝒔 𝒊𝒎𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆.
Hence our assumption that F is a bijection is false.
2. Using the Euclidean algorithm, find gcd(210, 45)
210 = 45*4 + 30
45 = 30*1 + 15
30 = 15*2 + 0
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gcd = 15
3. Using the extended Euclidian algorithm, find integers x and y such that
56x + 22y = gcd(56, 22).
56 = 22*2 + 12
22 = 12*1 + 10
12 = 10*1 + 2
10 = 5*2 – 0
Hence gcd(56, 22) = 2
So 2 = 1256 – 10
= 12 – (22 – 12*1)
= 2*12 – 22
= 2(56 – 2*22) – 22
= 2*56 – 4*22 – 22
= 2*56 – 5*22
Hence 56(2) – 22(-5) = 2
4. Odette believes that the set of all finite sequences using the alphabet {a, b, c} is countably infinite. Give proof or
counterexample. (For example: aab, cabcab, ccccccccccccccc are all finite sequences.)
Solution:
Odette is correct: Begin by listing all strings of length 1:
Then all 9 strings of length 2:
a, b, c
aa, ab, ac, ba, bb, bc, ca, cb, cc
Then all 27 strings of length 3, etc.
5. Swann believes that the set of all infinite sequences using the alphabet {a, b, c} is countably infinite. Give proof
or counterexample. (For example:
Solution:
abababab… , abcabccccbbaabbbbb… are infinite strings.)
Swann is mistaken. Just use Cantor’s diagonal argument on the set of infinite strings:
First string:
E11, E12, E13, E14, …
Second string: E21, E22, E23, E24,
Third string:
E31, E32, E33, E34,
Etc.……
Define the string e* as follows:
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𝑎 𝑖𝑓 𝐸𝑖 𝑖 = 𝑏 𝑜𝑟 𝑐
e*i = {
𝑐 𝑖𝑓 𝐸𝑖 𝑖 = 𝑎
Now notice that e*i cannot appear in our infinite list of strings
6. Explain briefly, using only a combinatorial interpretation, why
1776  1776 

  

 99  1677 
(Use complete sentences. Do not use any formulas. Do not use more general results
about combinations.)
Solution:
Let S be a set of 1776 objects.
Then there is a bijection between the set of all subsets of S of
size 99 and the set of all subsets of S of size 1677. This is true since selecting a subset of size 99 corresponds to
selecting a subset of size 1776 – 99 = 1677. It is not difficult to see that this correspondence is a bijection.
7.
Explain briefly, using only a combinatorial interpretation, why
1789  1788  1788 

  
  

25
25
24

 
 

(Use complete sentences. Do not use any formulas. Do not use more general
results about combinations.)
Solution:
Let S be a set of 1789 items. Call one of them a*. Then S = (S ~ {a*})  {a*}. Note that
(S ~ {a*}) has 1788 items. Two cases (mutually exclusive):
Case I: Choose 25 items from S ~ {a*}. Next, if we include {a*}). then we must choose the
remaining 24 items from S.
Case II: Choose all 25 items from (S ~ {a*})
Hence
1789  1788  1788 

  
  

 25   25   24 
8. Given a group of 13 married couples. The number of ways we can choose a subset of 5 individuals from this
group which contains no married couple is
Solution:
13 
choose 5 couples:  
5 
13 

5 
From each of the 5 chosen couples, choose 1 of the 2 partners. Hence the number of ways = 25 
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9. Let A, B be non-empty finite sets. Suppose that |A|=1789 and |B| = 2016. Then
(a) The number of surjections from A into B is __zero_____. Why?
The range of any mapping from A into B cannot contain more than 1789 elements.
(b) The number of injections from A into B is P(2016, 1789). Why?
Each element of A can “choose” one of the remaining elements of B. In particular, the first element of A can select from
1789 elements; the second element of A can choose from 1788 elements; etc.
10. Let Q be the set of all rational numbers. Define f: Q  Q as follows:
3
 x if x  0

f ( x)  
0 if x  0


(a) Is f well-defined? Justify your answer.
Solution: Yes.
For any non-zero rational number, x = p/q, we have f(x) = 3/x = 3q/p which is also rational.
And, if x = 0, f(0) = 0 is also rational.
(b)
Is f injective? Why?
Solution:
Yes. Let x and y be non-zero rational numbers. Now assume that f(x) = f(y). Then 3/x = 3/y, which implies
that x = y.
(c) Is f surjective? Why?
Solution: Yes: Let y be any non-zero rational number. Then 1/(3y) is also rational and f(1/(3y)) = y.
11. For any real numbers, c and d, let us define the binary operation  as follows:
c  d = c 2 + d2 – 1
Give either a brief justification or counterexample for each of the following assertions:
(a) The set of integers is closed under the operation  .
Solution: Yes. Integers are closed under taking squares as well as under addition and subtraction.
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(b) The set of even integers is closed under the operation .
Solution:
No. For a counter-example, let c = 2 amd d = 4. Then c  d = c2 + d2 – 1 = 4 + 16 – 1 = 19, an odd
number.
(c) The set of odd integers is closed under .
Solution: Yes
If c and d are odd, then c2 and d2 are also odd (proven earlier in class). Since the sum of two odd numbers
is even, we have c2 + d2 is even.
Now c2 + d2 – 1 is odd.
(d) The set of positive integers is closed under .
Solution: Yes.
If c and d are positive integers, then c2 + d2 must be at least 2. Thus c  d = c2 + d2 – 1 is at
least 1.
(e) The set of rational numbers is closed under .
Solution: Yes. If c and d are rational, then each of c2 and d2 must be rational. Since the sum and difference
of two rationals is rational, it follows that c  d = c2 + d2 – 1 is also rational.
(f) The set of irrational numbers is closed under .
Solution:
No. Let c = d = √2. Then c and d are irrational (as proven in class). But c  d = c2 + d2 – 1 = 5,
which clearly is rational.
12. Let X = {0, 2, 5, 8, 13, 15} and Y = P(X). Define T: X →P(X) as follows:
{ j} if j is prime
j  X T ( j )  
{0, 8, 13} if j is not prime
Find D* (as defined in Part II, problem 1)
Answer: D* = {15}
13. Give an example of three sets, X, Y, S, and two mappings, F: X →Y, G: Y →S such that 𝐺 is not injective yet 𝐺 ∘ 𝐹
is injective.
Solution:
Let X = {a}, Y = {b, c} and S = {d}.
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F: X → Y by: F(a) = c and
Define
G: Y → S by G(b) = G(c) = d.
Then G is clearly not injective, yet 𝐺 ∘ 𝐹 is injective since the domain of 𝐺 ∘ 𝐹 consists of a single point.
14.
Let S be the set of all polynomials for which odd powers of x (i.e., 1, 3, 5, 7,…) do not appear.
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For example, 4 + ½ x2 –  x8 ∈ 𝑆, 𝑏𝑢𝑡 𝑥 2 − 4 𝑥 81 ∉ 𝑆.
(a) Is S closed under differentiation? Why?
Solution: No. Counter-example.
Let p(x) = x4 ; this is a member of S but dp/dx = 3x3 is not a member of S.
(b) Is S closed under the operation of taking the second derivative?
Solution: Yes.
even integer ≥ 2
Let p be a finite sum of terms of the form k xn. where k is any real number and n is any
Now d2/dx2 p = kn(n-1) xn-2. But since n is even, n – 2 must be even as well.
(c) Is S closed under multiplication by x3?
Solution:
No. Counter-example:
Let p(x) = x4. Then x4 p(x) = x7 ∉ 𝑆.
(d) Is S closed under multiplication by x4 ?
Solution: Yes, x4p(x) consists of a finite sum of terms of the form xn+4. Since n is even, so is n+4.
(e) Is S closed under the following operation *?
For all 𝑝 ∈ 𝑆 𝑝 ∗ = p(1 + 𝑥 4 )
Solution: Yes. Let n be a non-negative even integer. Then one must verify that (1 + x4)n consists only of even
powers of x. This follows directly from the binomial theorem.
(f) Is S closed under the following operation :
For all 𝑝 ∈ 𝑆 𝑝 = p(1 + 𝑥)
Solution:
No. Counter-example.
Let p(x) = x2. Then p(1 + x) = (1 + x)2 = 1 + 2x + x2 ∉ S.
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Extra Credit: When 200! Is expanded, how many consecutive trailing zeroes appear at the end of the number?
Solution:
200! may be written (uniquely) as the product abc where a is a power of 2, b is a power of 5 and c
is neither divisible by 5 nor by 2. Now the number of occurrences of 10 is equal to the number of 5s in 2001!
since the 2s are more abundant than the 5s.
Now we must count:
There are 40 numbers in the product divisible by 5
8 numbers divisible by 25
1 number divisible by 125
Hence the number of 5s (and thus the number of terminating 0s) in 200! Is 40 + 8 + 1 = 49.