How many Z+ ≤ 1000 have distinct digits and are even?

Solution.

Single digit numbers:        _____

                                    {2, 4, 6, 8} = 4 digits

Double digit numbers:       _____   _____

                                    4 digits  {0, 2, 4, 6, 8}

                                                5-1=4 digits

                                                exclude the digit used in the first position

Triple digit numbers:        _____   _____   _____

                                    4 digits 4 digits 5-2=3 digits

                                                           exclude the 2 digits already used in previous positions

The only four digit number we have is 1000, but it does not meet our conditions.

So we have, 4 + (4 x 4) + (4 x 4 x 3) = 68 distinct even digits ≤ 1000.

Don't you feel like this sometimes when you write proofs?! Ha, writing proofs can seem nutty <:-)

Math Help Forum

Check out this great forum. It contains threads discussing actual textbook problems. It is the best online source I have found.

www.mathhelpforum.com

1.4-6: Use a direct proof to show that the product of 2 odd integers is odd.

Section 1.4 Problem 6.

Proof. Assume that x and y are odd integers. By definition of odd, there are integers a and b such that x = 2a + 1 and y = 2b + 1. By substitution

xy = (2a + 1) (2b + 1)

= 4ab + 2a + 2b + 1

= 2(2ab + a + b) + 1.

Since 2, a, and b are integers and integers have closure with respect to multiplication and addition, then 2ab + a + b is an integer. By definition of odd, it follows that the product of xand y is odd. ☐

1.4-18a: Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition.

Section 1.4 Problem 18a.

Proof. Let n be an integer. Assume that n is odd. By definition of odd, there exists an integer a such that n = 2a + 1. By substitution,

3n + 2 = 3(2a + 1) + 2

= 6a + 3 + 2

= 6a + 5

= 2(3a + 2) + 1.

Since 3, a, and 2 are integers and integers have closure with respect to multiplication, then 3a + 2 is an integer. Therefore by the definition of odd, 3n + 2 is odd. ☐

1.4-26: Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even.

Proof. Let n be a positive integer. Assume that n is even. By definition of even, this means that there exists an integer a such that n = 2a. By substitution

7n + 4 = 7(2a) + 4

= 14a + 4

= 2(7a + 2).

Since 7, a, and 2 are integers and integers have closure under addition and multiplication, then 7a + 2 is an integer. Therefore by the definition of even, 7n + 4 is even.

Now assume that n is odd. By the definition of odd, this means there is an integer b such that n = 2b + 1. By substitution

7n + 4 = 7(2b + 1) + 4

= 14b + 7 + 4

= 14b + 11

= 2(7b + 5) + 1.

Since 7, b, and 5 are integers and integers have closure under addition and multiplication, then 7b + 5 is an integer. Therefore by the definition of odd, it follows that 7n + 4 is odd.

Since n is odd implies that 7n + 4 is also odd, then clearly 7n + 4 is even implies n is even must be equivalently true.

Furthermore, since n is even implies 7n + 4 is even, and conversely 7n + 4 is even implies n is even are true, then by the definition of a biconditional statement it follows that the original proposition n is even if and only if 7n + 4 is even must also be true. ☐

The Set of Real Numbers

Real Numbers:

{1, 2, 3, 4, ... } Natural numbers, counting numbers, and positive integers

{0, 1, 2, 3, ...} Whole numbers, nonnegative integers

{..., -3, -2, -1} Negative integers

{..., -1, 0} Nonpositive integers

{..., -1, 0, 1, ...}   Integers

a/b Let a and b are integers, then a/b is a Rational Number (can be either a terminating or repeating decimal).

i.e. √2, π Irrational numbers are nonrepeating and nonterminating decimals.

Terms:

Repeating decimal or recurring decimal: a fraction with a prime denominator; a number, that when expressed in the decimal numeric system, at some point the same sequence of digits repeat infinitely-many times. i.e. 1/3=0.333... said as "0.3 repeating"

Terminating decimal represents a rational number that when represented in the decimal numeric system, its decimal at some point ends. i.e. 0.25

Truth value? ∀x∃y (x=y²)

Section 1.4
Determine the truth value and state why.
28b. ∀x∃y: x=y²

hmmm?

If you have worked with me, then you know what I'm going to say. Let's rewrite this to help our visual memory. Since most of us are accustomed to seeing the "y" on the left-hand side of the equations, let's write this:

from: ∀x∃y: x=y²

to: ∀x∃y: y²=x

Solution:

y²=x

√y² = √x
y = √x.

For this to be true, then x must be only non-negative integers. Therefore, our original statement is false.

Show that ∃n∃m (n² + m² = 6) is False

Section 1.4
27f. Prove that ∃n∃m (n² + m² = 6) is False.

The only possible pairs of integers whose sum is 6 are 1+5, 2+4, and 3+3. Since none of 2, 3, or 5 are squares of integers, then clearly the pairs of integers whose sum is 6 are not solutions that satisfy ∃n∃m (n² + m² = 6).

∀x (P(x) → Q(x)) and ∀xP(x) → ∀xQ(x)

Section 1.3 #43.
Determine whether ∀x (P(x) → Q(x)) and ∀xP(x) → ∀xQ(x) are logically equivalent. Justify your answer.

Proof: ∀x (P(x) → Q(x))≡ ∀x (¬P(x) ∨ Q(x)) by OULE (p→q). Since the universal quantifier, namely ∀x, cannot be distributed over a disjunction, it is not logically equivalent to ∀xP(x) → ∀xQ(x).

1.2 12D In Progress

1 [(p v q) ^ (p => r) ^ (q => r)] => r
2 [(p v q) ^ (~p v r) ^ (~q v r)] => r by OULE
3 ~[(p v q) ^ (~p v r) ^ (~q v r)] v r by OULE
[~(p v q) ^ ~(~p v r) v ~(~q v r)] v r by De Morgan's Law (I think this line should replace step 4.)
[(~p ^ ~q) ^ (~p ^ ~r) v (q ^ ~r)] v r by De Morgan's Law and Double Negation
[(~p ^ ~p) ^ (~q ^ ~r) v (q ^ ~r)] v r by Commutative and Association Laws
[~p ^ (~q ^ ~r) v (q ^ ~r)] v r by Idempotent Law
[~p ^ ((~q ^ ~r) v q) ^ ((~q ^ ~r) v ~r))] v r by Distributive Law
[~p ^ ((~q v q) ^ (~r v q)) ^ ((~q v ~r) ^ (~r v ~r))] v r by Distributive Law
[~p ^ ((T) ^ (~r v q)) ^ ((~q v ~r) ^ ~r)] v r by Commutative and Negation Law
[~p ^ (~r v q) ^ ((~q v ~r) ^ ~r)] v r by Identity Law
[((~p ^ ~r) v (~p ^ q)) ^ ((~q v ~r) ^ ~r)] v r by Distributive Law
(r v ((~p ^ ~r) v (~p ^ q))) ^ (r v ((~q v ~r) ^ ~r)) by Commutative and Distributive Laws (Review this line.)
(r v (~p ^ ~r))
4 [(~p v ~q) v (p v ~r) v (q v ~r)] v r by DeMorgans
5 [(~p v p) v (~q v ~r) v (q v ~r)] v r by Associative
6 [(T) v (~q v ~r) v (q v ~r)] v r by Domination
7 [T v (q v ~r)] v r by Domination
8 T v r = T by Domination

Section 1.2 Problem 12 b (my solution)

Prove [(p=>q) ^ (q=>r)] => (p=>r) is a Tautology

Proof: " "

1 = [(~pvq) ^ (~qvr)] => (~pvr) by OULE p=>q = ~pvr
2 =~[(~pvq) ^ (~qvr)} v (~pvr) by OULE " "
3 =[~(~pvq) v ~(~qvr)] v (~pvr) by De Morgan's Law
4 =[(~~p^~q) v (~~q^~r)] v (~pvr) by De Morgan's Law
5 =[(p^~q) v (q^~r)] v (~pvr) by Double Negation Law
6 =[((p^~q) v q ^ ((p^~q) v ~r)] v (~pvr) by Distributive Law
7 =[((pvq) ^ (~qvq)) ^ ((p^~q) v ~r))] v (~pvr) by Distributive Law
8 =[((pvq) ^ T) ^ ((p^~q) v ~r)] v (~pvr) by Negation law
9 =[(pvq) ^ ((p^~q) v ~r)] v (~pvr) by Identity law
10 =(~pvr) v [(pvq) ^ ((p^~q) v ~r)] by Commutative Law
11 =[(~pvr) v (pvq)[ ^ [(~pvr) v ((p^~q) v ~r] by Distributive Law
12 =[(~pvp) v (rvq)] ^ [(rv~r) v (~pv(p^~q))] by Commutative Law
13 =[T v (rvq)] ^ [Tv(~pv(p^~q))] by Negation Law
14 =T^T by Domination Law
15 =T by Idempotent Law or Definition of Conjunction

Section 1.2 Problem 12 b

Prove that [(p -> q) ^ (q -> r)] -> (p -> r) is a tautology.

1 = ~[(~p v q) ^ (~q v r)] v (~p v r) by OULE - (p => q) = ~p v q
2 = [~(~p v q) v ~(~q v r)] v (~p v r) by DeMorgans
3 = [(p ^ ~q) v (q ^ ~r)] v (~p v r) by deMorgans and Double Negation
4 = [((p ^ ~q) v q) ^ ((p ^ ~q) v ~r)] v (~p v r) by commutative and distributive
5 = [((q v p) ^ (q v ~q)) ^ ((p ^ ~q) v ~r)] v (~p v r) by Negation and Identity
6 = [(q v p) ^ ((p ^ ~q) v ~r)] v (~p v r) by commutative and distributive
7 = ((~p v r) v (q v p)) ^ ((~p v r) v ((p ^ ~q) v ~r)) by commutative and associative
8 = (p v ~p) v (q v r) ^ ((r v ~r) v (~p v (p ^ ~q))) by Negation
9 = (T v (q v r)) ^ (T v (~p v (p ^ ~q))) by commutative and domination
10 = T ^ T = T

(taken from another student in class)