Discrete test 1 study

Question 1

conjunction of p and q

Answer 1

p ^ q p and q

true when both p and q are true

Question 2

disjunction of p and q

Answer 2

p v q, p or q

true when at least p or q is true

Question 3

exclusive or of p and q

Answer 3

true when p and q are different signals

Question 4

if p then q

Answer 4

p -> q,

true in all cases except true implies false.

Question 5

bi-conditional statement of p and q

Answer 5

pq, p if and only if q.

true when p and q have the same value. true or false

Question 6

Converse of p -> q

Answer 6

q -> p

Question 7

contrapositive of p -> q

Answer 7

~q -> ~p

Question 8

inverse of p -> q

Answer 8

~p -> ~q

Question 9

the converse is equivalent to

Answer 9

the inverse

Question 10

the original is equivalent to the

Answer 10

contrapositive

Question 11

how to tell if something is logically equivalent in 2 ways

Answer 11

use truth tables and match columns or use the table of logical equivalences to get to what you need

Question 12

tautology

Answer 12

all is true

Question 13

contradiciton

Answer 13

all is false

Question 14

contingancy

Answer 14

neither always true nor always false

Question 15

Axpx

Answer 15

for all x in px

Question 16

Expx

Answer 16

there exists an x that satisfies px

Question 17

E!xpx

Answer 17

there exists exactly one x that satisfys p(x)

Question 18

how to negate A and E

Answer 18

flip them and move it in

Question 19

even

Answer 19

The integer n is even if there exists an integer k such that n = 2k.

Question 20

odd

Answer 20

The integer n is odd if there exists an integer k such that n = 2k +1.

Question 21

rational number

Answer 21

A rational number p has the form 𝑎
𝑏
, where a and b are
integers and b ≠0.

Question 22

Direct proof

Answer 22

To prove 𝑝 → 𝑞 using a direct proof, you assume

that p is true and show that q must also be true

Question 23

Indirect proof

Answer 23

(proof by contraposition). To prove 𝑝 → 𝑞 using an

indirect proof, you prove ¬𝑞 → ¬𝑝 (the contrapositive) directly.

Question 24

Proof by contradiction

Answer 24

To prove that a proposition p is true, we
assume that it is false and attain a logical contradiction, such as
𝑟 ⋀ ¬𝑟. Then it must be the case that p was true

Question 25

Proof of 𝒑 ↔ 𝒒.contradiction

Answer 25

To prove 𝑝 ↔ 𝑞, we may prove both 𝑝 → 𝑞

and 𝑞 → 𝑝

Question 26

Vacuous proof.

Answer 26

We know that when p is false, 𝑝 → 𝑞 is true. For
example, if P(n) (n is an integer) is the statement “if n is odd, then
n
2
is odd,” then the proposition P(2) is vacuously true since ‘2 is
odd’ is false.

Question 27

Trivial proof

Answer 27

We know that when q is true, 𝑝 → 𝑞 is true. For
example, if P(n) (n is an integer) is the statement “if n is even, then
n + 1 > n,” then it P(n) is trivially true since n + 1 > n is true for all
integers, regardless of whether or not n is even.

Question 28

Proof by cases.

Answer 28

. To prove (𝑝1⋁𝑝2⋁ … ⋁𝑝𝑛) → 𝑞, we show that

𝑝𝑖 → 𝑞 for i = 1, 2, …, n.

Question 29

Existence Proofs

Answer 29

To prove ∃𝑥𝑃(𝑥) we may either construct an x
for which P(x) is true (constructive existence proof), or we may
demonstrate that P(x) is true for at least one x without actually
finding a specific x (nonconstructive existence proof).

Question 30

Unique Existence Proof

Answer 30

To prove ∃! 𝑥𝑃(𝑥) we must show:
∃𝑥(𝑃(𝑥) ∧ ∀𝑦(𝑃(𝑦) → 𝑦 = 𝑥)) [Existence and Uniqueness]
Outline: We can use either a constructive or nonconstructive
existence proof to existence To prove uniqueness, suppose P(x)
and P(y) are both true and show that we must have x = y.