Mathematical Thinking Stanford, W1 Assignment 2
1. Simplify the symbolic statements.
(b) (p≥7)∧(p<12), 7≤p<12
(c) (x>5)∧(x<7), 5<x<7
(d) (x<4)∧(x<6), x<4
(e) $(y<4)∧(y^2 <9), y^2<9$
(f) (x≥0)∧(x≤0), x=0
2. Express each of your simplified statements from question 1 in natural English.
(b) p is greater than or equal to 7 and less than 12.
(c) x is greater than 5, and less than 7.
(d) x is less than 4
(e) y squared is less than 9
(f) x is equal to 0
3. What strategy would you adopt to show that the conjunction φ1 ∧ φ2 ∧ . . . ∧ φn is true?
show that all of φ1 , φ2 , . . . , φn are true
4. What strategy would you adopt to show that the conjunction φ1 ∧ φ2 ∧ . . . ∧ φn is false?
show that one of φ1 , φ2 , . . . , φn is false.
5. Simplify the symbolic statements
(a) (π>3)∨(π>10), π>3 (b) (x<0)∨(x>0), x≠0 (c) (x=0)∨(x>0), x≥0 (d) (x>0)∨(x≥0), x≥0 (e) (x>3)∨($x^2$ >9), $x^2$ >9
6. Express the above symbolic statements in natural English.
(a) π is greater than 3
(b) x is not equal to 0
(c) x is equal to or greater than 0
(d) x is equal to or greater than 0
(e) x squared is greater than 9
7. What strategy would you adopt to show that the disjunction φ1 ∨ φ2 ∨ . . . ∨ φn is true?
show that one of φ1 , φ2 , . . . , φn is true
8. What strategy would you adopt to show that the disjunction φ1 ∨ φ2 ∨ . . . ∨ φn is false?
show that all φ1 , φ2 , . . . , φn are false
9. Simplify the symbolic statements
(a) ¬(π > 3.2), π ≤ 3.2 (b) ¬(x < 0), x ≥ 0 (c) ¬($x^2$ > 0), x = 0 (d) ¬(x = 1), x≠1 (e) ¬¬ψ, ψ
10. Express the above symbolic statements in natural English.
(a) negation of / not the case that Pai is greater than 3.2, Pai is less than and equal to 3.2 (b) negation of / not the case that x is less than 0, x is greater than and equal to 0 (c) negation of / not the case that x squared is greater than 0, x is equal to 0 (d) negation of / not the case that x equal to 1, x not equal to 1 (e) double negation of / not the case that ψ(Psi), ψ
11. Express in logical notation
(a) D ∧ Y
(b) ¬Y ∧ T ∧ D
(c) ¬(Y ∧ D)
(d) T ∧ ¬Y ∧ ¬D
(e) ¬T ∧ Y ∧ D
DISCUSS
Guilty and Proven
Truth Table, Binary logic, True or False
G | P | G ∧ P | description | |
---|---|---|---|---|
T | T | T | ✔︎ | Correctly Proven Guilty |
T | F | F | ✘ | Incorrectly Proven Innocent |
F | T | F | ✘ | Incorrectly Proven Guilty |
F | F | F | ✔︎ | Correctly Proven Innocent |
Not disPleased
Not disPleased is neutral, this is not a binary logic but a triary logic
triary logic
Positive | neutral | Negative |
---|---|---|
Pleased | not displeased | disPleased |
Pleased | ¬(¬ pleased) | |
¬ pleased | ||
¬ pleased = displeased
¬ displeased ≠ pleased
¬ pleased < ¬ ( ¬ pleased) < pleased
displeased < ¬ displeased < pleased