Mathematical Thinking Stanford, W3 Assignment 5

(∀𝑚∈ ℕ)(∃𝑛 ∈ ℕ)(𝑛>𝑚), True Express the existence assertions a. ($∃x$ ∈ ℕ)($x^3 = 27$ ) b. ($∃𝑛$ ∈ ℕ)(𝑛>10000) c. natural number n is not a prime ($∃p$ ∈ ℕ)($∃m$ ∈ ℕ)($p$>1 ∧ $m$>1 ∧ $n=pm$) Express the ‘for all’ assertions a. ($∀x$ ∉ ℕ)($x^3$ = 28) ¬($∃x$ ∈ $ℕ$)($x^3$ = 28) ($∀x$ ∈ ℕ)($x^3$ ≠ 28) ($∀x$ ∈ ℕ)¬($x^3$ = 28) b. ($∀n$ ∈ ℕ)($n>0$ ) c. ($∀p$ ∈ ℕ)($∀q$ ∈ ℕ)[( $n=pq$) ⇒ ($p=1$ V $q=1$)] ...

2023-9-29 · 4 min · Atom.X

Mathematical Thinking Stanford, W3

the most difficult lecture Analysis of language - quantifiers irrational numbers ∀ for all ∃ there exists express an existence assertion. a confident and forceful statement of fact or belief. Combination of quantifiers there is no largest natural number. (∀𝑚∈ ℕ)(∃𝑛 ∈ ℕ)(𝑛>𝑚), True (∃𝑛 ∈ ℕ)(∀𝑚∈ ℕ)(𝑛>𝑚), False American Melanoma Foundation: “One American dies of Melanoma almost every hour.” ∃A∀H(A dies in hour H), False, misunderstanding: An American dies once every hour, ridiculous. ...

2023-9-28 · 1 min · Atom.X

Mathematical Thinking Stanford, W2 Assignment 4

1. Building truth table φ ⇔ ψ φ ψ φ ⇒ ψ ψ ⇒ φ φ ⇔ ψ T T T T ✔︎ T F F T F T T F F F T T ✔︎ a. φ ⇔ ψ is true if φ and ψ are both true or both false (φ ⇒ ψ)=(ψ ⇒ φ) , (φ ⇒ ψ)∧(ψ ⇒ φ) = φ ⇔ ψ, φ = ψ ...

2023-9-25 · 8 min · Atom.X

Mathematical Thinking Stanford, W2 QUIZ

Math Foundation of computing, Stanford university. Preliminary Course Notes - Keith Schwarz implication has a truth part(conditional) and a causation part. implication = conditional + causation conditional means ⇒ φ ⇒ ψ is the truth part of “ φ implies ψ ”. φ is the antecedent ψ is the consequent define the truth of φ⇒ψ in terms of the truth/falsity of φ and ψ. Equivalence Quiz Which of the following conditions is necessary and sufficient for the natural number $n$ to be multiple of 10 ? ...

2023-9-25 · 5 min · Atom.X

Mathematical Thinking Stanford, W2 Assignment 3

1. D, Y, T a. T ⇒ [D ∧ Y] b. D ⇒ ¬ Y c. ¬ D ⇒ ¬ T d. T ⇒ ¬ (D ∧ Y) e. ¬ D ∧ Y ∧ T , T ⇒ [¬ D ∧ Y], following means “then or and”, they are all conjunctions relationship. f. (T ∧ Y) ⇒ ¬ D , T ⇒ [Y ⇒ ¬ D] g. T ⇒ [D ⇔ Y], T⇒ [(D ⇒Y) ∧ (Y⇒D)] ...

2023-9-22 · 1 min · Atom.X