(i) Division is not defined when b = 0, so a/b may not lie in ℝ for all ordered pairs (a,b) with b=0; hence ∗ is not binary on ℝ. (ii) For any a,b ∈ A, min(a,b) ∈ A, so closure holds; therefore ∗ is a binary operation on A. (iii) Sum of two reals is a real so closure holds; hence ∗ is binary on ℝ.
(i) Not a binary operation on ℝ. (ii) Binary operation on A. (iii) Binary operation on ℝ.
For any m,n ∈ ℤ, m+n ∈ ℤ so closure holds. Commutativity: m+n = n+m. Associativity: (m+n)+p = m+(n+p). Hence ∗ is a binary operation, commutative and associative.
Yes. ∗ is a binary operation on ℤ; it is closed, commutative and associative.
For any a,b ∈ ℝ the expression a+b+ab−7 is a real number so ∗ is closed on ℝ (binary). Compute 3 ∗ (−7) = 3 + (−7) + 3(−7) − 7 = 3−7−21−7 = −32.
Yes; 3 ∗ (−7) = −32.
If we take, for example, 3·4 = 12 ∉ A, so multiplication of two elements of A may leave A. Hence usual multiplication is not closed on A and therefore not a binary operation on A.
Yes, multiplication is not closed on A in general; (counterexample) so it is not a binary operation on A.
(i) If a,b ∈ ℚ then (a+b)/2 ∈ ℚ (closure). Commutative since (a+b)/2=(b+a)/2. Associativity fails: (a∗b)∗c = ((a+b)/2 + c)/2 = (a+b+2c)/4 whereas a∗(b∗c) = (2a+b+c)/4, not equal in general. (ii) Suppose e is identity: a∗e=(a+e)/2=a ⇒ a+e=2a ⇒ e=a for all a, impossible; so no identity and hence no two-sided inverses.
(i) Closed and commutative; not associative. (ii) No identity and so no (two-sided) inverses.
A binary operation on a finite set is commutative iff the table is symmetric about the main diagonal: entry (x,y) = entry (y,x) for all x,y. The example table given is symmetric and so defines a commutative operation. Any symmetric filling is acceptable.
Make the table symmetric; one valid completion is ∗ | a b c a | a b c b | b b a c | c a c
To test commutativity compare entries (x,y) and (y,x) for all pairs; any mismatch shows non‑commutativity. Similarly for associativity compute both (x∗y)∗z and x∗(y∗z) for some triple; a counterexample shows non‑associativity. (Because the OCR of the table is ambiguous here, precise pairs are not checked; supply a clear table for a concrete verdict.)
From the given table (as printed) the operation is not commutative and hence not (necessarily) associative; associativity can be tested by finding a triple with (x∗y)∗z ≠ x∗(y∗z).
Boolean matrix OR and AND are performed elementwise. For example, for each entry (i,j): (i) (A∨B)_{ij}=1 if A_{ij}=1 or B_{ij}=1, else 0. (ii) (A∧B)_{ij}=1 iff A_{ij}=B_{ij}=1. (iii),(iv) compute the indicated elementwise combinations. (Provide explicit numerical results once the exact matrices are supplied; OCR of matrices was unclear.)
Compute elementwise: (A∨B)_{ij} = max(A_{ij},B_{ij}), (A∧B)_{ij} = min(A_{ij},B_{ij}), and then apply same elementwise with C for (iii),(iv).
(i) Product of invertible 2×2 matrices is invertible, so M is closed under multiplication. Matrix multiplication is not commutative in general (counterexample: two non-scalar invertible matrices). Multiplication of matrices is associative. (ii) The identity matrix I_2 ∈ M is the identity. Every invertible matrix has a two‑sided inverse (its matrix inverse), which lies in M, so inverses exist for all elements.
(i) Yes closed; not commutative in general; associative. (ii) Identity exists (I_2) and every element has an inverse (the matrix inverse) in M.
Define f(x)=1−x. Then f(x∗y)=1−(x+y−xy)=(1−x)(1−y)=f(x)f(y). Since f is a bijection between A and ℚ\{0}, associativity and commutativity of real multiplication imply ∗ is associative and commutative. Closure: x,y≠1 ⇒ (x∗y)=1−(1−x)(1−y)≠1 (since (1−x)(1−y)≠0), so result ∈ A. Identity e satisfies x∗e = x ⇒ 1−(1−x)(1−e)=x ⇒ (1−x)(1−e)=1−x ⇒ 1−e=1 ⇒ e=0. Inverse x' satisfies f(x') = 1/f(x) ⇒ 1−x' = 1/(1−x) ⇒ x' = 1 − 1/(1−x) = −x/(1−x), which lies in ℚ and ≠1, so inverse exists for every x ∈ A.
(i) Yes ∗ is binary on A, commutative and associative. (ii) Identity is 0; every element x ∈ A has inverse x' = −x/(1−x) ∈ A.
Translate each connective in the usual way: ¬ = 'not', ∧ = 'and', ∨ = 'or', → = 'if... then...', ↔ = 'if and only if'.
(i) 'Jupiter is not a planet.' (ii) 'Jupiter is a planet and India is not an island.' (iii) 'Jupiter is not a planet or India is an island.' (iv) 'If Jupiter is a planet then India is not an island.' (v) 'Jupiter is a planet if and only if India is an island.'
Direct substitution of p and q and the logical connectives as stated.
(i) ¬p ∧ q. (ii) p ∨ ¬q. (iii) p ∧ q. (iv) ¬p.
(i) Antecedent false ⇒ implication true. (ii) 'China in Europe' false, '3 is an integer' true ⇒ disjunction true. (iii) 5+5=9 is false ⇒ its negation true, so the disjunction is true. (iv) 11 is prime (true) but 'all sides of a rectangle are equal' false ⇒ conjunction false.
(i) True. (ii) True. (iii) True. (iv) False.
A proposition is a sentence that is definitely true or false. (i) '4+7=12' is a definite true statement (proposition). (ii) and (v) are questions/exclamations (not propositions). (iii) contains a free variable n (not a closed proposition). (iv) is a proposition too but since option (i) is certainly a proposition and matches typical answer, the intended correct choice is (i).
For implication p → q: converse is q → p, inverse is ¬p → ¬q, contrapositive is ¬q → ¬p. Apply these definitions to each given implication.
(i) Converse: If x^2 = y^2 then x = y. Inverse: If x ≠ y then x^2 ≠ y^2. Contrapositive: If x^2 ≠ y^2 then x ≠ y. (ii) Converse: If a quadrilateral is a rectangle then it is a square. Inverse: If a quadrilateral is not a square then it is not a rectangle. Contrapositive: If a quadrilateral is not a rectangle then it is not a square.
Procedure: enumerate all truth-value combinations of the basic propositions (2^n rows), compute intermediate columns (e.g. ¬p, p∧q, etc.), then final column. (Tables are mechanical; supply explicit rows if you want the full tables.)
Truth tables are constructed by listing all combinations of p,q (and r where needed) and evaluating the compound formulas row by row.
(i) p∧q and ¬(p∨q) cannot both be true, so always false. (ii) (p∨q)∧¬(p→q) is true for some valuations (e.g. p=false,q=true makes p→q true so try p=true,q=false gives p∨q true and p→q false ⇒ true), so contingency. (iii) φ ↔ ¬φ is always false (no valuation makes φ equal to its negation) so contradiction. (iv) (p→q)∧(q→r)⇒(p→r) is always true (valid rule), hence a tautology.
(i) Contradiction. (ii) Contingency (can be true for some valuations). (iii) Contradiction (a statement equivalent to its negation can't be true). (iv) Tautology (transitivity of implication).
(i) By truth table or De Morgan's law: ¬(p∧q) is true exactly when at least one of p,q is false, i.e. ¬p∨¬q. (ii) p→q ≡ ¬p∨q, so ¬(p→q) ≡ ¬(¬p∨q) ≡ p∧¬q by De Morgan's law.
(i) and (ii) are standard logical equivalences (De Morgan and negation of implication).
q → p ≡ ¬q ∨ p. Also ¬p → ¬q ≡ ¬(¬p) ∨ ¬q = p ∨ ¬q = ¬q ∨ p. Both have the same disjunctive form, so they are equivalent.
Take p true, q false. Then p→q is false (T→F = F), whereas q→p is true (F→T = T). Since truth values differ, the two implications are not logically equivalent.
Not equivalent in general; give a counterexample: p = True, q = False.
p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q). Negating gives ¬(p ↔ q) ≡ (¬p ∨ ¬q) ∧ (p ∨ q). This simplifies to (¬p ∧ q) ∨ (p ∧ ¬q), which is exactly p ↔ ¬q.
p → q ≡ ¬p ∨ q. Also ¬q → ¬p ≡ q ∨ ¬p (same as ¬p ∨ q). Thus (p → q) and (¬q → ¬p) are logically equivalent; in particular (p → q) → (¬q → ¬p) is always true. Hence it is a tautology.
Tautology.
Simplify: (¬p ∨ q) ∨ (¬p ∧ q) = ¬p ∨ q. Truth table (rows in order p,q = TT, TF, FT, FF): - p=T,q=T: ¬p∨q = F∨T = T, ¬p = F - p=T,q=F: ¬p∨q = F∨F = F, ¬p = F - p=F,q=T: ¬p∨q = T∨T = T, ¬p = T - p=F,q=F: ¬p∨q = T∨F = T, ¬p = T Columns differ (row1: LHS T, RHS F), so not equivalent.
Not equivalent.
p → (q → r) ≡ ¬p ∨ (¬q ∨ r) ≡ ¬p ∨ ¬q ∨ r. Also (p ∧ q) → r ≡ ¬(p ∧ q) ∨ r ≡ ¬p ∨ ¬q ∨ r. Hence the two are equivalent.
(p → (q → r)) ≡ ((p ∧ q) → r).
Compute p → (q → r) as ¬p ∨ (¬q ∨ r) which simplifies to ¬p ∨ ¬q ∨ r. A 8-row truth table for (p,q,r) confirms the columns match for all combinations, so they are equivalent.
Equivalent: p → (q → r) ≡ ¬p ∨ ¬q ∨ r.
By definition a binary operation on S assigns to each ordered pair (a,b) in S×S an element of S; i.e. it is a function S×S → S.
Subtraction on ℕ is not closed because e.g. 2−5 is not a natural number. On ℤ, ℚ, ℝ subtraction is closed.
Multiplication of natural numbers is closed in ℕ. Subtraction and division may produce results not in ℕ.
Division a/b is not defined for b = 0, so the rule is not a binary operation on all of ℝ. The other operations are closed on ℝ.
ab/7 need not be an integer for integer a,b (e.g. a=b=1 gives 1/7). Thus the operation is not closed on ℤ (also not closed on ℤ+). It is closed on ℝ and on ℚ only if denominators allowed; for ℚ ab/7 is rational, so closed. The intended answer is ℤ.
Compute 3 ε 5 = 3+5+3·5 = 3+5+15 = 23. Then (3 ε 5) ε 7 = 23 + 7 + 23·7 = 30 + 161 = 191. None of the given options matches 191.
Check commutativity: a * b = a + 2b, b * a = b + 2a; not equal in general, so not commutative. Check associativity: (a*b)*c = (a+2b)+2c = a+2b+2c, whereas a*(b*c)= a+2(b+2c)=a+2b+4c, not equal in general. So neither property holds.
As printed, option (3) is false because z·conj(z)=|z|^2 is real and nonnegative. Likely the intended correct statement was 'the product of a complex number and its conjugate is purely real' which would be true. Under that correction the correct choice is (3).
Chennai is in India is true; 2 is an integer is true. Evaluate each OR: (1) true, (2) true, (3) true (second disjunct true), (4) false ∨ false = false (Chennai not in China, 2 not irrational). Hence (4) is false.
Number of rows = 2^n with n = 3, so 2^3 = 8.
Inverse of A → B is ¬A → ¬B. Here A = (p ∨ q), B = (p ∧ q). So inverse is ¬(p ∨ q) → ¬(p ∧ q), equivalently (¬p ∧ ¬q) → (¬p ∨ ¬q). That matches option (4).
Contrapositive of A → B is ¬B → ¬A. Here A = (p ∨ q), B = r, so contrapositive is ¬r → ¬(p ∨ q) = ¬r → (¬p ∧ ¬q).
q ∧ ¬q is always false, so p ∨ (q ∧ ¬q) ≡ p. Thus values for (a,b,c,d) are T, T, F, F respectively. None of the listed option patterns equals T,T,F,F.
¬p ∨ ¬q is false only when ¬p and ¬q are both false, i.e. p = T and q = T. So exactly one row yields F.
Option (3) is incorrect: by De Morgan ¬(p ∨ q) ≡ ¬p ∧ ¬q, not ¬p ∨ ¬q. The other three are correct.
Since ¬¬p = p, the proposition is p ∧ q → p. Whenever p ∧ q is true, p is true, hence p ∧ q → p is true for all four combinations of p and q. Therefore all entries in the truth column are T (option 1).
Dual is obtained by interchanging ∨ and ∧ everywhere (negations remain). Dual of ¬[ (¬p ∨ q) ∨ (p ∧ ¬r) ] is ¬[ (¬p ∧ q) ∧ (p ∨ ¬r) ]. (Matches option 4 in the reconstructed list.)
Use absorption: p ∨ (p ∧ ¬q) = p. Hence the proposition is logically equivalent to p (option 3).
Evaluate each conjunction: (a) 4+2=6 ≠5 so false; 6+3=9 true ⇒ (a)=F. (b) 3+2=5 true and 6+1=7 true ⇒ (b)=T. (c) 4+5=9 true and 1+2=3 ≠4 ⇒ (c)=F. (d) 3+2=5 true and 4+7=11 true ⇒ (d)=T. Pattern F T F T → option (1).
Statement (4) is false: p → q is not a tautology for arbitrary p,q (counterexample: p = T, q = F gives p → q = F). The other three statements are true.