Educational

Composition Table Multiplication Modulo 5 on Z5 Question: Construct the composition table for \( \times_5 \) on \( Z_5 = \{0,1,2,3,4\} \). Concept: In modular arithmetic, \[ a \times_5 b = (a \times b) \mod 5 \] i.e., multiply two numbers and take the remainder after division by 5. :contentReference[oaicite:0]{index=0} Solution: Step 1: Compute values […]

Composition Table Multiplication Modulo 6 on Set {0,1,2,3,4,5} Question: Construct the composition table for \( \times_6 \) on the set \( S = \{0,1,2,3,4,5\} \). Concept: The operation \( a \times_6 b \) means multiplication modulo 6: \[ a \times_6 b = (a \times b) \mod 6 \] Solution: Step 1: Multiply the numbers and

Composition Table Addition Modulo 5 on Set {0,1,2,3,4} Question: Construct the composition table for \( +_5 \) on the set \( S = \{0,1,2,3,4\} \). Concept: The operation \( a +_5 b \) means addition modulo 5, i.e., \[ a +_5 b = (a + b) \mod 5 \] Solution: Step 1: Add the numbers

Composition Table Modulo 4 on Set {0,1,2,3} Question: Construct the composition table for \( \times_4 \) on the set \( S = \{0, 1, 2, 3\} \). Concept: The operation \( a \times_4 b \) means multiplication modulo 4, i.e., \[ a \times_4 b = (a \times b) \mod 4 \] Solution: Step 1: Compute

Binary Operation on R x R 📺 Watch Video Explanation: Given: \( (a,b)*(c,d) = (a+c, b+d), \quad (a,b),(c,d)\in \mathbb{R}\times\mathbb{R} \) Commutativity: \( (a,b)*(c,d) = (a+c, b+d) = (c+a, d+b) = (c,d)*(a,b) \) ✔ Commutative Associativity: LHS: \( [(a,b)*(c,d)]*(e,f) = (a+c, b+d)*(e,f) = (a+c+e, b+d+f) \) RHS: \( (a,b)*[(c,d)*(e,f)] = (a,b)*(c+e, d+f) = (a+c+e, b+d+f) \)

Identity Element for HCF 📺 Watch Video Explanation: Find identity element Given: \( a * b = \mathrm{HCF}(a,b), \quad a,b \in \mathbb{N} \) Identity Definition: \( a * e = a \Rightarrow \mathrm{HCF}(a,e) = a \) Analysis: This implies: \( a \mid e \quad \text{for all } a \in \mathbb{N} \) So, \( e \)

Binary Operation on Ordered Pairs 📺 Watch Video Explanation: Given: \( (a,b)\circ(c,d) = (ac, bd), \quad (a,b),(c,d)\in \mathbb{R}_0 \times \mathbb{R}_0 \) i. Commutativity: \( (a,b)\circ(c,d) = (ac, bd) = (ca, db) = (c,d)\circ(a,b) \) ✔ Commutative Associativity: LHS: \( [(a,b)\circ(c,d)]\circ(e,f) = (ac,bd)\circ(e,f) = (ace, bdf) \) RHS: \( (a,b)\circ[(c,d)\circ(e,f)] = (a,b)\circ(ce,df) = (ace, bdf) \)

Binary Operation Full Solution 📺 Watch Video Explanation: Given: \( a*b = a + b – ab, \quad a,b \in \mathbb{R} \setminus \{1\} \) Commutativity: \( a*b = a + b – ab = b + a – ba = b*a \) ✔ Commutative Associativity: \( (a*b)*c = a + b + c – ab

Binary Operation Full Solution 📺 Watch Video Explanation: Given: \( a \circ b = \frac{ab}{2}, \quad a,b \in \mathbb{Q}_0 \) i. Commutativity: \( a \circ b = \frac{ab}{2} = \frac{ba}{2} = b \circ a \) ✔ Commutative Associativity: LHS: \( (a \circ b)\circ c = \left(\frac{ab}{2}\right)\circ c = \frac{\frac{ab}{2} \cdot c}{2} = \frac{abc}{4} \) RHS:

Binary Operation on Ordered Pairs 📺 Watch Video Explanation: Given: \( (a,b) \circ (c,d) = (ac, bc + d) \) i. Commutativity: \( (a,b)\circ(c,d) = (ac, bc + d) \) \( (c,d)\circ(a,b) = (ca, da + b) \) Since: \( bc + d \neq da + b \ \text{(in general)} \) ❌ NOT commutative Associativity: