Binary operations

Step 1: (|a-b|=|b-a|), so the operation is commutative. Step 2: ((1*3)*6=2*6=4), while (1*(3*6)=1*3=2), so it is not associative. Step 3: Small counterexamples work well for absolute value operations.

Step 1: If (e) is identity, then (|a-e|=a) for every real (a). Step 2: Put (a=-1); the left side is always non-negative, but the right side is (-1). Step 3: Hence no identity exists; negative values often disprove such claims quickly.

Step 1: For identity (e), (\operatorname{lcm}(a,e)=a). Step 2: (\operatorname{lcm}(a,1)=a) for every natural (a). Step 3: For an LCM operation, check (1) immediately.

Step 1: For (\gcd(a,e)=a), (e) must be a multiple of every (a). Step 2: There is no single natural number that is a multiple of all natural numbers. Step 3: If (0) were included, the situation would differ, but in (\mathbb{N}) there is no identity here.

Step 1: For identity (e), (\gcd(a,e)=a). Step 2: (6) is a multiple of every element of this set, so (\gcd(a,6)=a). Step 3: In a finite set, the largest suitable multiple can become the identity for a GCD operation.

Step 1: For identity (e), (\operatorname{lcm}(a,e)=a). Step 2: (1) leaves every element unchanged because (\operatorname{lcm}(a,1)=a). Step 3: In LCM and GCD operations, (1) and the top common multiple play different roles.

Step 1: This is multiplication modulo (3), and the identity is (1). Step 2: The inverse of (1) is (1), and the inverse of (2) is (2) because (2\cdot 2=4) has remainder (1). Step 3: Multiplying by (0) always gives remainder (0), so (0) has no inverse.

Step 1: Products of the given elements modulo (6) again lie in (1,2,4,5); for example, (2*4=2), (5*5=1). Step 2: (1*a=a) and (a*1=a), so (1) is the identity. Step 3: For a finite set, sample products reveal the closure pattern.

Step 1: The identity is (0) because (a+0-0=a). Step 2: For inverse (x), (a+x-2ax=0), so (x(1-2a)=-a). Step 3: At (a=\frac{1}{2}), the coefficient becomes zero and the equation is impossible, so it has no inverse.

Step 1: The identity is (0). Step 2: From (a*x=0), (a+x-2ax=0). Step 3: Thus (x(1-2a)=-a), so (x=\frac{-a}{1-2a}); the excluded element prevents a zero denominator.

Step 1: If (5) is the identity, then (a*5=a). Step 2: (a+5+\lambda=a) gives (\lambda=-5). Step 3: Checking (5*a=a) gives the same result, so the verification is complete.

Step 1: This type of operation has identity (0). Step 2: From (2*x=0), (2+x+2\lambda x=0), so (x(1+2\lambda)=-2). Step 3: No inverse exists when (1+2\lambda=0), hence (\lambda=-\frac{1}{2}).

Step 1: The identity for this operation is (0). Step 2: If (2) and (-1) are inverses, then (2*(-1)=0). Step 3: (2-1-2m=0) gives (1-2m=0), so (m=\frac{1}{2}).

Step 1: This operation is addition modulo (5). Step 2: (3+4=7), and the remainder on division by (5) is (2). Step 3: For remainder operations, first do the usual calculation, then take the remainder.

Step 1: In addition modulo (5), the identity is (0). Step 2: For (4*x=0), (4+x) must have remainder (0) on division by (5). Step 3: Since (4+1=5), the inverse of (4) is (1).

Step 1: A binary operation is a function from (A\times A) to (A). Step 2: (A\times A) has (n^2) ordered pairs, and each pair has (n) possible outputs. Step 3: Therefore the total number is (n^{n^2}); understand it through function counting, not rote memory.

Step 1: In a commutative operation, (a*b=b*a), so ((a,b)) and ((b,a)) must have the same value. Step 2: The number of independent positions is (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}). Step 3: Each independent position has (n) choices, so the total number of operations is (n^{\frac{n(n+1)}{2}}).

Step 1: For closure, the result must again lie in (\mathbb{Z}). Step 2: In (a+b+ab), sums and products of integers remain integers. Step 3: In closure questions, focus only on the set of the result.

Step 1: For identity (e), (a*e=a) must hold. Step 2: From (a+e+4=a), we get (e=-4). Step 3: In identity questions, (a) may be any value, so (e) must be fixed.

Step 1: First find the identity: (a+e+4=a), so (e=-4). Step 2: If (x) is the inverse of (6), then (6+x+4=-4). Step 3: This gives (x=-14), so always find identity before inverse.