Binary operations

Step 1: Put (a*e=a), giving (a+e-ae=a). Step 2: Thus (e(1-a)=0), and (e=0) works for every (a). Step 3: Also (e*a=e+a-ea=a), so (0) is the identity.

Step 1: (1-(a*b)=1-a-b+ab=(1-a)(1-b)). Step 2: Since (a\neq 1) and (b\neq 1), this is non-zero, so (a*b\neq 1); hence closure holds. Step 3: Identity is (0), and for (a\neq 1), the inverse is (x=\frac{a}{a-1}).

Step 1: Sums and products of integers are still integers. Step 2: (a+b+ab) is a sum of three integers, so it belongs to (\mathbb{Z}). Step 3: Closure requires the result to remain in the same set, not necessarily positive.

Step 1: For a binary operation, the result must again belong to (A). Step 2: (2) and (4) are even, but (\frac{2+4}{2}=3), which is not even. Step 3: Once closure fails, it is not a binary operation on that set.

Step 1: Closure requires every result to lie in (A). Step 2: (1+7=8), whose remainder on division by (8) is (0). Step 3: Since (0\notin A), one counterexample is enough.

Step 1: Commutativity checks whether (a*b=b*a). Step 2: (a*b=a^2+b^2) and (b*a=b^2+a^2), equal by commutativity of usual addition. Step 3: Do not confuse commutativity with associativity or inverse.

Step 1: Associativity requires ((a*b)*c=a*(b*c)). Step 2: ((1*2)*3=5*3=25+9=34), while (1*(2*3)=1*13=1+169=170). Step 3: Being commutative does not guarantee associativity.

Step 1: Expanding ((a*b)*c) gives (a+b+c+kab+kc(a+b+kab)). Step 2: Expanding (a*(b*c)) gives (a+b+c+kbc+ka(b+c+kbc)). Step 3: Both equal (a+b+c+kab+kac+kbc+k^2abc), so it is associative for every (k).

Step 1: Put (a*e=a), giving (a+e+kae=a). Step 2: Thus (e(1+ka)=0). In the given set (1+ka\neq 0), so (e=0). Step 3: The excluded element is important for closure and inverse behavior.

Step 1: The identity is (0). Step 2: From (a*x=0), we get (a+x+2ax=0). Step 3: Hence (x(1+2a)=-a), so (x=\frac{-a}{1+2a}); the excluded value avoids zero denominator.

Step 1: If (a,b) are real, (a+2b) is real, so closure holds. Step 2: (1*2=5) and (2*1=4), so it is not commutative. Step 3: ((1*2)*3=11), but (1*(2*3)=17), so it is not associative.

Step 1: A right identity would require (a*e=2a+e=a), giving (e=-a), dependent on (a). Step 2: A left identity would require (e*a=2e+a=a), giving (e=0). Step 3: Since there is no single two-sided identity, no identity exists.

Step 1: Write (a+b-ab=1) as (1-a-b+ab=0). Step 2: This factors as ((1-a)(1-b)=0). Step 3: Hence (a=1) or (b=1); factoring saves time in such problems.

Step 1: The quotient of two non-zero rationals is again a non-zero rational, so closure holds. Step 2: But (2*4=\frac{1}{2}), while (4*2=2), so commutativity fails. Step 3: Check closure and commutativity separately.

Step 1: ((8*4)*2=(2)*2=1). Step 2: (8*(4*2)=8*2=4). Step 3: They are not equal, so associativity fails; a clear counterexample is the strongest argument.

Step 1: Under ordinary multiplication, the identity is (1). Step 2: The inverse of (1) is (1) because (1\cdot 1=1). Step 3: (0\cdot x) can never become (1), so (0) has no inverse.

Step 1: The product of (1) and (-1) always remains in ({1,-1}), so closure holds. Step 2: (1) is the identity. Step 3: The inverse of (1) is (1), and the inverse of (-1) is (-1), so every element has an inverse.

Step 1: From (a+b+ab=b), we get (a+ab=0). Step 2: Factor it as (a(1+b)=0). Step 3: Hence (a=0) or (b=-1); factoring is very useful in such condition-based questions.

Step 1: Under ordinary multiplication, the identity is (1). Step 2: If (a>0), then its inverse (\frac{1}{a}) is also positive. Step 3: Thus the inverse remains in (\mathbb{R}^{+}), which is the next key check after closure.

Step 1: From (a*e=a), (\frac{a+e}{2}=a). Step 2: This gives (e=a), which changes with (a). Step 3: An identity must be fixed, so no identity exists here.