Binary operations

Step 1: Write (a*e=a). Step 2: From (a+e-2=a), we get (e=2). Step 3: When a fixed number is added or subtracted, identity balances that fixed number.

Step 1: The identity is (2). Step 2: From (-3*x=2), (-3+x-2=2), so (x=7). Step 3: With negative numbers, write every term clearly.

Step 1: (a*0=|a-0|=a) because (a) is only (0) or (1). Step 2: (0*a=|0-a|=a) also holds. Step 3: For small sets, check both directions.

Step 1: In multiplication, (a\cdot1=a). Step 2: (1\cdot a=a) also holds for all (a\in A). Step 3: Identity is the element that does not change an element from either side.

Step 1: For absorbing element (z), (a*z=z) and (z*a=z) must hold. Step 2: (a*1=a+1-a=1) and (1*a=1+a-a=1). Step 3: Hence (1) is absorbing; do not confuse it with identity.

Step 1: For absorbing element (z), (a*z=z) is needed. Step 2: (a*(-1)=a-1-a=-1) and ((-1)*a=-1+a-a=-1). Step 3: Hence (-1) is absorbing; it also becomes the element without inverse.

Step 1: Write (1-(a*b)=1-(a+b-ab)). Step 2: Simplifying gives (1-a-b+ab). Step 3: This equals ((1-a)(1-b)), which helps in understanding associativity.

Step 1: Since (a,b\in A), (a\neq1) and (b\neq1). Step 2: (1-(a*b)=(1-a)(1-b)), which is not zero. Step 3: Therefore (a*b\neq1), so the result again lies in (A).

Step 1: The formula (a+b-ab) is unchanged when (a,b) are interchanged, so it is commutative. Step 2: Since (1-(a*b)=(1-a)(1-b)), associativity follows like multiplication. Step 3: In hard operations, look for such simpler forms.

Step 1: ((8*4)*2=(\frac{8}{4})*2=1). Step 2: (8*(4*2)=8*(2)=4). Step 3: The two values are different, so the operation is not associative.

Step 1: For identity (e), (\sqrt{a^2+e^2}=a) must hold for every real (a), which fails when (a<0). Step 2: Even (a*0=|a|), not (a) for negative (a). Step 3: Therefore there is actually no identity on (\mathbb{R}); avoid the tempting option (0).

Step 1: This is addition based on remainders modulo (4). Step 2: Dividing (a+0) by (4) gives the same remainder (a). Step 3: Hence (0) is the identity; in remainder operations, understand the basic addition first.

Step 1: The identity for this operation is (0). Step 2: For inverse (x) of (3), the remainder of (3+x) on division by (7) must be (0). Step 3: Since (3+4=7), the inverse is (4).

Step 1: In multiplication with remainders, the identity is (1). Step 2: For inverse (x) of (2), the remainder of (2x) on division by (5) must be (1). Step 3: (2\cdot3=6), whose remainder is (1), so the answer is (3).

Step 1: First (2*3=2+3-6=-1). Step 2: Now ((-1)*4=-1+4-(-1)(4)=7). Step 3: Do not change the bracket order; evaluate the inner operation first.

Step 1: From (a*e=a), (\frac{a+e}{2}=a). Step 2: This gives (e=a), which is not a fixed element. Step 3: Identity must be the same for every (a), so it does not exist here.

Step 1: Put (a=x) and (b=3) in the given formula. Step 2: (x*3=x+3+1=7), so (x+4=7). Step 3: Hence (x=3); apply the definition directly in such calculations.

Step 1: In (x*2), put (a=x) and (b=2). Step 2: (2x+3(2)=16), so (2x+6=16). Step 3: This gives (x=5); first apply the operation, then solve the simple equation.

Step 1: First find the identity. From (a+e-2ae=a), we get (e=0). Step 2: For inverse (x) of (3), (3+x-6x=0), so (3-5x=0). Step 3: Hence (x=\frac{3}{5}); while finding inverse, equate the result to the identity.

Step 1: For absorbing element (z), both (a*z=z) and (z*a=z) must hold. Step 2: From (a*z=a+z-2az=z), we get (a(1-2z)=0). Step 3: This holds for every (a) only when (z=\frac{1}{2}); hence the absorbing element is (\frac{1}{2}).