Question:

Consider the operators \diamondsuit and \square defined by ab=a+2b a \diamondsuit b = a + 2b and ab=ab a \square b = ab , for positive integers. Which of the following statements is/are TRUE?

Updated On: Jan 22, 2025
  • Operator \diamondsuit obeys the associative law
  • Operator \square obeys the associative law
  • Operator \diamondsuit over the operator \square obeys the distributive law
  • Operator \square over the operator \diamondsuit obeys the distributive law
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The Correct Option is B

Solution and Explanation

Step 1: Associativity of \diamondsuit .
To check if \diamondsuit is associative: (ab)c=(a+2b)+2c=a+2b+2c, (a \diamondsuit b) \diamondsuit c = (a + 2b) + 2c = a + 2b + 2c, a(bc)=a+2(b+2c)=a+2b+4c. a \diamondsuit (b \diamondsuit c) = a + 2(b + 2c) = a + 2b + 4c. Since the results are not equal, \diamondsuit is not associative. Step 2: Associativity of \square .
To check if \square is associative: (ab)c=(ab)c=(ab)c=abc, (a \square b) \square c = (ab) \square c = (ab)c = abc, a(bc)=a(bc)=a(bc)=abc. a \square (b \square c) = a \square (bc) = a(bc) = abc. Since the results are equal, \square is associative. Hence, (2) is true. Step 3: Distributivity of \square over \diamondsuit .
To check if \square is distributive over \diamondsuit : a(bc)=a(b+2c)=ab+2ac, a \square (b \diamondsuit c) = a \cdot (b + 2c) = ab + 2ac, (ab)(ac)=(ab)(ac)=ab+2ac. (a \square b) \diamondsuit (a \square c) = (ab) \diamondsuit (ac) = ab + 2ac. Since the results are equal, \square is distributive over \diamondsuit . Hence, (3) is false. Step 4: Distributivity of \diamondsuit over \square .
To check if \diamondsuit is distributive over \square : a(bc)=a+2(bc), a \diamondsuit (b \square c) = a + 2(bc), (ab)(ac)=(a+2b)(a+2c)a+2(bc). (a \diamondsuit b) \square (a \diamondsuit c) = (a + 2b)(a + 2c) \neq a + 2(bc). Since the results are not equal, \diamondsuit is not distributive over \square . Hence, (4) is true. Final Answer: (2),(4) \boxed{(2), (4)}
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