Question 34 of 35intermediate🔍 AnalyzeShort Answer3 marks

Let ff be the subset of Z×ZZ \times Z defined by f={(ab,a+b):a,b∈Z}f = \{(ab, a + b) : a, b \in Z\}. Is ff a function from ZZ to ZZ? Justify your answer.

Correct Answer

A relation ff from a set AA to a set BB is said to be a function if every element of set AA has one and only one image in set BB. This means that no two distinct ordered pairs in ff can have the same first element.

To determine if f={(ab,a+b):a,b∈Z}f = \{(ab, a + b) : a, b \in Z\} is a function, we must check if a specific first element corresponds to a unique second element. Let us consider specific integer values for aa and bb.

For a=2a = 2 and b=3b = 3, the ordered pair is (ab,a+b)=(6,5)(ab, a + b) = (6, 5). For a=1a = 1 and b=6b = 6, the ordered pair is (ab,a+b)=(6,7)(ab, a + b) = (6, 7). In this case, the same first element, 66, is associated with two different second elements, 55 and 77. Since the element 66 has more than one image, the relation ff is not a function from ZZ to ZZ.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 11 | (Page 18)
For More Understanding

Explanation

The justification relies on Definition 5 provided in the context, which states that a relation is a function if every element has a unique image. By finding two pairs (6, 5) and (6, 7) within the relation derived from the rule (ab, a+b), we demonstrate that the element 6 maps to two different values, violating the definition of a function.

Solution Steps

  1. Step 1: Recall Definition 5, which states a relation is a function if every element has one and only one image (no two distinct ordered pairs have the same first element).

  2. Step 2: Test the rule f={(ab,a+b):a,b∈Z}f = \{(ab, a + b) : a, b \in Z\} with specific integers. For a=2,b=3a=2, b=3, the pair is (6,5)(6, 5).

  3. Step 3: Test the rule with different integers. For a=1,b=6a=1, b=6, the pair is (6,7)(6, 7).

  4. Step 4: Observe that the first element 66 maps to two distinct second elements, 55 and 77.

  5. Step 5: Conclude that since an element has more than one image, ff is not a function.