To solve the problem, we first need to understand the operation defined on the set A = Q × Q, where Q represents the set of all rational numbers.
Identity Element of the Operation
The identity element, e, for the operation '∗' must satisfy the condition:
- (a, b) '∗' e = (a, b) for any (a, b) in A.
Using the operation defined:
(a, b) '∗' (e1, e2) = (ae1, b + ae2).
For this to equal (a, b), we need:
- ae1 = a, which implies e1 = 1 (since a ≠ 0).
- b + ae2 = b, which implies ae2 = 0. Since a can be any rational number, we need e2 = 0.
Thus, the identity element is:
e = (1, 0)
Finding Invertible Elements
An element (a, b) in A is invertible if there exists an element (c, d) in A such that:
(a, b) '∗' (c, d) = (1, 0).
Using the operation:
(a, b) '∗' (c, d) = (ac, b + ad) = (1, 0).
This gives us two equations:
From the first equation, we find:
c = 1/a (assuming a ≠ 0). Now substituting c into the second equation:
b + a(1/a) = 0, which simplifies to:
b + 1 = 0 or
b = -1.
Therefore, the invertible elements in A are of the form:
(a, -1), where a is any non-zero rational number.
Finding Inverses of Specific Elements
Now, let's find the inverses of the given elements:
Inverse of (5, 3)
Here, a = 5 and b = 3. Since 5 is non-zero, we can use the form:
- c = 1/5
- b = -1, so the inverse is:
(1/5, -1)
Inverse of (1/2, 4)
For this element, a = 1/2 and b = 4. Again, since 1/2 is non-zero:
- c = 2 (since 1/(1/2) = 2)
- b = -1, so the inverse is:
(2, -1)
In summary, the identity element is (1, 0), the invertible elements are of the form (a, -1) with a ≠ 0, and the inverses of (5, 3) and (1/2, 4) are (1/5, -1) and (2, -1), respectively.
