Let A= -{-1}, that is, the set of all real numbers except -1. Define an operation on A by a*b=a+b+ab (usual addition and multiplication in R)
a) Show that * is closed on A
b) show that * is associative
c) Find the identity element for * and verify that it is the identity
d) for an element c of A, what is the inverse of c for *? Why?
To show that the operation * is closed on A, we need to demonstrate that for any two elements a and b in A, the result of a*b is also in A.
a) Showing closure:
Let's take two arbitrary elements a and b from A. We know that A is defined as the set of all real numbers except -1. Hence, a and b are real numbers, and the only condition is that -1 is not included.
Now, let's calculate the result of a*b using the given operation:
a * b = a + b + ab
Since a and b are real numbers, their sum a + b is also a real number. Similarly, ab is the product of two real numbers, so it is also a real number.
The only condition we need to check is that -1 is not produced as a result in a*b. For this, we need to consider all possible cases.
Case 1: a = -1
If a = -1, then a*b = -1*b = -b. Since -1 is not included in A, this means that -b should not be -1. Therefore, b is not -1.
Case 2: b = -1
If b = -1, then a*b = a*(-1) = -a. Similar to Case 1, -a should not be -1. Therefore, a is not -1.
Case 3: a and b are not -1
In this case, both a and b are real numbers other than -1. Hence, a + b + ab is a real number, and it is not -1.
Therefore, in any case, the result of a*b is a real number other than -1, which means it is in A. Thus, * is closed on A.
b) To show that * is associative, we need to verify that for any three elements a, b, and c in A, the equation (a * b) * c = a * (b * c) holds.
Let's compute each side of the equation and compare:
Left side: (a * b) * c = ((a + b + ab) + c + (a + b + ab)c)
Right side: a * (b * c) = (a + (b + c + bc) + a(b + c + bc))
Expanding both sides:
(a + b + ab) + c + (a + b + ab)c = a + b + c + ab + ac + bc + a(b + c) + abc
Comparing the terms on both sides, we can see that they are identical. Hence, (a * b) * c = a * (b * c), and * is associative.
c) To find the identity element for *, we need to determine an element e in A such that for any element a in A, a * e = e * a = a.
First, let's consider the equation a * e = a:
a + e + ae = a
Rearranging the terms:
e + ae = 0
Factorizing 'e':
e(1 + a) = 0
Since 'e' is an element of A, it cannot be equal to -1. So, for the above equation to hold, (1 + a) = 0.
Hence, the identity element e for * is -1.
Now, let's verify that -1 indeed acts as the identity element by checking a * -1 = -1 * a = a for any element 'a' in A.
For a * -1:
a + (-1) + a(-1) = a - 1 - a = -1
Similarly, for -1 * a:
-1 + a + (-1)a = -1 - a + a = -1
In both cases, the result is -1, which means -1 is the identity element for *.
d) To find the inverse of an element c in A for *, we need to determine an element c' in A such that c * c' = c' * c = e, where e is the identity element.
Let's solve the equation c * c' = -1:
c + c' + cc' = -1
We can rearrange the terms:
cc' + c' + c + 1 = 0
Factoring 'c':
c(c' + 1) + c' + 1 = 0
For this equation to hold, (c' + 1) must equal zero.
Therefore, c' = -1 - c.
Hence, the inverse of an element c in A for * is -1 minus c.