Let A ={ 1, 2, 3,4, 5}, a*b = L.C.M of a and b is a binary operation on A, explain ?

Answer

To determine whether the operation defined by \( a * b = \text{L.C.M.}(a, b) \) is a binary operation on the set \( A = \{ 1, 2, 3, 4, 5 \} \), we need to follow these steps:

### Step 1: Understand the Definition of a Binary Operation
A binary operation on a set \( A \) is a function that takes two elements from \( A \) and produces another element from \( A \). This means that for any \( a, b \in A \), the result of the operation \( a * b \) must also be an element of \( A \).

### Step 2: Check the Closure Property
We need to check if the operation \( * \) is closed in the set \( A \). This means we need to verify if \( a * b \) (which is \( \text{L.C.M.}(a, b) \)) is always in \( A \) for all \( a, b \in A \).

### Step 3: Calculate L.C.M. for Elements in \( A \)
Let's calculate the L.C.M. for some pairs of elements in \( A \):

- For \( a = 2 \) and \( b = 3 \):
\[ \text{L.C.M.}(2, 3) = 6 \]
Here, \( 6 \notin A \).

- For \( a = 3 \) and \( b = 4 \):
\[ \text{L.C.M.}(3, 4) = 12 \]
Here, \( 12 \notin A \).

- For \( a = 4 \) and \( b = 5 \):
\[ \text{L.C.M.}(4, 5) = 20 \]
Here, \( 20 \notin A \).

### Step 4: Conclusion
Since we found that the L.C.M. of some pairs of elements in \( A \) results in values that are not in \( A \), we conclude that the operation \( * \) is not closed in \( A \).

Therefore, the operation \( a * b = \text{L.C.M.}(a, b) \) is **not a binary operation** on the set \( A \).

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