Prove that the greatest integer function `f:RtoR`, given by `f(x)=[x]` is a many-one function.

Answer

To prove that the greatest integer function \( f: \mathbb{R} \to \mathbb{R} \), defined by \( f(x) = \lfloor x \rfloor \) (where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \)), is a many-one function, we will follow these steps:

### Step 1: Definition of Many-One Function
A function \( f \) is called many-one if there exist at least two different elements in the domain that map to the same element in the codomain. In other words, if \( f(a) = f(b) \) for \( a \neq b \), then \( f \) is many-one.

### Step 2: Analyze the Greatest Integer Function
The greatest integer function \( f(x) = \lfloor x \rfloor \) takes a real number \( x \) and returns the largest integer less than or equal to \( x \).

### Step 3: Identify Intervals
Consider any two real numbers \( a \) and \( b \) such that \( \lfloor a \rfloor = \lfloor b \rfloor \). This means that both \( a \) and \( b \) fall within the same interval of integers. Specifically, if \( n = \lfloor a \rfloor = \lfloor b \rfloor \), then:
\[ n \leq a < n + 1 \quad \text{and} \quad n \leq b < n + 1 \]

### Step 4: Example of Different Inputs Mapping to Same Output
For example, let’s take \( a = 1.5 \) and \( b = 1.9 \):
\[ \lfloor 1.5 \rfloor = 1 \quad \text{and} \quad \lfloor 1.9 \rfloor = 1 \]
Here, \( a \neq b \) but \( f(a) = f(b) \).

### Step 5: Generalization
More generally, for any integer \( n \):
- Any \( x \) in the interval \( [n, n+1) \) will satisfy \( f(x) = n \).
- This means that there are infinitely many \( x \) values (like \( n, n+0.1, n+0.5, n+0.9 \), etc.) that all map to the same integer \( n \).

### Conclusion
Since we can find multiple distinct values \( a \) and \( b \) such that \( f(a) = f(b) \), we conclude that the greatest integer function \( f(x) = \lfloor x \rfloor \) is indeed a many-one function.