Find whether the following functions are one-one or not.
`f:ZtoZ`, defined by `f(x)=x^(2)+5` for all `x inZ`

Answer

To determine whether the function \( f: \mathbb{Z} \to \mathbb{Z} \) defined by \( f(x) = x^2 + 5 \) is one-one, we will follow these steps:

### Step 1: Understand the definition of a one-one function
A function \( f \) is called one-one (or injective) if for every pair of distinct elements \( x_1 \) and \( x_2 \) in the domain, \( f(x_1) \neq f(x_2) \). In other words, if \( f(x_1) = f(x_2) \), then it must follow that \( x_1 = x_2 \).

### Step 2: Assume \( f(x_1) = f(x_2) \)
Let \( x_1, x_2 \in \mathbb{Z} \) be such that:
\[ f(x_1) = f(x_2) \]
This implies:
\[ x_1^2 + 5 = x_2^2 + 5 \]

### Step 3: Simplify the equation
We can simplify the equation by subtracting 5 from both sides:
\[ x_1^2 = x_2^2 \]

### Step 4: Analyze the implications of \( x_1^2 = x_2^2 \)
The equation \( x_1^2 = x_2^2 \) can be factored as:
\[ (x_1 - x_2)(x_1 + x_2) = 0 \]
This gives us two cases:
1. \( x_1 - x_2 = 0 \) which implies \( x_1 = x_2 \)
2. \( x_1 + x_2 = 0 \) which implies \( x_1 = -x_2 \)

### Step 5: Conclusion
From the second case, we see that if \( x_1 \) is a positive integer, \( x_2 \) can be its negative counterpart (e.g., \( 3 \) and \( -3 \) both yield \( f(3) = f(-3) = 14 \)). Therefore, we have found distinct values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \).

Thus, the function \( f(x) = x^2 + 5 \) is **not one-one**.