Find whether the following are into functions (surjections) or not.
`f:ZtoZ` defined by `f(x)=5x-9` for all `x inZ`.

Answer

To determine whether the function \( f: \mathbb{Z} \to \mathbb{Z} \) defined by \( f(x) = 5x - 9 \) is a surjection (onto function), we will follow these steps:

### Step 1: Check if the function is injective (one-to-one)
To check if \( f \) is injective, we assume that \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in \mathbb{Z} \) and show that this implies \( x_1 = x_2 \).

1. Start with the equation:
\[ f(x_1) = f(x_2) \]
This means:
\[ 5x_1 - 9 = 5x_2 - 9 \]

2. Simplify the equation:
\[ 5x_1 = 5x_2 \]

3. Divide both sides by 5:
\[ x_1 = x_2 \]

Since we have shown that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is injective.

### Step 2: Check if the function is surjective (onto)
To check if \( f \) is surjective, we need to show that for every \( y \in \mathbb{Z} \), there exists an \( x \in \mathbb{Z} \) such that \( f(x) = y \).

1. Let \( y \) be any element in the codomain \( \mathbb{Z} \). We need to find \( x \) such that:
\[ f(x) = y \]
This means:
\[ 5x - 9 = y \]

2. Rearranging the equation to solve for \( x \):
\[ 5x = y + 9 \]

3. Dividing both sides by 5 gives:
\[ x = \frac{y + 9}{5} \]

4. For \( x \) to be an integer, \( y + 9 \) must be divisible by 5. This means that \( y \) can take values such that \( y \equiv 1 \mod 5 \) (i.e., \( y \) can be expressed as \( 5k - 9 \) for some integer \( k \)).

Since not every integer \( y \) satisfies this condition, there are integers in \( \mathbb{Z} \) that cannot be expressed in the form \( 5x - 9 \). Therefore, the function \( f \) is not surjective.

### Conclusion
The function \( f(x) = 5x - 9 \) is an injective function but not a surjective function.