If Q is the set of rational numbers and a function `f:Q to Q ` is defined as `f(x)=5x-4, x in Q`, then show that f is one-one and onto.

Answer

To show that the function \( f: Q \to Q \) defined by \( f(x) = 5x - 4 \) is one-one and onto, we will go through the following steps:

### Step 1: Show that \( f \) is one-one

To prove that \( f \) is one-one, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \).
\[ f(x_1) = 5x_1 - 4 \quad \text{and} \quad f(x_2) = 5x_2 - 4 \]
Therefore, we have:
\[ 5x_1 - 4 = 5x_2 - 4 \]

2. Simplifying this equation:
\[ 5x_1 - 4 + 4 = 5x_2 - 4 + 4 \]
\[ 5x_1 = 5x_2 \]

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

Since \( x_1 = x_2 \) whenever \( f(x_1) = f(x_2) \), the function \( f \) is one-one.

### Step 2: Show that \( f \) is onto

To prove that \( f \) is onto, we need to show that for every \( y \in Q \), there exists an \( x \in Q \) such that \( f(x) = y \).

1. Let \( y \) be any rational number. We want to find \( x \) such that:
\[ f(x) = y \]
This gives us:
\[ 5x - 4 = y \]

2. Solving for \( x \):
\[ 5x = y + 4 \]
\[ x = \frac{y + 4}{5} \]

3. Since \( y \) is a rational number, \( y + 4 \) is also a rational number. The division of a rational number by a non-zero rational number (5 in this case) results in another rational number. Therefore, \( x \) is rational.

Since for every \( y \in Q \), we can find an \( x \in Q \) such that \( f(x) = y \), the function \( f \) is onto.

### Conclusion

Since we have shown that \( f \) is both one-one and onto, we conclude that the function \( f(x) = 5x - 4 \) is a bijection from \( Q \) to \( Q \).

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