To prove that for all rational numbers $x, y \in \mathbb{Q}$ (with $x < y$), there exists a rational number $z \in \mathbb{Q}$ such that $x < z < y$, we proceed with a constructive argument.

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Proof:

We are asked to find a rational number $z \in \mathbb{Q}$ such that $x < z < y$, given that $x$ and $y$ are rational numbers, and $x < y$.

Step 1: Define a candidate $z$

One way to ensure that $x < z < y$ is to use the arithmetic mean (average) of $x$ and $y$: $z = \frac{x + y}{2}$ This $z$ is guaranteed to lie between $x$ and $y$.

Step 2: Verify that $x < z < y$

Since $x < y$, their average will fall strictly between them. We prove this below:

1. First, show that $x < z$: $x < \frac{x + y}{2} \quad \text{if and only if} \quad 2x < x + y \quad \text{or} \quad x < y$ Since $x < y$ by assumption, this inequality holds.

2. Now, show that $z < y$: $\frac{x + y}{2} < y \quad \text{if and only if} \quad x + y < 2y \quad \text{or} \quad x < y$ Again, since $x < y$ by assumption, this inequality holds.

Thus, we have: $x < \frac{x + y}{2} < y$ Therefore, $z = \frac{x + y}{2}$ is a rational number that lies strictly between $x$ and $y$.

Step 3: Conclusion

Since the arithmetic mean of two rational numbers is also a rational number, $z = \frac{x + y}{2} \in \mathbb{Q}$. Therefore, for any two rational numbers $x < y$, we have found a rational number $z$ such that $x < z < y$.

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Conclusion

For any two rational numbers $x, y \in \mathbb{Q}$ with $x < y$, there exists a rational number $z \in \mathbb{Q}$ such that $x < z < y$. The choice $z = \frac{x + y}{2}$ satisfies this condition.

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Do you want further details or have any questions?
Here are 5 related questions that expand on this concept:

1. Can this method be generalized to real numbers $x, y \in \mathbb{R}$?
2. Is the set of rational numbers $\mathbb{Q}$ dense in $\mathbb{R}$?
3. How can we show that there are infinitely many rational numbers between any two distinct rational numbers?
4. Can you prove that there exists an irrational number between two distinct rational numbers?
5. How can this idea be extended to sequences of rational numbers converging to a real number?

Tip: The density of rational numbers states that between any two real numbers (whether rational or not), there always exists a rational number.