Prove that every positive real numbers y > 0 there is a rational number z e Q so that 0 < z < y

Solution:
1. Let y be an arbitrary positive real number, i.e., y > 0.

2. Since y is positive, we can consider the interval (0, y).

3. Recall that the rational numbers \mathbb{Q} are dense in the real numbers \mathbb{R}. This means that between any two real numbers, there exists at least one rational number.

4. Given that density, for any positive real number y, there exists a rational number z \in \mathbb{Q} such that
0 < z < y.

5. To concretize the argument, consider the number \frac{y}{2} which is positive since y is positive. The number \frac{y}{2} is less than y and can be made rational if y itself is rational.

6. However, if y is irrational, we can always find rational estimating numbers such as \frac{m}{n} where \frac{m}{n} < y and \frac{m}{n} is sufficiently close to y but positive.

7. Thus, there definitely exists a rational number z such that 0 < z < y.

Therefore, we have proven that for every y > 0, there exists a rational number z \in \mathbb{Q} such that 0 < z < y.