The problem in the image asks to show that the given values in Mindscapes 10–16 are irrational. Let's break down how to prove a number is irrational for the two examples we can see:

1. Mindscape 10: $\sqrt{5}$
2. Mindscape 11: $2 \cdot \sqrt{3}$

How to Prove a Number is Irrational

To prove a number is irrational, we need to demonstrate that it cannot be written as a ratio of two integers. This is done through a proof by contradiction, which assumes the number is rational and leads to a logical inconsistency.

Example 1: $\sqrt{5}$

1. Assume $\sqrt{5}$ is rational.
2. That means $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are integers with no common factors, and $q \neq 0$.
3. Squaring both sides: $5 = \frac{p^2}{q^2}$, which gives $p^2 = 5q^2$.
4. Therefore, $p^2$ is divisible by 5, meaning $p$ must be divisible by 5 (because 5 is prime).
5. Let $p = 5k$ for some integer $k$. Substituting into the equation:
   $(5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies q^2 = 5k^2.$
6. This implies $q^2$ is divisible by 5, so $q$ must also be divisible by 5.
7. But if both $p$ and $q$ are divisible by 5, this contradicts the assumption that $p$ and $q$ have no common factors. Thus, $\sqrt{5}$ is irrational.

Example 2: $2 \cdot \sqrt{3}$

1. Assume $2 \cdot \sqrt{3}$ is rational.
2. Let $2 \cdot \sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are integers with no common factors, and $q \neq 0$.
3. Solving for $\sqrt{3}$, we get:
   $\sqrt{3} = \frac{p}{2q}.$
4. This implies $\sqrt{3}$ is rational, which is a contradiction because it is well-known that $\sqrt{3}$ is irrational.
5. Hence, $2 \cdot \sqrt{3}$ must also be irrational.

These steps show that both $\sqrt{5}$ and $2 \cdot \sqrt{3}$ are irrational.

Would you like further details or any clarifications?
Here are 5 related questions you might consider:

1. How would you prove that $\sqrt{2}$ is irrational?
2. Can every square root of a non-perfect square be irrational?
3. What is the difference between rational and irrational numbers in terms of their decimal expansions?
4. How does the proof for $\sqrt{5}$ change if you consider $\sqrt{6}$?
5. Why does the irrationality of $\sqrt{n}$ depend on $n$ being prime or not?

Tip: Proof by contradiction is a powerful method in mathematics, especially when proving something like irrationality. Make sure you're comfortable with setting up logical contradictions.