To prove that $\sqrt{17}$ is irrational, let's proceed with a proof by contradiction:

Step 1: Assume $\sqrt{17}$ is rational.

If $\sqrt{17}$ is rational, it can be expressed as a fraction $\frac{p}{q}$, where:

1. $p$ and $q$ are integers,
2. $\gcd(p, q) = 1$ (i.e., the fraction is in its simplest form),
3. $q \neq 0$.

Thus, we can write: $\sqrt{17} = \frac{p}{q}.$

Step 2: Square both sides.

$17 = \frac{p^2}{q^2}.$

Multiply through by $q^2$ to eliminate the fraction: $17q^2 = p^2.$

Step 3: Analyze the equation.

This equation implies that $p^2$ is divisible by $17$. Therefore, $p$ must also be divisible by $17$ (since $17$ is prime, if it divides $p^2$, it must divide $p$).

Let $p = 17k$, where $k$ is an integer.

Step 4: Substitute $p = 17k$ into the equation.

$17q^2 = (17k)^2.$

Simplify: $17q^2 = 289k^2.$

Divide through by $17$: $q^2 = 17k^2.$

Step 5: Analyze the new equation.

This equation implies that $q^2$ is divisible by $17$. Therefore, $q$ must also be divisible by $17$.

Step 6: Contradiction.

If both $p$ and $q$ are divisible by $17$, this contradicts the assumption that $\frac{p}{q}$ is in its simplest form ($\gcd(p, q) = 1$).

Conclusion:

The assumption that $\sqrt{17}$ is rational leads to a contradiction. Therefore, $\sqrt{17}$ must be irrational.

Would you like me to explain a specific step in more detail?
Here are 5 related questions for further exploration:

1. How can we generalize this proof for any non-perfect square integer $n$?
2. What are the properties of prime numbers that make this proof work?
3. Can a similar method be used to prove $\sqrt{18}$ is irrational?
4. How do we prove that other roots, like $\sqrt[3]{2}$, are irrational?
5. Is there a geometric interpretation of why $\sqrt{17}$ is irrational?

Tip: When tackling proof questions, breaking the problem into smaller logical steps is key to finding contradictions or verifying claims!