To prove that $3\sqrt{5} + \sqrt{7}$ is an irrational number, we can use proof by contradiction. Here's how we approach the proof:

Step 1: Assume the opposite (i.e., assume the number is rational)

Suppose that $3\sqrt{5} + \sqrt{7}$ is rational. A rational number can be written as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. So, we assume:

$3\sqrt{5} + \sqrt{7} = \frac{p}{q}$

Step 2: Isolate one of the square roots

We want to isolate one of the square roots to manipulate the equation. Let's isolate $\sqrt{7}$:

$\sqrt{7} = \frac{p}{q} - 3\sqrt{5}$

Step 3: Square both sides of the equation

Next, we'll square both sides to eliminate the square root on the left-hand side:

$7 = \left( \frac{p}{q} - 3\sqrt{5} \right)^2$

Expanding the right-hand side:

$7 = \left( \frac{p}{q} \right)^2 - 2 \cdot \frac{p}{q} \cdot 3\sqrt{5} + (3\sqrt{5})^2$ $7 = \frac{p^2}{q^2} - \frac{6p\sqrt{5}}{q} + 45$

Simplify this:

$7 = \frac{p^2}{q^2} + 45 - \frac{6p\sqrt{5}}{q}$

Step 4: Rearranging the equation

Now, isolate the term involving $\sqrt{5}$:

$7 - 45 - \frac{p^2}{q^2} = - \frac{6p\sqrt{5}}{q}$

Simplify:

$-38 - \frac{p^2}{q^2} = - \frac{6p\sqrt{5}}{q}$

$38 + \frac{p^2}{q^2} = \frac{6p\sqrt{5}}{q}$

Step 5: Conclusion from the form of the equation

The right-hand side of this equation contains $\sqrt{5}$, which is irrational. For the equation to hold true, the left-hand side must also contain $\sqrt{5}$, but the left side is entirely rational. This is a contradiction because a rational number cannot equal an irrational number.

Step 6: Conclusion

Since our assumption that $3\sqrt{5} + \sqrt{7}$ is rational led to a contradiction, it follows that $3\sqrt{5} + \sqrt{7}$ must be irrational.

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Would you like any additional clarifications on the proof? Here are some related questions:

1. What does it mean for a number to be irrational?
2. How do we prove that $\sqrt{5}$ is irrational?
3. What are some examples of rational and irrational numbers?
4. Can the sum of a rational and an irrational number be rational?
5. How can we prove that a number like $\sqrt{2} + \sqrt{3}$ is irrational?

Tip: Always remember that any time you are dealing with sums or products of irrational numbers, proving irrationality often involves isolating one term and finding a contradiction!