$$  $$prove that $$ \sqrt7 $$ is irrational number. Use that result to prove 2$$ +3\sqrt7 $$ is an irrational number

### Proof that $ \sqrt{7} $ is an Irrational Number:

We will use **proof by contradiction**.

1. **Assume $ \sqrt{7} $ is rational.**
   This means we can express $ \sqrt{7} $ as a fraction of two integers $ \frac{p}{q} $, where $ p $ and $ q $ are coprime integers (i.e., $ \gcd(p, q) = 1 $ and $ q \neq 0 $).

   \[
   \sqrt{7} = \frac{p}{q}
   \]

2. **Square both sides to eliminate the square root.**

   \[
   7 = \frac{p^2}{q^2}
   \]

   Multiply both sides by $ q^2 $ to clear the denominator:

   \[
   7q^2 = p^2
   \]

3. **Analyze the equation.**

   This equation implies that $ p^2 $ is divisible by 7, meaning that $ p $ must also be divisible by 7 (since 7 is a prime number and a property of primes is that if a prime divides a square, it must divide the original number).

   So, let $ p = 7k $ for some integer $ k $.

4. **Substitute $ p = 7k $ back into the equation.**

   \[
   7q^2 = (7k)^2
   \]

   Simplifying:

   \[
   7q^2 = 49k^2
   \]

   Divide both sides by 7:

   \[
   q^2 = 7k^2
   \]

   This implies that $ q^2 $ is divisible by 7, so $ q $ must also be divisible by 7.

5. **Contradiction:**

   Now we have shown that both $ p $ and $ q $ are divisible by 7, which contradicts our assumption that $ \gcd(p, q) = 1 $. Therefore, the assumption that $ \sqrt{7} $ is rational must be false.

   Thus, $ \sqrt{7} $ is irrational.

---

### Proof that $ 2 + 3\sqrt{7} $ is Irrational:

Next, we use the fact that $ \sqrt{7} $ is irrational to prove that $ 2 + 3\sqrt{7} $ is also irrational.

1. **Assume $ 2 + 3\sqrt{7} $ is rational.**
   This means $ 2 + 3\sqrt{7} = r $, where $ r $ is a rational number.

2. **Isolate $ 3\sqrt{7} $:**

   \[
   3\sqrt{7} = r - 2
   \]

   Since $ r $ is rational and $ 2 $ is rational, $ r - 2 $ is rational. Thus, $ 3\sqrt{7} $ is rational.

3. **Divide both sides by 3:**

   \[
   \sqrt{7} = \frac{r - 2}{3}
   \]

   Since $ r - 2 $ is rational and 3 is rational, $ \frac{r - 2}{3} $ is rational. This implies that $ \sqrt{7} $ is rational.

4. **Contradiction:**
   
   This contradicts the fact that $ \sqrt{7} $ is irrational (as proven earlier). Therefore, the assumption that $ 2 + 3\sqrt{7} $ is rational must be false.

   Hence, $ 2 + 3\sqrt{7} $ is irrational.

---

Do you need further clarification or more details on any part of the proof? 

Here are 5 related questions to further explore this topic:
1. How can we extend this proof technique to other square roots of non-perfect squares?
2. Why do we require the assumption that $ \gcd(p, q) = 1 $ in the proof by contradiction?
3. Can you generalize this result for numbers of the form $ a + b\sqrt{n} $ where $ \sqrt{n} $ is irrational?
4. How can you prove the irrationality of other numbers like $ \sqrt{2} $ or $ \sqrt{5} $?
5. Is it possible for the sum of a rational and an irrational number to be rational?

**Tip:** Always begin a proof by contradiction with a clear assumption that is opposite to what you want to prove, and then work logically until you reach a contradiction.

Prove that √7 is an irrational number. Use that result to prove 2 + 3√7 is an irrational number.

Discover a proof that √7 is irrational using the method of contradiction, and apply this result to demonstrate that 2 + 3√7 is also irrational. Explore key mathematical concepts including irrational numbers and prime properties. Suitable for high school students in Grades 9-12.

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