To prove that $\sqrt{7}$ is irrational, we can use a proof by contradiction. Here are the steps:

Step 1: Assume $\sqrt{7}$ is rational

• Suppose that $\sqrt{7}$ is a rational number.
• If $\sqrt{7}$ is rational, then we can express it as a fraction of two integers: $\sqrt{7} = \frac{a}{b}$ where $a$ and $b$ are integers, $b \neq 0$, and the fraction $\frac{a}{b}$ is in its simplest form (i.e., $a$ and $b$ have no common divisors other than 1).

Step 2: Square both sides

• Squaring both sides of the equation gives: $7 = \frac{a^2}{b^2}$
• Multiply both sides by $b^2$ to eliminate the denominator: $7b^2 = a^2$
• This equation implies that $a^2$ is divisible by 7.

Step 3: Analyze the divisibility of $a$

• If $a^2$ is divisible by 7, then $a$ must also be divisible by 7. This is because if a prime number divides a square, it must divide the base of that square.
• Therefore, we can write $a = 7k$ for some integer $k$.

Step 4: Substitute $a = 7k$ into the equation

• Substitute $a = 7k$ into the equation $7b^2 = a^2$: $7b^2 = (7k)^2$ $7b^2 = 49k^2$
• Divide both sides by 7: $b^2 = 7k^2$
• This equation implies that $b^2$ is divisible by 7, and thus $b$ must also be divisible by 7.

Step 5: Contradiction

• We now have that both $a$ and $b$ are divisible by 7. However, this contradicts our original assumption that $\frac{a}{b}$ is in its simplest form, as $a$ and $b$ have a common divisor (7).

Conclusion

• Since assuming $\sqrt{7}$ is rational leads to a contradiction, the original assumption must be false. Therefore, $\sqrt{7}$ is irrational.

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How can we generalize this proof to other square roots of non-perfect squares?
2. What is the difference between rational and irrational numbers?
3. How does the Fundamental Theorem of Arithmetic support this proof?
4. Can $\sqrt{49}$ be considered irrational, and why?
5. How does this proof relate to the proof of the irrationality of $\sqrt{2}$?

Tip: For any non-perfect square $n$, the square root $\sqrt{n}$ is irrational.