To prove that $\sqrt{7}$ is irrational, we use a proof by contradiction. Here's the step-by-step proof:

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

If $\sqrt{7}$ is rational, then it can be expressed as a fraction: $\sqrt{7} = \frac{p}{q}$ where $p$ and $q$ are integers with no common factors (i.e., the fraction is in its simplest form) and $q \neq 0$.

Step 2: Square both sides.

Squaring both sides gives: $7 = \frac{p^2}{q^2}$ Multiply through by $q^2$ to eliminate the denominator: $7q^2 = p^2$

Step 3: Analyze divisibility.

This equation shows that $p^2$ is divisible by $7$. From a fundamental property of integers, if $7$ divides $p^2$, then $7$ must also divide $p$.

Thus, we can write $p = 7k$ for some integer $k$.

Step 4: Substitute back.

Substitute $p = 7k$ into the equation $7q^2 = p^2$: $7q^2 = (7k)^2$ $7q^2 = 49k^2$ Divide through by $7$: $q^2 = 7k^2$

Step 5: Contradiction.

This shows that $q^2$ is divisible by $7$, which means $q$ is also divisible by $7$.

Step 6: Simplest form contradiction.

If both $p$ and $q$ are divisible by $7$, this contradicts the initial assumption that $\frac{p}{q}$ is in simplest form.

Conclusion:

Since assuming $\sqrt{7}$ is rational leads to a contradiction, we conclude that $\sqrt{7}$ is irrational.

Would you like me to elaborate further on any step?

Questions to Expand Learning:

1. Why does $7 \mid p^2 \implies 7 \mid p$?
2. Can this proof be generalized for other primes like $\sqrt{11}$ or $\sqrt{13}$?
3. How does this proof relate to the Fundamental Theorem of Arithmetic?
4. What are examples of numbers that are rational?
5. Can a similar contradiction be used to prove $\sqrt{2}$ is irrational?

Tip:

When proving irrationality, always start by assuming the number is rational and use properties of divisibility to find contradictions.