To prove that $\sqrt{5}$ is irrational using a parity argument, we'll assume, for contradiction, that $\sqrt{5}$ is rational. That is, assume:

$x = \frac{m}{n}, \quad \text{where} \quad m, n \in \mathbb{N} \quad \text{and} \quad \gcd(m,n) = 1$

This means $m$ and $n$ have no common factors other than 1. Also, $x^2 = 5$, so squaring both sides gives:

$\left(\frac{m}{n}\right)^2 = 5 \implies \frac{m^2}{n^2} = 5 \implies m^2 = 5n^2$

Now, $m^2 = 5n^2$ implies that $m^2$ is divisible by 5. Thus, $m$ must also be divisible by 5 (since any square divisible by a prime implies the original number is divisible by that prime). So, we can write:

$m = 5k \quad \text{for some integer} \quad k$

Substitute this into the equation $m^2 = 5n^2$:

$(5k)^2 = 5n^2 \implies 25k^2 = 5n^2 \implies n^2 = 5k^2$

This shows that $n^2$ is also divisible by 5, which means $n$ must be divisible by 5.

Contradiction:

Now, both $m$ and $n$ are divisible by 5, which contradicts our assumption that $\gcd(m,n) = 1$ (since $m$ and $n$ would share a factor of 5). Therefore, our original assumption that $\sqrt{5}$ is rational must be false.

Thus, $\sqrt{5}$ is irrational.

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Exploring the four cases (odd/even combinations):

• Case 1: $m$ and $n$ are both even:
  If both $m$ and $n$ are even, then $\frac{m}{n}$ is not in its lowest terms, contradicting the assumption that $\gcd(m,n) = 1$.

• Case 2: $m$ is even, $n$ is odd:
  If $m$ is even, then $m^2$ is divisible by 4. However, since $m^2 = 5n^2$, $m^2$ must also be divisible by 5. This creates a contradiction, as no number can be divisible by both 4 and 5 unless it's divisible by 20, which would imply that both $m$ and $n$ are divisible by 5, contradicting our assumption.

• Case 3: $m$ is odd, $n$ is even:
  If $n$ is even, $n^2$ is divisible by 4. But $m^2 = 5n^2$, so $m^2$ must be divisible by 20, which implies that $m$ must also be divisible by 5, again contradicting the assumption that $m$ and $n$ have no common factors.

• Case 4: $m$ and $n$ are both odd:
  If both $m$ and $n$ are odd, then $m^2$ is odd and $n^2$ is odd. But $m^2 = 5n^2$ implies that $m^2$ is divisible by 5, so $m$ must be divisible by 5, leading to the same contradiction that $m$ and $n$ are not coprime.

In all cases, a contradiction arises, which proves that $\sqrt{5}$ cannot be rational.

Would you like more details or have any further questions?

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Here are five related questions:

1. Can this method be extended to prove the irrationality of $\sqrt{p}$ for any prime $p$?
2. How does the method of infinite descent relate to the proof of irrationality?
3. What are other common proofs used to show irrationality of square roots of primes?
4. How can parity arguments be applied in number theory outside irrationality proofs?
5. What is the generalization of this parity argument to higher roots (e.g., $\sqrt[3]{p}$)?

Tip: Always check if an assumption leads to a contradiction in proofs involving rationality or coprimeness.