Question

determine if \\(\sqrt{2}\\) is rational or irrational and give a reason for your answer.
answer attempt 2 out of 3
the number \\(\sqrt{2}\\) is rational because it is the square root of a non - perfect square

Explanation:

Step1: Recall definition of rational numbers

A rational number can be expressed as $\frac{p}{q}$ where $p, q$ are integers and $q
eq0$, and $p$ and $q$ have no common factors other than 1 (in lowest terms).

Step2: Assume $\sqrt{2}$ is rational

Suppose $\sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are coprime integers ($\gcd(p,q) = 1$) and $q>0$. Then squaring both sides, we get $2=\frac{p^{2}}{q^{2}}$, so $p^{2}=2q^{2}$.

Step3: Analyze parity of $p$

From $p^{2}=2q^{2}$, $p^{2}$ is even (since it is 2 times an integer). If a square of an integer is even, the integer itself is even. So let $p = 2k$ for some integer $k$.

Step4: Substitute $p = 2k$ into equation

Substituting $p = 2k$ into $p^{2}=2q^{2}$, we get $(2k)^{2}=2q^{2}$, which simplifies to $4k^{2}=2q^{2}$, and then $2k^{2}=q^{2}$.

Step5: Analyze parity of $q$

From $2k^{2}=q^{2}$, $q^{2}$ is even, so $q$ must be even (similar to step 3). But if both $p$ and $q$ are even, they have a common factor of 2, which contradicts the assumption that $\gcd(p,q)=1$. So our initial assumption that $\sqrt{2}$ is rational is false.

Step6: Conclusion about $\sqrt{2}$

Since we can't express $\sqrt{2}$ as a fraction of two coprime integers, $\sqrt{2}$ is irrational. Also, a rational number's square root of a non - perfect square is not the reason for being rational; in fact, the square root of a non - perfect square positive integer is irrational.

Answer:

The number $\sqrt{2}$ is irrational because it cannot be expressed as a fraction $\frac{p}{q}$ (where $p,q$ are integers, $q
eq0$, and $\gcd(p,q) = 1$) as shown by the contradiction in the above steps, and the square root of a non - perfect square positive integer is irrational.