I think I have a valid proof for the irrationality of $\sqrt{n}$, when $n$ is not the square of a positive integer. My teacher taught us a different method to prove $\sqrt{2}$ is irrational, which is perfectly valid, but my method just seems a little neater to me and it works in general for any value of n that isn't a perfect square .
Firstly, lets assume $\sqrt{n}$, when $n$ is not the square of a positive integer, is rational.
Hence:
$\sqrt{n}$ = $\displaystyle\frac{a}{b}$, where $a, b\in\mathbb{N}$ $b\neq0$
Therefore, by squaring both sides we find:
${n}$ = $\displaystyle\frac{a^2}{b^2}$
Hence, we see $a^2$ = $nb^2$ $(1)$
Let $a$ have a prime factorisation of $a= \prod\limits_{i=1}^n {p_i}^{e_i}$, where ${p_i}$ is a prime, and ${e_i}$ is the multiplicity of the prime power ${p_i}$. Hence, $a^2$ will have a prime factorisation of $a^2= \prod\limits_{i=1}^n {p_i}^{2e_i}$.
Also, we can state $b$ in the form of $b= \prod\limits_{i=1}^n {q_i}^{f_i}$, hence $b^2$ can be expressed as $b^2= \prod\limits_{i=1}^n {q_i}^{2f_i}$, where ${q_i}$ is a prime, and $f_i$ is its corresponding multiplicity.
Now consider equation $(1)$. We know that $n$ is not a perfect square of a positive integer, so the multiplicities in its prime factorisation must not all be even, as a number is a perfect square if, and only if, (iff), the multiplicities of its prime factors are all even. Hence, it is impossible for equation $(1)$ to hold, as the multiplicities on the LHS must all be even, and on the RHS, it is impossible for this to be the case, as the multiplicities in $b^2$ are all even, but in $n$ they are not. Hence, the product of these two can never have even multiplicities for all prime factors, hence equality can never hold.
Hence, this results in a contradiction, so our initial assumption that $\sqrt{n}$, when n is not the square of a positive integer, being rational was false, hence $\sqrt{n}$ must be irrational, for $n$ not being a perfect square.
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