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Proof that $\sqrt{2}$ is irrational using the unique prime factorization theorem.

My proof:

Assume for the purpose of contradiction that $\sqrt{2}$ is rational.

By the unique prime fact. the., we now that $\sqrt{2} = x^y $ where $x$ is a prime and $y$ is a positive integer. $$\sqrt{2} = x^y$$ $$2 = x^{2y}$$ $$2 = x^y x^y$$

We know that 2 is prime number whose unique factorization is $2^1$, thus having 2 equal $x^y x^y$ is a contradiction.

Joel
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    Note: the prime factorization theorem applies to integers, not rationals. Hence $\sqrt{2}=x^y$ is invalid. – vadim123 May 01 '15 at 04:31
  • Sorry, not even close. Start by saying suppose to the contrary that $\sqrt{2}$ is rational. Then there exist integers $a$ and $b$, with $b\ne 0$, such that $\sqrt{2}=\frac{a}{b}$. From this equation it follows that $a^2=2b^2$. Continue, using unique factorization to derive a contradiction. – André Nicolas May 01 '15 at 04:32
  • Hint: After you take it as $\frac{a}{b}$ where $\gcd(a,b)=1$, a few manipulations will later show that $\gcd(a,b)\neq 1$ resulting in a contradiction. – Prasun Biswas May 01 '15 at 04:37
  • Let $2^s$ be the power of $2$ in the prime power factorization of $a$. (We allow $s=0$.) Then the power of $2$ in the prime power factorization of $a^2$ is $2^{2s}$. Let $2^t$ be the power of $2$ in the prime power factorization of $b$. (We allow $t=0$.) Then the power of $2$ in the factorization of $2b^2$ is $2^{2t+1}$. If $a^2=2b^2$, then $2^{2s}=2^{2t+1}$, giving $2s=2t+1$. This is impossible, since $2s$ is even and $2t+1$ is odd. – André Nicolas May 01 '15 at 04:53
  • @vadim It seems a bit disingenuous to claim that prime factorization doesn't apply to rationals. It isn't quite the same, but every rational admits a unique representation $p_1^{s_1}...p_n^{s_n}$ where $p_i$ are distinct primes and $s^i$ are non-zero integers. – jxnh May 01 '15 at 06:00
  • @JHance, indeed this is true, but it is not the prime factorization theorem but a corollary. – vadim123 May 01 '15 at 13:45

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No, the proof is not rigorous--as a commenter pointed out, the prime factorization theorem only applies to integers. Moreover, the prime factorization theorem would not imply that any given integer has the form $x^y$ for $x,y$ both integers. It implies that every integer can be decomposed into a product of prime powers.

So to attempt a proof of this form, you would have to assume that $\sqrt{2}$ is rational with least rational expression $a/b$ where $a,b$ both integers and $b\ne 0$. From there you can use the prime factorization theorem to say that $a=p_1^{a_1}\cdot\cdot\cdot p_n^{a_n}$, and $b=q_1^{b_1}\cdot \cdot\cdot q_m^{a_m}$ where $p_i,q_j$ are all prime for each $i=1,...,n$ and $j=1,...,m$ and $a_k,b_\ell$ are all positive integers $k=1,...,n$ and $\ell=1,...,m$.

However, where you'd go from here isn't exactly clear--or at the least, this doesn't seem to be the best proof method. Better to just assume $\sqrt{2}=a/b$ where $a/b$ is in least form, therefore they share no common factors, and then proceed to scrutinize the fact that $2=a^2/b^2$ implying that the right-hand side is even.

Addem
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  • Found this great method on Khan Academy: https://www.khanacademy.org/math/algebra/ratio-proportion-topic/alg-rational-irrational-numbers/v/proof-that-square-root-of-2-is-irrational – Joel May 01 '15 at 04:52
  • I don't need the prime factorization theorem to see that every integer $n$ has the form $x^y$, just set $x$ to be $n$ and $y$ to be $1$ ;) – Sebastian Bechtel May 01 '15 at 05:31