# Alternative proof that $\sqrt{2}$ is irrational

I'm trying to understand an alternative to the standard proof that $$\sqrt{2}$$ is irrational, which uses the fundamental theorem of arithmetic. It doesn't look overly crisp to me at this moment.

Suppose for the sake of contradiction that there exists $$x \in \mathbb{Q}$$ satisfying $$x^2 = 2$$, and write $$x = \frac{p}{q}$$ where $$p,q \in \mathbb{Z}$$, $$q \neq 0$$, and $$p,q$$ have no common factors. Without loss of generality, we can assume $$x > 0$$ because $$x^2 = (-x)^2$$ for any $$x \in \mathbb{Q}$$, so we can assume $$p,q \in \mathbb{N}$$. Then $$x^2 = \left(\frac{p}{q}\right)^2 = \frac{p^2}{q^2} = 2$$, so $$p^2 = 2q^2$$.

I want to know say that the LHS has an even number of prime factors and the RHS has an odd number of prime factors, but for the fundamental theorem of arithmetic requires that the elements be greater than $$1$$. Is there a way to do this without casework?

• The fundamental theorem of arithmetics does not require a $>1$ condition. The number $1$ has $0$ prime divisors and uniquely factors as the empty product of primes. Oct 22, 2021 at 5:00
• @QiZhu One standard formulation of the FTA is exactly that. Oct 22, 2021 at 5:01
• If $q=3$, then the $RHS$ has an even number of prime factors. Oct 22, 2021 at 5:03
• You’ve copied the proof, except for the part with which you had a problem? Oct 22, 2021 at 5:03
• Brad, what you are looking for in the simplest manner is that the $LHS$ will have an even number of $2$'s, while the $RHS$ will not, contradicting the $FTA$. Oct 22, 2021 at 5:11

There is no need for casework: FTA holds for $$1$$, namely $$1$$ has no prime factors and factors uniquely into the empty product of primes.

I'm quite shocked when I googled the FTA that so many sources give the "wrong" statement of FTA. There is no need to demand some $$> 1$$ condition, but unfortunately some authors disregard "trivial" cases. This bad habit can make the theory uglier than it needs to be, as seen in this question.

For readers who understand German, I recommend the article Über die Null, den leeren Raum und andere triviale Fälle by Martin Brandenburg in which he talks about many such problems ($$0^0$$, trivial products, trivial vector spaces, empty manifolds, ...). Related is also MO/45951 on which Martin's article is based on.

In general, for any positive integer $$n$$, including $$1$$, we can write $$n$$ as

$$n=2^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_k^{e_k},$$ where each $$p_i$$ is a distinct prime not equal to $$2$$ and $$e_i\in\mathbb N\cup\{0\}$$ for all $$i=2,3,...,k$$. Indeed, if $$n=1$$, then each $$e_i=0$$. This follows directly from FTA.

Now, we can write $$p=2^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_k^{e_k}$$ and $$q=2^{f_1}\cdot q_2^{f_2}\cdot ...\cdot q_j^{f_j}$$. Therefore, we have

\begin{align*} (2^{e_1}\cdot p_2^{e_2}...p_k^{e_k})^2&=2(2^{f_1}\cdot q_2^{f_2}\cdot...\cdot q_j^{f_j})^2\\ 2^{2e_1}\cdot p_2^{2e_2}\cdot ...\cdot p_k^{2e_k}&=2^{2f_1+1}\cdot q_2^{2f_2}\cdot...\cdot q_j^{2f_j} \end{align*} Therefore, $$2e_1 = 2f_1+1$$, but $$e_1$$ and $$f_1$$ are integers, which leads to a contradiction.