Square Roots of Primes are Irrational

In the spirit of this blog, I will post proofs of theorems and other little lemmas that I find intriguing or just fun. Today I want to talk about a fun little problem I encountered in my early high school days, but didn't know how to tackle it at the time. But I figured it out a few years later when I came back to it. What was the problem, you ask? To prove that the square root of any prime $p$ is irrational. Most people are familiar with the proof that $\sqrt{2}$ is irrational. It goes like this. Assume that $\sqrt{2}$ is rational, that is, that there exists relatively prime integers $m, n$ such that $$ \sqrt{2} = \frac{m}{n}$$ Rearranging the equation, we obtain $$2n^2 = m^2$$. This implies that $m$ is even, and so we can write $m = 2r, r \in \mathbb Z$. Substituting into the equation, we have $$2n^2 = (2r)^2 = 4r^2$$ This implies that $n$ is also even. Now that we know that both $m$ and $n$ are even, we have a contradiction. $m$ and $n$ were assumed to be relatively prime, so $\sqrt{2} \notin \mathbb Q.$ Now let $p$ be any prime and assume $$\sqrt{p} = \frac{m}{n}$$ for any $(m, n) = 1, m, n \in \mathbb Z$. Then we can proceed as before. $pn^2 = m^2$, which implies that $p$ must divide $m^2$. By the fundamental theorem of arithmetic, $m^2$ has the two sets of identical prime factors, and so $p$ divides $m$. Thus $m = pk, k \in \mathbb Z$. Then we have $$pn^2 = p^2k^2 \rightarrow n^2 = pk^2$$ And there you have it. $p$ divides $n$ and $p$ divides $m$, which contradicts our assumption of $(m, n) = 1$. Thus the square root of any prime is irrational.

Hard Lefschetz Theorem

Let $X$ be a Kähler manifold of complex dimension $dim_{\mathbb C} = n$. Let $[\omega]$ be the cohomology class of a Kähler metric on $X$. Then powers of the class $[\omega]$ define a linear morphism between groups $$ L^k: H^{n-k}(X, {\mathbb C}) \longrightarrow H^{n+k}(X, \mathbb C)$$ Hard Lefschetz asserts that this is, in fact, an isomorphism of vector spaces. Proof to come soon!