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Euclid's proof that the primes never run out is over two thousand years old.
\begin{theorem}
There are infinitely many primes.
\end{theorem}
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\noindent\textit{Proof idea.} If there were finitely many, their product
plus one would have a prime factor that is none of them.
\webnextpane[See the proof]
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\begin{proof}[\proofname\webpanelink{prev}{ (hide)}]
Suppose $p_1, \dots, p_k$ were all of them, and let
\[ N = p_1 p_2 \cdots p_k + 1 . \]
No $p_i$ divides $N$, since each leaves remainder $1$. Yet $N > 1$ has a
prime factor, and it is none of the $p_i$ -- a contradiction.
\end{proof}
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The same idea shows that there are infinitely many primes of the form
$4k + 3$, though not, without more work, of the form $4k + 1$.

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\noindent\textbf{Why is $\sqrt 2$ irrational?} Because a fraction for it
would have to be reducible forever. \webpanelink{longer}{Say more}
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\noindent\textbf{Why is $\sqrt 2$ irrational?} If $\sqrt 2 = p/q$ in lowest
terms, then $p^2 = 2q^2$, so $p$ is even; then $q$ is even as well, and the
fraction was not in lowest terms after all.
\webpanelink{full}{The full argument} \textperiodcentered{} \webpanelink{short}{Less}
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\noindent\textbf{Why is $\sqrt 2$ irrational?} Suppose $\sqrt 2 = p/q$ with
$p, q$ positive integers sharing no factor. Squaring,
\[ p^2 = 2q^2 , \]
so $p^2$ is even, hence $p$ is even: $p = 2r$. Then $4r^2 = 2q^2$, so
$q^2 = 2r^2$ is even and $q$ is even too -- contradicting that $p$ and $q$
share no factor. \webpanelink{first}{Back to the short answer}
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\noindent\textbf{Is every even number above 2 a sum of two primes?}
Nobody knows. \webnextpane[What is known]
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\noindent\textbf{Is every even number above 2 a sum of two primes?}
This is Goldbach's conjecture, checked by computer up to $4 \cdot 10^{18}$.
Every odd number above 5 is a sum of three primes (Helfgott, 2013).
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