We start with two of the most famous proofs ever. Both use the results about prime numbers and factorization that we proved last time. They also both wield contradiction, further demonstrating the surprising power and potentially broad nature of proofs in the spaace.
There exist infinitely many prime numbers.
Assume for contradiction that there are a finite number of primes. Let $p_1,\ldots,p_k$ be the $k$ primes. To find a contradiction, it is enough to show that some prime is not included on this list. Consider $x=p_1 p_2\cdots p_k+1$. Observe that $x$ is not divisible by any of $p_1,\ldots, p_k$ (as this would imply that $p_i\mid 1$, which is impossible). On the other hand, by the Fundamental Theorem of Arithmetic, $x$ can be factored into primes, and in particular some prime $p$ divides $x$. Since this prime can't be any of $p_1,\ldots,p_k$, we have found an additional prime beyond $p_1,\ldots,p_k$.
Any square root of $2$ is irrational. More explicitly, for all $a,b\in\mathbb{Z}$ with $b\neq 0$, $(a/b)^2 \neq 2$.
Let's continue avoiding division, and prove the equivalent statement that $a^2 \neq 2b^2$ for all $a,b\in\mathbb{Z}$ with $b\neq 0$. (Note that the assumption that $b$ is non-zero is necessary, since otherwise we can take $a=0$ and $b=0$.)
Now suppose for the purpose of contradiction that there exist $a,b\in\mathbb{Z}$, $b\neq 0$, such that $a^2 = 2 b^2$. By negating $b$ if necessary, we may assume that $b$ is positive. Since there exists at least one such pair with positive $b$, we can select a pair that minimizes the value of $b$. We are going to contradict our choice of $a,b$ by showing that we can find another pair with a smaller $b$.
Since $a^2=2b^2$, we have that $2|a^2$. Since $2$ is prime, we have that $2|a$ by Euclid's Lemma from the last lecture. Thus there exists $k\in\mathbb{Z}$ such that $a=2k$.
Next we have that $2b^2 = a^2 = (2k)^2 = 4k^2$. Thus $b^2 = 2k^2$. We can repeat the argument from the previous paragraph to show that $2|b$ as well, so there is $\ell\in\mathbb{Z}$ such that $b=2\ell$.
Since $a^2=2b^2$, we have that $(2k)^2=2(2\ell)^2$, which implies that $k,\ell$ is another pair satisfying $k^2=2\ell^2$. Moreover $\ell$ is positive and $\ell \lt b$, which contradicts our choice of $a$ and $b$. This completes the proof.