What do you think about my first proof employing the pigeonhole principle? What mark/grade would you give me? Besides, I am curious about whether you like the style.
Theorem
Among three elements of the unit interval, there are two whose distance is less than or equal to $\frac{1}{2}$.
Proof
Let $I = [0,1]$ denote the unit interval. Let $x = (x_1, x_2, x_3) \in I^3$. The components of $x$ denote our three elements of the unit interval. It remains to prove that for two of them - the two will be denoted by $x_i$ and $x_j$, $i \ne j$ - we have $\left| x_i - x_j \right| \le \frac{1}{2}$.
For all components of $x$, the condition holds that the component is an element of $I$. We partition this condition into two cases, introdcuing the subintervals $I_1 = \left[ 0,\frac{1}{2} \right]$ and $I_2 = \left( \frac{1}{2}, 1 \right]$, as follows:
Case 1: The component is an element of $I_1$. By this theorem, a difference of elements of $I_1$ is less than or equal to $\frac{1}{2}$.
Case 2: The component is an element of $I_2$. We see that $I_2 \subset \left[ \frac{1}{2}, 1 \right]$. Hence, by the same theorem, a difference of elements of $I_2$ is less than or equal to $\frac{1}{2}$.
By the definitions of $I$, $I_1$, $I_2$, we see that the cases are collectively exhaustive; by the definitions of $I_1$, $I_2$, we see that the cases are mutually exclusive. Thus for each component, we have precisely one case. Hence, we may apply the pigeon-hole principle: We infer that there are two components, for which we have the same case. We denote these two components by $x_i$ and $x_j$ - where of course $i \ne j$.
Since for $x_i$ and $x_j$, we have the same case, $x_i$ and $x_j$ belong to the same subinterval. Hence, by the deductions that concluded the cases, introducing $d_1$ and $d_2$, we infer that
- $d_1 = x_i - x_j \le \frac{1}{2} \quad$ and
- $d_2 = x_j - x_i \le \frac{1}{2}$ .
Introducing yet another symbol $d = |x_i - x_j|$, we see that it remains to prove that $d \le \frac{1}{2}$. By the definition of absolute values, we have
$$d = d_1\quad \text{or} \quad d = d_2\text{.}$$
For each of the two possibilites, using the above inequalities, we deduce that $d \le \frac{1}{2}$. By disjunction elimination, this proof is concluded.
You can shorten it though, to this.
Note that among the two intervals $[0, 0.5]$ and $(0.5, 1]$, there must exist an interval with at least $2$ numbers. The difference between these numbers is clearly $\le 0.5$.