4.1 Example 1

Given that $i$ is an integer greater than 1, we claim the following:

$$\forall i \in I \wedge i \ge 2, \forall j \in I \wedge 1 \l... ...mes k) \rightarrow ((j \le \sqrt{i}) \vee (k \le \sqrt{i}))$$ (1)

In this proposition, $I$ is the set of all integers. In English, the proposition states that for all positive integers $i$ greater than 1, if $j$ and $k$ are matching factors of $i$, then at least one of $j$ or $k$ must be less than or equal to the square root of $i$.

The proposition seems to make sense, but how do we prove this? We can prove by contradiction. In other words, we assume that proposition 1 is not true. The negation of proposition 1 states the following:

$$\exists i \in I \wedge i \ge 2, \exists j \in I \wedge 1 \l... ...= j \times k) \wedge ((j > \sqrt{i}) \wedge (k > \sqrt{i}))$$ (2)

In other words, we are saying that there is at least an integer, say $i$, such that it has positive matching factors $j$ and $k$, and both $j$ and $k$ are greater than $\sqrt{i}$. Is this possible?

Given proposition 2, let ue examine the product of $j$ and $k$. In algebra, there is a rule stating that $x > y \wedge k > 0 \rightarrow x \times k > y \times k$. Applying this rule in our case, we have the following:

$j > \sqrt{i}$: this is given in proposition 2.

$j \times \sqrt{i} > \sqrt{i} \times \sqrt{i}$: this is derived from the inequality rule stated above.

$j \times \sqrt{i} \times \frac{k}{\sqrt{i}} > \sqrt{i} \times \sqrt{i} \times 1$: this is also using the same inequality rule. One side is multiplied by $\frac{k}{\sqrt{i}}$, while the other side is multiplied only by 1.

$j \times k > i$: this follows because simplification on the left hand side. At the same time, $i = \sqrt{i} \times \sqrt{i}$.

$i > i$: this follows because $i = j \times k$, as stated in proposition 2.

$i > i$ is a contradiction because $i = i$ is true for all numbers. However, when $i = i$, $i$ cannot be greater than $i$.

Because we arrive at this contradiction, we know that proposition 2 cannot be true. Because proposition 2 is the negation of proposition 1, proposition 1 must be true, then!