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Let $L:{\Bbb N} \to {\Bbb N}$ such that $L(n) = {\text{least prime factor p of n}}$. Let $g:{\Bbb N} \to {\Bbb N}$ such that $g(n) = {\text{biggest positive integer d such that d|n and 1}} \leqslant {\text{d < n}}$.

Show that:$$g(n) = \frac{n}{{L(n)}},\forall n \in {\Bbb N}{\text{ such that }}n \geqslant 2$$

My proof:

Since $n \geqslant 2$, $n$ is either composite or prime.

If $n$ is prime(i.e $n = p$ where $p$ is prime), then $$g(n) = 1 = \frac{p}{p} = \frac{n}{{L(n)}}$$ This is because $L(n)$ is defined as least prime factor of n.

If$n $is composite, by the Fundamental Theorem of Arithmetic: "Any positive integer bigger than 1 can be expressed as a product of primes." Let $p$ be the smallest prime of the product $$n = {p_1}^{{\alpha _1}}p_2^{{\alpha _2}}...{p_m}^{{\alpha _m}},{\text{ }}i < j,{\text{ }}{p_i} < {p_j}{\text{ and }}i,j \in {{\Bbb Z}^ + }{\text{ and }}{\alpha _i} \in {{\Bbb Z}^{ \geqslant 0}}$$

That is, pick $p = p_1$. Notice that $p$ is necessarily $L(n)$ because since $p = p_1$ implies that $p|n$ and is the smallest prime that divides n. Therefore, $$p_1 = p = L(n)$$

Then, if $x = p = L(n)$, then $1 < x < n$ and $x|n$ (by definition). This implies that $$\exists y \in {{\Bbb Z}^ + }:y = \frac{n}{x} = \frac{n}{{L(n)}} = {p_1}^{{\alpha _1} - 1}{p_2}^{{\alpha _2}}...{p_m}^{{\alpha _m}}$$ But $y|n$ such that $y < n$. Implies that $y$ is the greatest integer less than$n $such that $y|n$.

Now let $d = g(n)$. Since $$\frac{n}{d} = \frac{n}{{g(n)}} = m$$ then $m$ is prime and $m = p$.

BWOC, If $n=md$ and $m = ab$ where $1 < a,b < m \Rightarrow n = abd \Rightarrow ad|n$. Since $1. But we found another factor of $n$ ($ad$) that divides $n$! This contradicts the definition of $d=g(n)$. Also, since $d = g(n)$ is the biggest factor of n, implies that $m = \frac{n}{d} = \frac{n}{g(n)}$ is the smallest prime factor of n. So $m = L(n) = p$. Hence, we have that $$x = m = p = L(n) = \frac{n}{g(n)} \Rightarrow g(n) = \frac{n}{p} =\frac{n}{L(n)} = y$$ or just $$g(n) = \frac{n}{L(n)}$$

Q.E.D.