A proof of the infinitude of prime numbers

Assume—for the sake of contradiction—that the elements in the set of all prime numbers are finite, and thus define the following sets, Here, ${\mathbb{N}}^{*}\setminus \mathbb{P}\subset {\mathbb{N}}^{*}$ and $\mathbb{C}\subseteq {\mathbb{N}}^{*}\setminus \mathbb{P}$. To show that ${\mathbb{N}}^{*}\setminus \mathbb{P}\ne \mathbb{C}$, notice that $\mathbb{C}$ is the set of all natural numbers of the form $np$, where $n\in {\mathbb{N}}^{*}$ and $p\in \{p_1,p_2,p_3,\cdots ,p_k\}$; whereas ${\mathbb{N}}^{*}\setminus \mathbb{P}$ includes all natural numbers of the form $nk$, where $n,k\in {\mathbb{N}}^{*}$. It will now be shown that for any sufficiently large range $\epsilon \in {\mathbb{N}}^{*}$, the total amount of natural numbers of the form $nk\le \epsilon$ is greater than the total amount of natural numbers of the form $np\le \epsilon$. By doing so, it would be concludable that $|{\mathbb{N}}^{*}\setminus \mathbb{P}|\ne |\mathbb{C}|$, for any sufficiently large range $\epsilon \in {\mathbb{N}}^{*}$ and thus the set ${\mathbb{N}}^{*}\setminus \mathbb{P}$ always contains more elements than $\mathbb{C}$, up to that range $\epsilon \in {\mathbb{N}}^{*}$. Which would then be used to form a contradiction.

Now, since $\mathbb{C}$ is the set of all natural numbers of the form $np$, where $n\in {\mathbb{N}}^{*}$ and $p\in \{p_1,p_2,p_3,\cdots ,p_k\}$. Therefore, for the ith prime $p_i\in \{p_1,p_2,p_3,\cdots ,p_k\}$, the natural numbers of the form $n{p}_i$, where $n\in {\mathbb{N}}^{*}$, are as follows, Now, the number of natural numbers of the form $n{p}_i$, where $n\in {\mathbb{N}}^{*}$, less than or equal to a large range $\epsilon \in {\mathbb{N}}^{*}$ yields, where, $c_i{p}_i\le \epsilon$ or $c_i\le {\displaystyle \frac{\epsilon }{p_i}}$. Since $n\ge 2$ in the expression $n{p}_i$, therefore $(c_i-1)$ in the expression $c_i{p}_i$ represents a count of natural numbers of the form $n{p}_i$ less than or equal to the given range $\epsilon$. Now, to count the number of natural numbers of the general form of $np$ less than or equal to the given range $\epsilon$, the following is done. Here, $p_n$ is such a prime number for which $n{p}_n\le \epsilon$ and $n{p}_{n+1}\neg \le \epsilon$, for all $n\in {\mathbb{N}}^{*}$. If $\epsilon$ is large enough, then $n{p}_n=n{p}_k\le \epsilon$, where $p_k=max\left(\mathbb{P}\right)$. This implies that the amount of natural numbers of the form $n{p}_1$, $n{p}_2$, $n{p}_3$, ⋯, $n{p}_n$ less than or equal to $\epsilon$ are $(c_1-1),(c_2-1),(c_3-1),\cdots ,(c_n-1)$, respectively. Now, summing these quantities results in the total amount of natural numbers of the general form $np$, less than or equal to $\epsilon$. Let such a total amount be $\alpha$. Now, note that $c_i$ in the expression $c_i{p}_i$ for all $i\in \{1,2,3,\cdots ,k\}$ is greater than 1, as each n in the expression $n{p}_i$ is an element of ${\mathbb{N}}^{*}$ (i.e., greater than 1). As a result, $\alpha =\left(c_1-1\right)+\left(c_2-1\right)+\left(c_3-1\right)+\cdots +\left(c_n-1\right)>1$, as shown below. where, $\alpha$ represents the total amount of natural numbers of the form $np$, where $n\in {\mathbb{N}}^{*}$ and $p\in \{p_1,p_2,p_3,\cdots ,p_k\}$, less than or equal to $\epsilon \in {\mathbb{N}}^{*}$.

Now, on the other hand, since the set ${\mathbb{N}}^{*}\setminus \mathbb{P}$ contains all natural numbers of the form $nk$, where $n,k\in {\mathbb{N}}^{*}$. Therefore, for the ith natural number $k_i\in {\mathbb{N}}^{*}$, the natural numbers of the form $n{k}_i$, where $n\in {\mathbb{N}}^{*}$, are as follows, Now, the number of natural numbers of the form $n{k}_i$ less than or equal to a large range $\epsilon \in {\mathbb{N}}^{*}$ yields, where, $c_i{k}_i\le \epsilon$ or $c_i\le {\displaystyle \frac{\epsilon }{k_i}}$. Since $n\ge 2$ in the expression $n{k}_i$, therefore $(c_i-1)$ in the expression $c_i{k}_i$ represents a count of natural numbers of the form $n{k}_i$ less than or equal to the given range $\epsilon$. Now, to count the amount of natural numbers of the general form $nk$ less than or equal to the given range $\epsilon$, the following is done. Here, $k_n$ is such a natural number for which $n{k}_n\le \epsilon$ and $n{k}_{n+1}\neg \le \epsilon$, for all $n\in {\mathbb{N}}^{*}$. This implies that the amount of natural numbers of the form $n{k}_1,n{k}_2,n{k}_3,\cdots ,n{k}_n$ less than or equal to $\epsilon$ are $(c_1-1),(c_2-1),(c_3-1),\cdots ,(c_n-1)$, respectively. Now, summing these quantities results in the total amount of natural numbers of the general form $nk$, less than or equal to $\epsilon$. Let such a total quantity be $\beta$. Now, note that $c_i$ in the expression $c_i{k}_i$ for all $i\in \{1,2,3,\cdots ,n\}$ is greater than 1, as each n in the expression $nk$ is an element of ${\mathbb{N}}^{*}$ (i.e., greater than 1). As a result, $\beta =\left(c_1-1\right)+\left(c_2-1\right)+\left(c_3-1\right)+\cdots +\left(c_n-1\right)>1$, as shown below. where, $\beta$ represents the total amount of natural numbers of the form $nk$, where $n,k\in {\mathbb{N}}^{*}$, less than or equal to $\epsilon \in {\mathbb{N}}^{*}$

Now, comparing Equation (1) and (2), we get, Since, in the expression $nk$, $k\in {\mathbb{N}}^{*}$, therefore ${\displaystyle \frac{1}{k_1}}+{\displaystyle \frac{1}{k_2}}+{\displaystyle \frac{1}{k_3}}+\cdots +{\displaystyle \frac{1}{k_n}}$ represents ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}+\cdots +{\displaystyle \frac{1}{n}}$. Which is the Harmonic number $H_n$ minus 1. Therefore, ${\displaystyle {\displaystyle \frac{1}{k_1}}+{\displaystyle \frac{1}{k_2}}+{\displaystyle \frac{1}{k_3}}+\cdots +{\displaystyle \frac{1}{k_n}}=\sum _{k=1}^n{\displaystyle \frac{1}{k}}-1}$. Similarly, ${\displaystyle \frac{1}{p_1}}+{\displaystyle \frac{1}{p_2}}+{\displaystyle \frac{1}{p_3}}+\cdots +{\displaystyle \frac{1}{p_n}}$ represents the sum of the reciprocals of the prime numbers. Therefore, ${\displaystyle {\displaystyle \frac{1}{p_1}}+{\displaystyle \frac{1}{p_2}}+{\displaystyle \frac{1}{p_3}}+\cdots +{\displaystyle \frac{1}{p_n}}=\sum _{i=1}^n{\displaystyle \frac{1}{p_i}}}$. As a result, we have the following. Now, since ${\displaystyle \sum _{k=1}^n{\displaystyle \frac{1}{k}}-1\sim ln\left(n\right)}$ and ${\displaystyle \sum _{i=1}^n{\displaystyle \frac{1}{p_i}}\sim ln\left(ln\left(n\right)\right)}$[9] for sufficiently large values of the terms n obtained by choosing a sufficiently large range $\epsilon$, and $ln\left(n\right)>ln\left(ln\left(n\right)\right)$; therefore, ${\displaystyle \sum _{k=1}^n{\displaystyle \frac{1}{k}}-1>\sum _{i=1}^n{\displaystyle \frac{1}{p_i}}}$ for sufficiently large values of n. As a result, ${\displaystyle \frac{{\displaystyle \sum _{k=1}^n{\displaystyle \frac{1}{k}}-1}}{{\displaystyle \sum _{i=1}^n{\displaystyle \frac{1}{p_i}}}}}=L>1$, for sufficiently large n. This implies that

This implies that for any sufficiently large range $\epsilon \in {\mathbb{N}}^{*}$, the total amount of natural numbers of the form $np\le \epsilon$ is always less than that of the form $nk\le \epsilon$. In other words, the number of elements in ${\mathbb{N}}^{*}\setminus \mathbb{P}$ is always greater than the number of elements in $\mathbb{C}$, up to any sufficiently large range $\epsilon \in {\mathbb{N}}^{*}$. As a result, for any sufficiently large range $\epsilon$, since $\alpha <\beta$ for $\epsilon$, therefore $\exists \delta =nk\in {\mathbb{N}}^{*}\setminus \mathbb{P}$ such that $\delta \notin \mathbb{C}$. Since $\delta \notin \mathbb{C}$, therefore, $\delta \ne np$, $\forall p\in \mathbb{P}$, $\forall n\in {\mathbb{N}}^{*}$, where $p,n<\delta$. However, if $\delta \ne np$, then, $\forall p\in \mathbb{P}$, ${\displaystyle \frac{\delta }{p}}\ne n$, where $n\in {\mathbb{N}}^{*}$. This implies that $\forall p\in \mathbb{P}$, $p\nmid \delta$, therefore, by the definition of a prime number, $\delta \in \mathbb{P}$. However, this contradicts the premise that $\delta \in {\mathbb{N}}^{*}\setminus \mathbb{P}$.

Should $\delta$ be in $\mathbb{C}$ for even large range ${\epsilon }_1>\epsilon$, then, since $\alpha <\beta$ also for ${\epsilon }_1$, therefore $\exists {\delta }_1=nk\in {\mathbb{N}}^{*}\setminus \mathbb{P}$ such that ${\delta }_1\notin \mathbb{C}$ and the rest follows a similar contradiction! Thus, the argument can be repeated endlessly with every new chosen large range $\epsilon \in {\mathbb{N}}^{*}$, proving the infinitude of the prime numbers each time by contradiction. This completes the proof. □