Prove that Any positive integer can be written as a product as a product of primes in one and only one way.

 This statement is known as the Fundamental Theorem of Arithmetic. It states that every integer greater than 1 can be uniquely represented as a product of prime numbers, up to the order of the factors. Here's a proof:

Proof of the Fundamental Theorem of Arithmetic

1. Existence:

   • We need to show that any positive integer $n > 1$ can be written as a product of primes.
   • We proceed by induction on $n$.

   Base case: $n = 2$. Since 2 is a prime number, it is trivially a product of primes (itself).

   Inductive step: Assume that any integer $k$ with $2 \leq k \leq n$ can be written as a product of primes. We need to show that $n+1$ can also be written as a product of primes.

   • If $n+1$ is prime, then it is already a product of a single prime (itself).
   • If $n+1$ is not prime, then it can be written as a product of two smaller integers, say $a$ and $b$, where $2 \leq a, b < n+1$.
   • By the induction hypothesis, both $a$ and $b$ can be written as a product of primes.
   • Therefore, $n+1 = a \times b$ can be written as a product of the primes that constitute $a$ and $b$.

   Thus, by induction, any positive integer greater than 1 can be written as a product of primes.

2. Uniqueness:

   • Suppose there are two different factorizations of $n$ into primes: $$n = p_1 p_2 \cdots p_k = q_1 q_2 \cdots q_m$$ where $p_1, p_2, \ldots, p_k$ and $q_1, q_2, \ldots, q_m$ are prime numbers.
   • Assume without loss of generality that $p_1 \leq p_2 \leq \cdots \leq p_k$ and $q_1 \leq q_2 \leq \cdots \leq q_m$.

   Step 1: We show that $p_1$ must be equal to one of the $q_i$. Since $p_1$ is a prime, it must divide the product $q_1 q_2 \cdots q_m$. By the definition of a prime, $p_1$ must divide one of the $q_i$. Without loss of generality, suppose $p_1$ divides $q_1$. Since $q_1$ is prime, the only divisors of $q_1$ are 1 and $q_1$ itself. Therefore, $p_1 = q_1$.

   Step 2: Cancel $p_1$ from both sides of the equation:

   $$p_2 p_3 \cdots p_k = q_2 q_3 \cdots q_m$$

   By repeating the same argument, we can show that $p_2$ must be equal to one of the remaining $q_i$, and so on.

   Step 3: Continuing this process, we eventually show that each $p_i$ is equal to a corresponding $q_i$, proving that the original factorizations were identical except for the order of the factors.

Thus, every positive integer greater than 1 can be written as a product of primes in one and only one way, up to the order of the factors. This completes the proof of the Fundamental Theorem of Arithmetic.