# Can a number have a prime factor that isn't a part of the number's prime factorization?

If $p|x$, after that write the prime factorization of $\frac{x}{p} = p_1p_2....p_n$. Yet after that $x=pp_1p_2...p_n$ is a prime factorization of $x$, and also one-of-a-kind factorization reveals it is the only prime factorization of $x$.

If we are operating in an one-of-a-kind factorisation domain name, after that this is difficult: if $p$ separates $x$, after that $x = p y$ for some $y$, therefore $p$ (or an associate) have to show up in the prime factorisation of $x$, which is one-of-a-kind by theory.

Actually, if we pass the rigorous definition of prime, after that if $p$ is prime and also separates $x$, it (or an associate) have to show up in *any kind of * factorisation of $x$ right into irreducibles, no matter whether the factorisation is one-of-a-kind.

Nonetheless, there are rings which are not one-of-a-kind factorisation domain names. As an example, in the ring $\mathbb{Z}[\sqrt{-5}]$, $(1 - \sqrt{-5})(1 + \sqrt{-5}) = 6 = 2 \cdot 3$, yet $2$, $3$, and also $1 \pm \sqrt{-5}$ are all irreducible and also non - associate. (Of training course, this additionally reveals that none are prime.)

This can not take place in the all-natural numbers. You claimed your self that factorizations are **one-of-a-kind **. If prime $p$ separates $x$, yet $p$ does not show up in the one-of-a-kind factorization of $x$, after that write $x = kp$. After that variable $k$.

Currently you have a **2nd ** factorization of $x$ that does include $p$ breaching your individuality presumption.

allowed $p$ divide $x$, where $x = p_1p_2\cdots p_n$ so we have $\frac {x}{p} = \frac {p_1p_2\cdots p_n}{p}=k , k \in \mathbb N$ yet if $p \neq p_1 , p \neq p_2 \cdots p \neq p_n $ that suggests among the tops $p_1p_2\cdots p_n$ was divisble by p! negating the reality that it was a prime. So no it is not feasible.

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