if a composite number is divisible by a prime number , the prime number must be present in the prime factrisation of that composite number? I think this should be true since prime factorisation it self means stating all the primes which divide the composite number . Please correct me if I am wrong .
 A: You are correct. One way to see this is to let the composite number be $c$ and the prime number be $p$. Then $p \mid c$ means that there exists an integer $k$ such that $c = pk$. Now, since $c$ is composite, then $k$ can't be $1$, with it actually being $\gt 1$, so it must be a prime or a composite number. By the Fundamental theorem of arithmetic, $k$ has a unique, up to order, prime factorization, e.g.,
$$k = \prod_{i=1}^{j}p_i^{r_i} \tag{1}\label{eq1A}$$
Thus, $c$ would be
$$c = p\left(\prod_{i=1}^{j}p_i^{r_i}\right) \tag{2}\label{eq2A}$$
This is a prime factorization of $c$, but as $c$ itself has a unique (up to order) prime factorization, this shows that $p$ must be in any prime factorization of $c$.
A: Here is a proof without the unicity part of the fundamental theorem of arithmetic (the existence is supposed known because of the statement).
Let $a$ an integer and $a=\prod_{i=1}^n p_i^{\alpha_i}$ a prime factorization of $a$.
Suppose $p$ is prime and $p|a$ where $p \ne p_i$ for all $i \in \{1,\dots,n\}$.
Then $p$ divises $p_1\left(p_1^{\alpha_1-1}\prod_{i=1}^n p_i^{\alpha_i}\right)$. Since $p$ and $p_1$ are different primes, by Euclid's lemma, $p$ also divises $p_1^{\alpha_1-1}\prod_{i=1}^n p_i^{\alpha_i}$. Repeat this until there is no more $p_1$ (we substract a power of $p_1$ every time) and do the same for $p_2$. Applying the same reasoning again and again will yields that $p$ divises $1$, which is not possible for a prime number. Hence a contradiction and $p$ is in the prime factorization of $a$.
Note :
I just realized that proof is just a slightly modified version of the one of unicity part in fundamental theorem of arithmetic.
