Product of Two Distinct Primes has 4 Positive Divisors

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.


Then $n$ has exactly $4$ positive divisors.


Proof

Let $n = p \times q$ where $p$ and $q$ are primes.


We have by definition of divisor:

\(\ds 1\) \(\divides\) \(\ds n\) One Divides all Integers
\(\ds p\) \(\divides\) \(\ds n\) Definition of Divisor of Integer
\(\ds q\) \(\divides\) \(\ds n\) Definition of Divisor of Integer
\(\ds p \times q\) \(\divides\) \(\ds n\) Integer Divides Itself

where $\divides$ denotes divisibility.


Suppose $a \divides n$ such that $1 \le a < n$.

Suppose $a \ne p$.

By definition of prime number:

$a \perp p$

where $\perp$ denotes coprimality.

From Euclid's Lemma:

$a \divides q$

and so by definition of prime number:

$a = q$

or:

$a = 1$

Similarly, suppose $a \ne q$.

By the same reasoning:

$a = p$

or:

$a = 1$

Thus the only positive divisors are as above.

$\blacksquare$


The validity of the material on this page is questionable.
In particular: The above is of course rubbish.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Questionable}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Product_of_Two_Distinct_Primes_has_4_Positive_Divisors&oldid=454193"