Prime Decomposition of Integer is Unique

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Theorem

Let $n$ be an integer such that $n > 1$.


Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.


In the words of Euclid:

If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.

(The Elements: Book $\text{IX}$: Proposition $14$)


Proof 1

Aiming for a contradiction, suppose the supposition false.

That is, suppose there is at least one positive integer that can be expressed in more than one way as a product of primes.

Let the smallest of these be $m$.

Thus:

$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$

where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime.

By definition, $m$ is not itself prime.

Therefore:

$r, s \ge 2$

Let us arrange that the primes which compose $m$ are in order of size:

$p_1 \le p_2 \le \dots \le p_r$

and:

$q_1 \le q_2 \le \dots \le q_s$

Let us arrange that $p_1 \le q_1$.

Suppose $p_1 = q_1$.

Then:

$\dfrac m {p_1} = p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s = \dfrac m {q_1}$

But then we have the positive integer $\dfrac m {p_1}$ being expressible in two different ways.

This contradicts the fact that $m$ is the smallest positive integer that can be so expressed.

Therefore:

$p_1 \ne q_1 \implies p_1 < q_1 \implies p_1 < q_2, q_3, \ldots, q_s$

as we arranged them in order.

From Prime not Divisor implies Coprime:

$1 < p_1 < q_j: 1 < j < s \implies p_1 \nmid q_j$

But:

$p_1 \mathrel \backslash m \implies p_1 \mathrel \backslash q_1 q_2 \ldots q_s$

where $\backslash$ denotes divisibility.

Thus from Euclid's Lemma for Prime Divisors:

$\exists j: 1 \le j \le s: p_1 \mathrel \backslash q_j$

But $q_j$ was supposed to be a prime.

This is a contradiction.

Hence, by Proof by Contradiction, the supposition was false.

$\blacksquare$


Proof 2

Suppose $n$ has two prime factorizations:

$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$

where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.

Since $p_1 \mathrel \backslash q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 = q_j$ for some $1 \le j \le s$.

Thus:

$p_1 \ge q_1$

Similarly, since $q_1 \mathrel \backslash p_1 p_2 \dots p_r$, from Euclid's Lemma for Prime Divisors:

$q_1 \ge p_1$

Thus, $p_1 = q_1$, so we may cancel these common factors, which gives:

$p_2 p_3 \cdots p_r = q_2 q_3 \dots q_s$


This process is repeated to show that:

$p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r$

If $r < s$, we arrive at $1 = q_{r+1} q_{r+2} \cdots q_s$ after canceling all common factors.

But by Divisors of One, the only divisors $1$ are $1$ and $-1$.

Hence $q_{r+1}, q_{r+2}, \ldots, q_s$ cannot be prime numbers

From that contradiction it follows that $r = s$.

Thus:

$p_1 = q_1, p_2 = q_2, \ldots, p_r = q_s$

which means the two factorizations are identical.

Therefore, the prime factorizations of $n$ is unique.

$\blacksquare$


Also see


Historical Note

This theorem is Proposition $14$ of Book $\text{IX}$ of Euclid's The Elements.


Sources