Is Prime Factorisation Really Unique? Different Orderings Aren't Different Factorisations

Here is a doubt that trips up a lot of students the first time they meet the Fundamental Theorem of Arithmetic:

"You said prime factorisation is unique. But 12 = 2 \times 2 \times 3 and 12 = 3 \times 2 \times 2 are different. And 12 = 2 \times 3 \times 2 is another one. So there are three factorisations, not one. Isn't the theorem wrong?"

The confusion is real and natural. You are being told "unique" but staring at multiple-looking expressions. The fix is a small shift in what "unique" means — and once you see it, the theorem becomes crystal clear.

What "unique" actually promises

The Fundamental Theorem of Arithmetic says:

Every integer n \geq 2 can be written as a product of primes in exactly one way, up to the order of the factors.

Those five words at the end — up to the order of the factors — are doing all the work.

What "unique up to order" means: two products are counted as the same factorisation if they use the same primes with the same multiplicities, regardless of how you arrange them. So:

• 2 \times 2 \times 3
• 2 \times 3 \times 2
• 3 \times 2 \times 2

are one factorisation written three different ways, not three different factorisations. They all have two copies of 2 and one copy of 3. That is what makes them equal as factorisations.

Why we don't count orderings as different

Multiplication is commutative (ab = ba) and associative ((ab)c = a(bc)). Because of these two rules, you can rearrange and regroup the factors without changing the product. The expressions 2 \times 2 \times 3 and 3 \times 2 \times 2 are literally the same number, 12. They are not "two different factorisations equal to 12" — they are the same factorisation, notated differently.

Compare this to an equation like "write 5 as a sum of three positive integers." You could write 5 = 1 + 1 + 3 or 5 = 1 + 2 + 2. These are genuinely different — they use different numbers. But 1 + 1 + 3 and 3 + 1 + 1 and 1 + 3 + 1 are the same decomposition, just reordered.

Prime factorisation works the same way. The content (which primes, how many of each) is what matters. The arrangement (which order you write them in) is just cosmetic.

The standard form that removes the ambiguity

To make the factorisation look unique on paper (not just in spirit), write it in standard form: list the primes in increasing order, and use exponents to collect repeated primes.

• 12 = 2^2 \times 3
• 100 = 2^2 \times 5^2
• 360 = 2^3 \times 3^2 \times 5

Standard form gives you exactly one written expression per integer. The ambiguity of "which order" is gone because there is only one ordering allowed — smallest prime first.

So when a textbook says "the prime factorisation of 72 is 2^3 \times 3^2," it means the standard form. You could write it as 3^2 \times 2^3 or 2 \times 2 \times 2 \times 3 \times 3 or even 3 \times 2 \times 3 \times 2 \times 2 — all equivalent, all the same factorisation, all representing the same multiset of primes. Standard form just picks one canonical way.

Interactive: convince yourself

Why the multiset is what matters

The theorem is really saying: the multiset of prime factors of n is a property of n itself. It does not depend on how you found the factorisation, what tree you drew, or which order you wrote the primes in. Two people, working independently, will always end up with the same multiset — even if their written expressions look superficially different.

This is the uniqueness part of the theorem, and it is a strong statement. Compare it with a non-example: the number 6 can be written as 1 + 5, 2 + 4, 3 + 3 — three genuinely different ways to express 6 as a sum of two positive integers, all legitimate and none of them reducible to the others. Sums don't have the "unique decomposition" property. Products of primes do.

What the theorem rules out

The Fundamental Theorem does NOT let you do any of the following:

• Write 12 = 4 \times 3 = 6 \times 2 = 12 \times 1 and call these "three factorisations." These are not all prime factorisations — 4, 6, 12, and 1 are not prime. When you insist on primes only, 12 has exactly one decomposition (as a multiset): \{2, 2, 3\}.

• Write 12 = 2 \times 6 = 2 \times 2 \times 3 and call the two right-hand sides "different factorisations of 12." Only the fully prime-factorised form (2 \times 2 \times 3) is a prime factorisation. Anything containing composite factors is a partial factorisation, not a final one.

• Swap 2^2 \times 3 for 2 \times 6. These look similar, but 6 = 2 \times 3 is composite, so the expression 2 \times 6 is not in prime form. To get the true prime factorisation, you break 6 down further: 2 \times 6 = 2 \times 2 \times 3 = 2^2 \times 3.

The same doubt, in a different dress

Students sometimes think the theorem fails because:

• "I can pick different starting primes." (12 = 2 \times 6 vs. 12 = 3 \times 4.) But when you fully decompose both routes, you always land on 2 \times 2 \times 3.
• "I can group the primes differently." (12 = (2 \times 2) \times 3 vs. 12 = 2 \times (2 \times 3).) Grouping is associativity; it does not change the multiset.
• "I can write the primes in different orders." (2 \times 2 \times 3 vs. 3 \times 2 \times 2.) Ordering is commutativity; it does not change the multiset.

All three "different factorisations" collapse under commutativity and associativity to the same unordered collection of primes. The theorem fixes that collection.

Why this matters for bigger results

Uniqueness is not just a curious fact — it is what makes the GCD and LCM well-defined. When you compute \gcd(a, b) by taking \min of exponents at each prime, you need to know that "the exponents" are fixed numbers (independent of how you wrote the factorisation). Uniqueness guarantees that. If two people factor 60 and get \{2, 2, 3, 5\} and \{2, 3, 2, 5\}, they get the same GCD with 48 = \{2, 2, 2, 2, 3\}, because both produce the same exponent vector after sorting.

Without uniqueness, "the exponent of 2 in 60" would not be a well-defined concept — it would depend on which factorisation you chose. Uniqueness is what makes number theory possible as a systematic subject.

The one-line takeaway

"Unique" in the Fundamental Theorem means unique as a multiset of primes — unique up to the order you write the factors in. 2 \times 2 \times 3 and 3 \times 2 \times 2 are the same factorisation, dressed differently. Standard form (p_1 < p_2 < \dots with exponents) makes the uniqueness visible on paper.

Related: Number Theory Basics · Why is 1 Not Considered a Prime Number? · Divisibility Tree Explorer · Factor Clock — Divisors of n · Why √2 is Irrational