Some students understand how this works, like they know what the theorem means, but, say, imagine some student asks why, not how. Not really proving the theorem, rather why does it exist or why is it the way it is.

Is there a simple way to explain this?

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    $\begingroup$ Your statement is not the FTA, it's just multiplication of two numbers once you know their prime decomposition. Your "proof" is kind of a proof of the FTA, but it's missing some steps: how do you know that $p_1=q_1$ and so on? Lastly, you should clarify what kind of students you mean. $\endgroup$
    – Javier
    Commented Oct 1, 2016 at 14:13
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    $\begingroup$ It seems like you may have misread part of the Wikipedia article on the Fundamental Theorem of Arithmetic? The actual statement is that every natural number $n$ > 1 is a unique product of primes. A subsection under "Applications" notes that "The canonical representation, when it is known, is convenient for easily computing products..." and hence the formula in your question: en.wikipedia.org/wiki/… $\endgroup$ Commented Oct 1, 2016 at 17:27
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    $\begingroup$ Honestly, understanding how proofs work helped me understand points in linear algebra more than anything else. I suspect understanding how to prove the FTA would help me understand it far more than any other explanation. Of course, that requires understanding how proofs work. $\endgroup$ Commented Oct 1, 2016 at 21:44

2 Answers 2


The way I have explained the fundamental theorem of arithmetic in the past is by establishing what it means to be prime (has exactly two positive divisors) and then having students construct factor trees (where the prime factors at the end are circled).

The prime definition avoids some of the caveat-language otherwise seen ("a number only divisible by itself and 1, but not 1") and the factor trees are fun to do with a number like 72, after which you can ask different students to draw their trees on the board to compare and contrast.

One important observation to emerge is that when you create a factor tree for a natural number, although different numbers may appear along the way (and although some may not appear at all -- for example, the commonly used approach never reveals the factor of 1) the circled numbers at the bottom of the tree are always the same. And, in fact, this is the fundamental theorem of arithmetic:

Fix a whole number: All of its factor trees have the same numbers at the bottom.

The next approach I take is to give students a more formal definition of the fundamental theorem of arithmetic, and then ask them to think about different metaphors that capture this idea. The most common ones I have seen are DNA and fingerprints, but having them create analogies for a mathematical idea is a powerful tool for instruction and learning. Once students come to see prime numbers as something like building blocks that uniquely define each whole number, I think that the question of why it works like this can be quite successfully postponed till a much later course.

All of this is to say, I believe the why is best answered by developing one's intuition around the theorem, and that a more careful answer than that might be best staved off till a course that involves formal proof writing (at which point the existence of prime factorization is shown easily to students who can think inductively -- factor the number, if any number in its factorization is non-prime, then that means it can be written as the product of strictly smaller factors, so continue till you have all primes -- and then use something like Euclid's lemma to show uniqueness). Incidentally, the existence part of the theorem is also a very good example of when you want strong induction in your toolbox.

Lastly: You mention that some students "know what the theorem means," and so I thought I'd direct you to a nice paper that discusses this in the context of younger students (ages 11-13):

Griffiths, M. (2013). Intuiting the fundamental theorem of arithmetic. Educational Studies in Mathematics, 82(1), 75-96. Springer Link.

(You may be interested in some of its references, as well.)

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    $\begingroup$ +1 nice answer. However, I have a thing about factoring trees: I think it's a major lost opportunity to practice writing the equals sign in a non-operative context. $\endgroup$ Commented Oct 2, 2016 at 2:43
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    $\begingroup$ @DanielR.Collins Do you mean, e.g., writing something like: $72 = 6 \cdot 12 = 2 \cdot 3 \cdot 12 = 2 \cdot 3 \cdot 2 \cdot 6 = 2 \cdot 3 \cdot 2 \cdot 2 \cdot 3 = 2^3 3^2$? (Side-note: I sometimes have students compare and contrast "factor trees" and "factor pairs" and never lose the opportunity to make a "pear tree" joke! If illustrated right, it could even have a "factor rainbow" added to the scene...) $\endgroup$ Commented Oct 2, 2016 at 2:56
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    $\begingroup$ Yes, exactly (and thanks for the example). $\endgroup$ Commented Oct 2, 2016 at 3:03

To really understand why the integers $\mathbb{Z}$ have unique prime factorization, it helps to understand how unique prime factorization can fail in other settings. For example, $$(2 + \sqrt{10}) \cdot (2 - \sqrt{10}) = -6 = -2 \cdot 3,$$ so prime factorizations in $\mathbb{Z}[\sqrt{10}] = \{a + b\sqrt{10}: a, b \in \mathbb{Z}\}$ aren't unique. On the other hand, prime factorizations in $\mathbb{Z}[\sqrt{d}] = \{a + b \sqrt{d}: a, b \in \mathbb{Z}\}$ are unique for many values of $d$ (conjecturally, for infinitely many squarefree integers $d > 1$).

What's the reason for this difference? One perspective is, it's a matter of whether we have a good notion of "greatest common divisor", since this is what lets us compare any two prime factorizations of the same number and show they're the same (up to reordering and multiplication by units).

Indeed, if we have a gcd function, then we can recover the largest power of a prime $p$ dividing an integer $n$ as the largest $k$ such that $\gcd(p^k, n) = p^k$. Assuming we have a well-defined gcd function, a similar argument works in much more generality. (For those familiar with the terminology, a Noetherian commutative ring is a GCD domain if and only if it's a unique factorization domain.)

But it's easy to see that the integers have a gcd function: the set of all common divisors of two given integers is a bounded set of integers, and bounded sets of integers are finite, so we can literally pick the largest of all the common divisors with respect to the usual way of ordering the integers. (In more general contexts, any of these could fail: the set of common divisors might not be bounded, bounded subsets might not be finite, and there might not be a reasonable notion of "ordering".)

How to explain this to students depends on the level of the students. Anyway, like Javier said in the comment above, your statement and proof of the FTA aren't right, so you should probably make sure you understand those yourself before trying to explain the deeper connections to students.

  • $\begingroup$ (It would greatly improve your answer, if you could add a paragraph connecting GCD to UFD. I do not work in number theory, so I don't want to suggest anything – your intutions are problably much better than mine.) $\endgroup$
    – dtldarek
    Commented Oct 3, 2016 at 9:13
  • $\begingroup$ I was trying to avoid bringing in terminology from ring theory or algebraic number theory, but I can add a sentence or two mentioning that. $\endgroup$ Commented Oct 3, 2016 at 23:11

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