Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

3.1: Basics and the FTA

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    \(\newcommand{\NN}{\mathbb N}\)
    \(\newcommand{\RR}{\mathbb R}\)
    \(\newcommand{\QQ}{\mathbb Q}\)
    \(\newcommand{\ZZ}{\mathbb Z}\)
    \(\newcommand{\Cc}{\mathcal C}\)
    \(\newcommand{\Dd}{\mathcal D}\)
    \(\newcommand{\Ee}{\mathcal E}\)
    \(\newcommand{\Ff}{\mathcal F}\)
    \(\newcommand{\Kk}{\mathcal K}\)
    \(\newcommand{\Mm}{\mathcal M}\)
    \(\newcommand{\Pp}{\mathcal P}\)

    First of all, we make the following definition.

    Definition: Prime

    We say \(p\in\NN\) is prime if \(p>1\) and the only natural numbers which divide \(p\) are \(1\) and \(p\).

    Example \(\PageIndex{1}\)

    Some primes are \(2\), \(3\), \(5\), \(7\), \(11\), \(13\), and \(17\). Notice \(2\) is the only even prime (clearly – any other would be a multiple of \(2\) and hence could not be prime), and has some unusual properties – the joke is that “\(2\) is the oddest prime.”

    The largest prime known to humans at the time of this writing is \[2^{57,885,161}-1\] which was proven to be prime in January of 2013 by a distributed computer program called GIMPS [the Great Internet Mersenne Prime Search] running on hundreds of machines across the Internet. In contrast, we also use the following term

    Definition: Composite

    A number \(c\in\NN\) which is greater than \(1\) and not prime is called composite.

    How far does a naive, brute-force check have to go in order to see if a number is composite?

    Theorem \(\PageIndex{1}\)

    If \(n\) is a composite, then it has a positive divisor \(d\) satisfying \(d\le\sqrt{n}\).


    Suppose \(n\) is composite. Then it has some divisor \(a\in\NN\). Notice that \(n=a\cdot\dfrac{n}{a}\), so \(\dfrac{n}{a}\in\NN\) is also a divisor. But \(a\) and \(\dfrac{n}{a}\) cannot both be less than \(\sqrt{n}\), because if they were we would have \[n=a\cdot\dfrac{n}{a}<\sqrt{n}\cdot\sqrt{n}=n\] which would be a contradiction. Hence either \(a\) or \(\dfrac{n}{a}\) is the divisor \(d\) promised by the theorem statement.

    Euclid’s Lemma (Lemma 2.1.1) takes a particularly nice form if the divisor involved is prime:

    Proposition \(\PageIndex{1}\)

    Suppose \(p\) is a prime and \(a,b\in\ZZ\). If \(p\mid ab\) then \(p\mid\) or \(p\mid b\).


    Notice that \(\gcd(p,a)\mid p\), therefore \(\gcd(p,a)\) is either \(1\) or \(p\) since \(p\) is prime. But also \(\gcd(a,p)\mid a\), so either \(p\mid a\) or \(\gcd(a,p)=1\). If \(p\mid a\), we are done. If not, since therefore \(\gcd(a,p)=1\), Euclid’s Lemma 2.1.1 tells us that \(p\mid b\).

    A more general form of this is

    Corollary \(\PageIndex{1}\)

    Suppose \(p\) is a prime, \(k\in\NN\), and \(a_1,\dots,a_k\in\ZZ\). Then if \(p\mid a_1\dots a_k\), it follows that \(p\) divides at least one of the \(a_j\).


    Left to the reader (use induction on \(k\)).

    This leads to the aptly named

    Theorem \(\PageIndex{2}\): The Fundamental Theorem of Arithmetic

    Let \(n\in\NN\), \(n\ge2\). Then \(\exists k\in\NN\) and primes \(p_1,\dots,p_k\) such that \(n=p_1\dots p_k\). Furthermore, if \(l\in\NN\) and \(q_1,\dots,q_l\) are also primes such that \(n=q_1\dots q_l\), then \(l=k\) and the factorization in terms of the \(q\)’s is merely a reordering of that in term s of the \(p\)’s.


    We use the Second Principle of Mathematical induction for the existence part. The general statement we are proving is \(\forall n\in\ZZ,\ n>1\Rightarrow S(n)\), where \(S(n)\) is the statement “\(\exists k\in\NN\) and primes \(p_1,\dots,p_k\) such that \(n=p_1\dots p_k\).”

    As the base case, say \(n=2\). Then \(k=1\) and \(p_1=2\) works.

    Now assume \(S(k)\) is true for all \(k<n\). If \(n\) is prime, then \(k=1\) and \(p_1=n\) works. So suppose \(n\) is instead composite, with divisor \(d\neq1,n\). Then both \(d,\dfrac{n}{d}<n\), so by the inductive hypothesis \(\exists k,k^\prime\in\NN\) and primes \(p_1,\dots,p_k,p^\prime_1,\dots,p^\prime_{k^\prime}\) such that \(d=p_1\dots p_k\) and \(\dfrac{n}{d}=p^\prime_1\dots p^\prime_{k^\prime}\). So then \(n\) is the product \[n=p_1\dots p_k\cdot p^\prime_1\dots p^\prime_{k^\prime}\] of the \(k+k^\prime\) primes. Hence \(S(n)\) is true as well, and so prime factorizations always exist.

    Now suppose \(n\in\ZZ\) satisfies \(n>1\) and \(\exists k,l\in\NN\) and both primes \(p_1,\dots p_k\) and \(q_1,\dots,q_l\) such that \[p_1\dots p_k = n = q_1\dots q_l\ .\] Certainly \(p_1\) divides the left hand side of these dual expressions for \(n\). Then by Corollary 3.1.1, \(p_1\) divides one of the \(q_j\), which means it must be that \(p_1=q_j\) since they are prime. Removing the \(p_1\) from the left and the \(q_j\) from the right, we get \[p_2\dots p_k = n = q_1\dots q_{j-1}\cdot q_{j+1}\dots q_l\ .\] Continuing in this way, either we get the uniqueness statement in the theorem, or we run out of \(p\)’s or \(q\)’s. However, we cannot run of of primes on one side before the other, because that would make a product of primes on one side equal to \(1\), which is impossible.

    Exercise \(\PageIndex{1}\)

    1. Provide all the details of the proof of Corollary 3.1.1.
    2. State and prove a theorem about the prime factorizations of numbers \(a,b\in\NN\) and of their gcd.
    3. A number \(n\in\ZZ,\ n>1\) is called square-free if it is not divisible by the square of any natural number other than \(1\). Prove that an \(n\in\ZZ,\ n\ge1\) is square-free if and only if it is the product of distinct primes.