Def/Canonical decomposition into primes

Logical Connections

This definition logically relies on the following definitions and statements: State/Natural numbers can be factored into primes, State/Uniqueness of prime factorization, Def/Absolute value, State/Rational numbers can be expressed in lowest terms

The following statements and definitions logically rely on the material of this page: State/Canonical decompositions can be used to find GCD and LCM, State/Canonical decompositions of binomial coefficients, State/Canonical decompositions of factorials, State/Divisibility corresponds to inequalities of prime exponents, and State/Rational roots of integers are integers

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One may also refer to the canonical decomposition of nonzero integers into primes. Namely, if $z \in \ZZ$, and $z \neq 0$, then there exists a unique natural number $n$ and a unique element $e_{-1} \in \{0,1 \}$, such that: $$z = (-1)^{e_{-1}} n.$$ Namely, $e_{-1}$ equals $0$ or $1$, depending on whether $z>0$ or $z < 0$, respectively. The unique natural number $n$ may be defined simply as the absolute value of $z$.

Using the canonical decomposition of $n$, one obtains a canonical decomposition of $z$: $$z = (-1)^{e_{-1}} \prod_{p \in P} p^{e_p},$$ where,

• $e_{-1} \in \{0,1 \}$. $e_p \in \NN$.
• $e_p = 0$ for all sufficiently large prime numbers $p$ (so that the above product is a finite product).
• The natural numbers $e_p$, and the exponent $e_{-1} \in \{0,1 \}$ are uniquely determined by $z$.

One may also refer to the canonical decomposition of nonzero rational numbers into primes. Namely, if $q \in \QQ$, and $q \neq 0$, then there exists a unique ordered pair of natural numbers $(n,d)$ and a unique element $e_{-1} \in \{0,1 \}$, such that:

• $q = (-1)^{e_{-1}} \frac{n}{d}$.
• $GCD(n,d) = 1$. (since rational numbers can be expressed in lowest terms).

Consider the canonical decompositions of $n$ and $d$; there exist unique natural numbers $f_p$ and $g_p$ for every prime number, such that: $$n = \prod_{p \in P} p^{f_p}, \mbox{ and } d = \prod_{p \in P} p^{g_p}.$$ Furthermore, since $GCD(n,d) = 1$, we find that, for all prime numbers $p$:

• $f_p \neq 0 \Rightarrow g_p = 0$, and
• $g_p \neq 0 \Rightarrow f_p = 0$.

Indeed, if $f_p \neq 0$, then $p$ divides $n$, and hence cannot divide $d$, and thus $g_p = 0$. Similarly, if $g_p \neq 0$, then $p$ divides $d$, and hence cannot divide $n$, and thus $f_p = 0$.

Therefore, we arrive at a canonical decomposition of $q$: $$q = (-1)^{e_{-1}} \prod_{p \in P} p^{e_p},$$ where:

• $e_p = f_p - g_p \in \ZZ$. $e_{-1} \in \{0,1 \}$.
• $e_p = 0$ for all sufficiently large prime numbers $p$.
• The integers $e_p$ and the exponent $e_{-1} \in \{0,1 \}$ are uniquely determined by $q$.