The Integers include all natural numbers, their additive inverses, and zero. We use the symbol $\mathbb{Z}$ to refer to the set of integers.
Properties
- The integers are closed under addition, subtraction, and multiplication.
- Addition of integers is both associative & commutative.
- Multiplication of integers is both associative & commutative, and distributes over addition.
- Every integer has an additive inverse, i.e., an integer you can add to it to get zero.
Other Properties from Our Investigations
- For $a, b, c \in \mathbb{Z}$, if $a \mid b$ and $b \mid c$ then $a \mid c$
Proof: Since $a \mid b$ and $b \mid c$ we have $a(d_1) = b$ and $b(d_2) = c$ for $d_1,d_2 \in \mathbb{Z}$. Then by substitution we get $(a(d_1))d_2 = c$, which shows that $c$ is a multiple of $a$. In other words $a \mid c$ as desired! ${}_\square$
- For $a, b, c \in \mathbb{Z}$ if $a \mid b$ and $a \mid c$ then $a \mid (b+c)$.
Proof: From our definition of divides we have $a(n_1) = b,\ a(n_2) = c$ where $n_1, n_2 \in \mathbb{N}$. Let's add these quantities together:
(1)Since $(n_1+n_2)$ is an integer, we have shown that $b+c$ is a multiple of $a$. In other words, $a \mid (b+c)$ as desired! ${}_\square$
- For $a, b, q, r, k \in \mathbb{Z}$, if $a = bq + r$ and $k \mid a, k \mid b$ then $k \mid r$.
Proof: From our definition of divides we have that $kd_1 = a, kd_2 = b$ where $d_1, d_2 \in \mathbb{Z}$. Then by substitution we get
(2)and since $(d_1 - d_2q) \in \mathbb{Z}$ that gives us that $k \mid r$ as desired. ${}_\square$
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- Two consecutive integers are relatively prime (i.e., their GCD is one).
Proof: Let $n \in \mathbb{Z}$, then $n, n+1$ are consecutive integers. Suppose $k \in \mathbb{N}$ is a common divisor for $n, n+1$; we wish to show that $k = 1$.
Since $k$ is a common divisor, that means we have $d_1, d_2 \in \mathbb{Z}$ so that $kd_1 = n, kd_2 = n + 1$. By substitution we have
(3)so that $k \mid 1$. But the only natural number that divides one is $1$, so it must be the case that $k = 1$, which means that $n, n+1$ will always be relatively prime! ${}_\square$
- If $a,b,c \in \mathbb{Z}$, $a|bc$, and $\gcd(a,b)=1$, then $a|c$.
Proof: Let $a,b,c \in \mathbb{Z}$, $a|bc$, and $\gcd(a,b)=1$.
By definition, $a|bc$ means $a \cdot n_1 = bc$ where $n_1 \in \mathbb{Z}$.
If the $\gcd(a,b)=1$, then $ax+by=1$ where $x,y \in \mathbb{Z}$.
by substitution of $bc$ we get,
(5)Since addition and multiplication are closed in the $\mathbb{Z}$, we can let $(cx+n_{1}y) = n_2$ where $n_2 \in \mathbb{Z}$.
So, $a(n_2) = c$.
$\therefore a|c$
- For $a, b, c \in \mathbb{Z}$, if $c \mid a$ and $c \mid b$ then $c \mid (ax + by)$ for all $x,y \in \mathbb{Z}$.
Proof: From our definition of divides we have that $cn_1 = a, cn_2 = b$ where $n_1,n_2 \in \mathbb{Z}$. We would like to show that $ax + by$ is a multiple of $c$, so let's multiply $a, b$ by some arbitrary $x, y$, respectively:
(6)Now let's add these two together and see what we can say:
(7)And since $(n_1 + n_2)$ is an integer, we have shown that $c \mid (ax + by$ for any integers $x, y$! ${}_\square$