Integers

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)
\begin{align} an_1 + an_2 &= b + c\\ a(n_1 + n_2) &= b + c \end{align}

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)
\begin{align} kd_1 &= (kd_2)q + r\\ kd_1 - kd_2q &= r\\ k(d_1 - d_2q) & = r \end{align}

and since $(d_1 - d_2q) \in \mathbb{Z}$ that gives us that $k \mid r$ as desired. ${}_\square$
[[/math]]

  • 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)
\begin{align} kd_2 &= kd_1 + 1\\ kd_2 - kd_1 &= 1\\ k(d_2 - d_1) &=1 \end{align}

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 $x,y \in \mathbb{Z}$.

(4)
\begin{align} c(ax+by&=1)\\ acx+bcy&=c \end{align}

by substitution of $bc$ we get,

(5)
\begin{align} acx+(an_1)y&=c\\ a(cx+n_{1}y)&=c \end{align}

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)
\begin{align} (cn_1)x &= (a)x\\ (cn_2)y &= (b)y \end{align}

Now let's add these two together and see what we can say:

(7)
\begin{align} ax + by &= cn_1 + cn_2\\ ax + by &= c(n_1 + n_2) \end{align}

And since $(n_1 + n_2)$ is an integer, we have shown that $c \mid (ax + by$ for any integers $x, y$! ${}_\square$

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