Introduction to $\mathbb{Z}$

• rules
  1. $\mathbb{N}\subseteq \mathbb{Z}$
     • if $n\in \mathbb{N}⟹n\in \mathbb{Z}$
  2. N</span>{0}, then −n∈Z
     • 👨‍🏫 we don't need −0 :p
• order: an extension of $<$ on N:
  • if $a,b\in Na\underset{\mathbb{Z}}{<}b⟺a<Nb$
  • $\forall a,b,\in \mathbb{N},-a<b$
  • $\forall a,b,\in \mathbb{N},-a<-b⟺b<a$
• predecessor:
  • pN​:N</span>{0}→N
  • p:Z</span>{0}→Z
  • rules
    1. ∀a∈N</span>{0}, p(a)=pN​(a)
    2. $p(0)=-1=-s(0)$
    3. ∀a∈N</span>{0}, p(−a)=−(a+1)=−s(a)
• addition can be extended similarly
  • $\forall a\in \mathbb{Z},a+0=a$
  • ∀a∈Z, ∀b∈N</span>{0}, b=s(c)⟹a+b=s(a+p(b))
  • ∀a∈Z, ∀b∈N</span>{0}, b=s(c)⟹a+(−b)=p(a+s(−b))
• multiplication
  • $\forall a\in \mathbb{Z},a\cdot 0=0$
  • ∀a∈Z, ∀b∈N</span>{0}, a⋅b=a⋅p(b)+a
  • ∀a∈Z, ∀b∈N</span>{0}, a⋅(−b)=a⋅s(−b)+(−a)
• subtraction
  • $a-b:=a+(-b)$
  • $a-(-b):=a+b$
• absolute value
  • $|0|=0$
  • $\forall n\in \mathbb{N},|n|:=n$
  • $\forall n\in \mathbb{N}\{0\},|-n|:=-n$
  • $|\cdot |:\mathbb{Z}\mapsto \mathbb{N}$

Division

• theorem: let $a,b\in \mathbb{Z},b>0$
  • there exists unique $q,r$ s.t. $a=q\cdot b+r$
    • and $0\le r<b$
• let $r=a-q\cdot b$
• and $A:=\{a-q\cdot b|q\in \mathbb{Z}\}$
  • $a\subseteq \mathbb{Z}$
  • in $\mathbb{Z}$: well-ordering principle doesn't work
  • we want $A\subseteq \mathbb{N}$
    • s.t. we can argue well ordering principle
• $A$ itself is non empty
  • $a\ge 0$, then simply: $a\in A$ when $q=0$
  • if $a<0$: use following lemma
• lemma: $\forall a,b\in \mathbb{Z},b>0⟹\exists q\in \mathbb{Z}q\cdot b>a$
  • if $a<0$: use $q=0$
  • if $a=0$: use $q=1$
  • if $a>0$: use $q=a+1$
  • thus: we can make $a-q\cdot b\in \mathbb{N}$ for arbitrary $a,b$
• let $r^{\ast }$ be minimum element of $A$ (exists due to )
  • so $\exists q^{\ast }$ s.t. $a=b\cdot q^{\ast }+r^{\ast }$
• suppose: $r^{\ast }>b$
  • $a=b\cdot (q^{\ast }+1)+(r^{\ast }-b)$
  • however, in this case, we can make a even smaller remainder
    • $r^{\ast }-b$
  • by contradiction: this implies $r^{\ast }<b$
• proof of uniqueness
  • proof by contradiction, again
  • let following be two different solutions
    • $a=q_1\cdot b+r_1$
    • $a=q_2\cdot b+r_2$
  • if $r_1=r_2$, then $q_1=q_2$
  • if $r_1\ne r_2$ and $q_1\ne q_2$:
    • then $|(q_1-q_2)\cdot b|=|r_1-r_2|$
      • lemma: $|a\cdot b|\ge b$
      • 👨‍🏫 prove yourself!
    • but $|r_1-r_2|<r$
      • thus contradiction
• theorem: for every $a,b\in \mathbb{Z}$, where $b\ne 0$:
  • division by negative number
  • there exists unique $q,r$ s.t. $a=b\cdot q+r$
    • where $0\le r<|b|$
• definition: $b|a$ if ∃q∈Z</span>{0}, a=b⋅q
• theorem 2.2: $\forall a,b,c\in \mathbb{Z}$
  1. $a|0,1|a,a|a$
  2. $a|1⟺a\in \{1,-1\}$
  3. $a|b\wedge c|d⟹ac|bd$
     • $b=ma,d=nc$
     • $bd=manc=ac(mn)$
       • thus $ac|bd$
  4. $a|b\wedge b|c⟹a|c$
     • $b=ma,c=nb$
     • $c=(nm)a$
  5. $a|b\wedge b|a⟺a=\pm b$
     • use absolute value to prove
  6. $a|b\wedge a|c⟺a|(bx+cy)$ for every $x,y$

Greatest Common Divisor

• 👨‍🏫 you expected prime numbers? no! GCD first!
• GCD: let $a,b\in \mathbb{Z}$, not BOTH are $0$
  • unless: all number might be common divisor
• $d\in \mathbb{N}$: is $\gcd (a,b)$ iff:
  • $d|a\wedge d|b$
  • $\forall d^{\prime },d^{\prime }|a\wedge d^{\prime }|b⟹d^{\prime }|d$
    • 👨‍🏫 or equivalently: $d^{\prime }\le d$
      • but it will make some of our proof unnecessary
    • more interesting :)
• for $d^{\prime }\le d$, finding one is trivial
  • find common divisor / intersection of divisors
  • and find greatest & positive one
• unlike so: we must also prove existence of $d$
• theorem: for every $a,b\in \mathbb{Z}$, not both 0
  • $\exists x,y\in \mathbb{Z}$ s.t. $\gcd (a,b)=a\cdot x+b\cdot y$
• proof: let $A:=\{d|d>0\wedge \exists x,yd=a\cdot x+b\cdot y\}$
  • $A\subseteq \mathbb{N}$
  • $A\ne \varnothing$
    • if $a=0\vee b=0$: it can be any multiple of another
    • otherwise
  • thus: $A$ has a minimum element, $d$
  • $d$: has all properties of $\gcd$
    • $d$: divides both
    • suppose $d\nmid a$
      • $a=q\cdot d+r$, $0<r<d$
      • $a=q\cdot d+r=q\cdot (ax+by)+r$
      • $r=a-q\cdot (ax+by)=r=a(1-qn)+b(-qy)$
        • which: is smaller than $d$, thus contradiction
      • finally, $d|a$
    • similarly, prove $d|b$
    • thus we have shown $d|a\wedge d|b$
  • suppose: $d^{\prime }|a\wedge d^{\prime }|b$
    • $d=ax+by$
    • $d^{\prime }|a⟹d^{\prime }|ax$
      • $d^{\prime }|b⟹d^{\prime }|by$
      • $d^{\prime }|ax+by⟹d^{\prime }|d$
  • 👨‍🏫 furthermore, $\gcd$ can be written as linear combination of $a,b$
• corollary: $\forall a,b\in \mathbb{Z}$ where not both $a,b$ 0
  • $\{ax+by|x,y\in \mathbb{Z}\}=\{q\cdot \gcd (a,b)|q\in \mathbb{Z}\}$
• relatively prime: $a\perp b$ iff $\gcd (a,b=1)$
  • $a\perp b⟺\exists x,y,ax+by=1$
• theorem: if $\gcd (a,b)=d$ then $\gcd (a/d,b/d)=1$
  • $\exists x,y,a\cdot x+b\cdot y=d$
  • $a/d\cdot x+b/d\cdot y=1$
  • the only problem: $a/d,b/d\in \mathbb{Z}$, but it's given
• theorem: if $a|c$ and $b|c$ and $a\perp b$, then $a\cdot b|c$
  • 👨‍🏫 only if we had common prime factor, this would have been easier... (but not defined yet)
  • $a\cdot x+b\cdot y=1$
  • $c=a\cdot x\cdot c+b\cdot y\cdot c$
  • $c=a\cdot x\cdot bn+b\cdot y\cdot am$
  • $c=ab(xn+ym)$
  • thus proven

• $bc=aq+r$
• $1=ax+by$
• $c=cax+cby$
• as $a|cax$, $a|bc⟹a|cby⟹a|c$