Disprove by Counter Example or Prove each of the following :

1. $\forall a,b,c \in Z, a \vert b \wedge a \vert c \rightarrow a\vert bc$
   show: when $a\vert b \wedge a\vert c$ it must be true that $\exists k \in Z, ak=bc$
   proof:
   $a \vert b \rightarrow \exists m \in Z, am = b$
   $am = b$
   $(am)c = bc$
   $a(mc) = bc$ where $mc \in Z$
   THEREFORE $\exists k \in Z, ak=bc$
   THEREFORE $a \vert bc$
   
2. $\forall a,b,c \in Z, a \not \vert b \wedge a \not \vert c \rightarrow a \not \vert bc$
   False by Counter Example: $a = 4$ and $b = 10$ and $c=14$
   $a \not \vert b$ means $4 \not \vert 10$ this is true
   $a \not \vert c$ means $4 \not \vert 14$ this is true
   $a \not \vert bc$ means $4 \not \vert (10*14)$ means $4 \not \vert 140$ this is false
   
3. $\forall q,r \in Q, qr \in Q$
   show: $\exists a,b \in Z, qr = \frac{a}{b}$ where $b \neq 0$
   proof:
   $q \in Q$ means $\exists c,d \in Z, d \neq 0, q = \frac{c}{d}$
   $r \in Q$ means $\exists e,f \in Z, f \neq 0, r = \frac{e}{f}$
   
   $qr = (\frac{c}{d})(\frac{e}{f})$
   $qr = \frac{ce}{df}$
   $df \neq 0$
   $cd \in Z$
   $df \in Z$
   THEREFORE $qr \in Q$
   
4. $\forall a \in Q, a \not \in Z \rightarrow (\lfloor a \rfloor \neq \lceil a \rceil)$
   proof:
   $\lfloor a \rfloor \rightarrow \exists r \in Q, 0<r<1, \lfloor a \rfloor = a-r$
   $\lceil a \rceil \rightarrow \exists s \in Q, 0<s<1, \lceil a \rceil = a+s$
   ASSUME: $\lfloor a \rfloor = \lceil a \rceil$
   $a - r = a + s$ $-r = s$ where $0<r<1$ and $0<s<1$
   $\rightarrow \leftarrow$
   THEREFORE: $\lfloor a \rfloor \neq \lceil a \rceil$
   
5. $\forall a \in Q, \exists q \in Z, aq \in Z$
   proof:
   $a \in Q \rightarrow \exists c,d \in Z, d \neq 0, a = \frac{c}{d}$
   ASSUME: $\forall q \in Z, \frac{c}{d}q \not \in Z$
   let $q = d$
   $\frac{c}{d} q = \frac{c}{d} d = c$ where $c \in Z$
   $\rightarrow$$\leftarrow$
   THEREFORE: $\forall a \in Q, \exists q \in Z, aq \in Z$
6. $\forall a,b,c \in Z, a \vert b \vee a \vert c \rightarrow a\vert bc$
   show: $\exists k \in Z, ak=bc$
   proof:
   
   case 1:
   $a \vert b \rightarrow a\vert bc$
   $a \vert b \rightarrow \exists m \in Z, am = b$ $am = b$
   $(am)c = bc$
   $a(mc) = bc$ where $mc \in Z$
   THEREFORE: $a \vert bc$
   
   case 2:
   $a \vert c \rightarrow a\vert bc$
   $a \vert c \rightarrow \exists n \in Z, an = c$ $an = c$
   $(an)b = bc$
   $a(nb) = bc$ where $nb \in Z$
   THEREFORE: $a \vert bc$
   
   THEREFORE: $a\vert b \vee a\vert c \rightarrow a\vert bc$
7. $\forall a,b \in Z, a \not \vert b^2 \rightarrow a \not \vert b$
   proof:
   CONTRAPOSITIVE: $\forall a,b \in Z, a\vert b \rightarrow a\vert b^2$
   $a\vert b$ means $\exists k \in Z, ak = b$
   $ak = b$
   $(ak)^2 = b^2$
   $a^2k^2 = b^2$
   $a(ak^2) = b^2$ where $ak^2 \in Z$
   THEREFORE: $a\vert b^2$
   

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