SharedGroup Assignment 4.texOpen in CoCalc
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\title{Group Assignment 4}
\author{Mariana Sullivan, Jerry Law, and Joyce Connolly}

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Everyone worked on this problem and we did not talk to anyone but our own group members and Professor Sandefur, or use any online resources other than our text.

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2.3.12.

Given $j \in \mathbb{Z}^+ - \{1\}$.  Suppose that for every $n, m \in \mathbb{Z}^+$, if j divides $nm$ then $j$ divides $n$ or $j$ divides $m$. Then $j$ is prime.  With

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$p(j) \equiv \{ \forall n, m \in \mathbb{Z}^+$, if j divides $nm$, $j$ divides $n$ $\vee$ $j$ divides $m$ $\}$

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and $q(j) \equiv \{$j is prime$\}$, this can be written as an implication $p(j) \Rightarrow q(j)$ where statement $p(j)$ is an implication, that is, $p(j) \equiv \{r \Rightarrow s\}$.

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{\bf a)}  Let $j=3$, which is prime.  Since $\frac{nm}{j} \in \mathbb{Z}$, without loss of generality, we can say $n=3$ and $m=1$.  j does divide n but not m because $\frac{n}{j} \in \mathbb{Z}$ but $\frac{m}{j} \notin \mathbb{Z}$.  j does divide n or m.  For another example, let $j=5$, which is prime.  Since $\frac{nm}{j} \in \mathbb{Z}$, without loss of generality, we can say $n=1$ and $m=5$.  j does not divide n but does divide m because $\frac{n}{j} \notin \mathbb{Z}$ and $\frac{m}{j} \in \mathbb{Z}$.  Again, j does divide n or m.  This makes us believe that the implication $p \equiv \{r \Rightarrow s \}$ is true when j is prime.

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Let $j=4$, which is composite.  Since $\frac{nm}{j} \in \mathbb{Z}$, we can say $n=2$ and $m=2$.  j does not divide n and does not divide m because $\frac{n}{j} \notin \mathbb{Z}$ and $\frac{m}{j} \notin \mathbb{Z}$.  j does not divide n or m.  This shows that the implication $p \equiv \{r \Rightarrow s \}$ is not true when j is composite.

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{\bf b)} $p \equiv \{\forall$ n, m $\in \mathbb{Z}^+$ st $j \in \mathbb{Z}^+ - \{1\}$ divides $nm$, j divides n $\vee$ j divides m$\}$

$p \equiv \{\forall$ n, m $\in \mathbb{Z}^+$ st $j \in \mathbb{Z}^+ - \{1\}$ divides $nm$, $\frac{n}{j} \in \mathbb{Z}^+$ $\vee$ $\frac{m}{j} \in \mathbb{Z}^+ \}$

$\neg p \equiv \{{\exists n, m \in \mathbb{Z}^+}$ where $j \in \mathbb{Z}^+ - \{1\}$ divides $nm$ st $\frac{n}{j} \notin \mathbb{Z}^+$ and $\frac{m}{j} \notin \mathbb{Z}^+ \}$

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{\bf c)} $q(j) \equiv \{$j is prime$\}$

$\neg q \equiv \{$j is composite$\}$

$\neg q \equiv \{\exists a,b \in \mathbb{Z}^+$ such that $ab=j$ and $1<a,b<j$

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{\bf d) Proof:} We will find the contrapositive of $p \Rightarrow q$ and prove that it is true, ultimately proving that $p \Rightarrow q$ is true. The contrapositive of $p \Rightarrow q$ is $\lnot q \Rightarrow \lnot p$. 

$\forall j \in \mathbb{Z}^+ - \{1\}$ s.t. $\exists a,b \in \mathbb{Z}^+$ s.t. $ab=j$ and $1<a,b<j, {\exists n, m \in \mathbb{Z}^+}$ where $j$ divides $nm$ and $\frac{n}{j} \notin \mathbb{Z}^+$ and $\frac{m}{j} \notin \mathbb{Z}^+ \}$

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Set $m_0 = a_0$

Set $n_0 = b_0$

Because $a_0 b_0 =n_0 a_0$=j, j divides $nm$.

For any $\frac {a_0}{j}$ to be $\in \mathbb{Z}^+$, $a_0 \geq j$.  It is given that $a_0 < j$.  Therefore, $\frac {a_0}{j} \notin \mathbb{Z}^+$ and j does not divide $a_0$.

For any $\frac {b_0}{j}$ to be $\in \mathbb{Z}^+$, $b_0 \geq j$. It is given that $b_0 < j$. Therefore, $\frac{b_0}{j} \notin \mathbb{Z}^+$ and j does not divide $b_0$

Thus, the contradiction statement is true, so $p \Rightarrow q$ is true.

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