Question 19 Post Write

19. a. Using our definition of the integers $\mathbb{Z}$ as a quotient of $\mathbb{N}\times\mathbb{N}$, propose an appropriate definition of $<_{\mathbb{Z}}$.
$(m, n)<_{\mathbb{Z}}(p, q){\leftrightarrow}m+q<_{\mathbb{N}}n+p$

b. Prove that your proposal is well defined and extends the linear order $<_{\mathbb{N}}$ on the natural numbers.
The way we define integers is through ${\mathbb{N}}$. If $a, b {\in}{\mathbb{N}}, then (a, 0), (b,0){\in}{\mathbb{Z}}$. It follows that if $a<b {\in}{\mathbb{N}}, then (a, 0)<(b,0){\in}{\mathbb{Z}}$ for any two ${\mathbb{N}}$ that are < each other shows $<_{\mathbb{Z}}$.
To show that the proposal is well-defined, we want to show that if $(a, b)<(c, d)$ then $(a{\sim}, b{\sim})<(c{\sim}, d{\sim})$.
$(a, b){\sim}, (a, {\sim}, b{\sim}){\rightarrow} a+b{\sim}=b+a{\sim}$
$a+b{\sim}+(-b{\sim})=b+a{\sim}+(-b{\sim})$ add $-b{\sim}$ to both sides
$a=b+a{\sim}+(-b{\sim})$ inverse property of addition
$(c, d){\sim}, (c, {\sim}, d{\sim}){\rightarrow} c+d{\sim}=d+c{\sim}$
$c+d{\sim}+(-d{\sim})=d+c{\sim}+(-d{\sim})$ add $-d{\sim}$ to both sides
$c=d+c{\sim}+(-d{\sim})$ inverse property of addition
$(a, b)<(c, d){\rightarrow}a+d<b+c$
$a+d<b+c = b+a{\sim}+(-b{\sim})+d<b+d+c{\sim}+(-d{\sim})$ by subsituting the above
$b+a{\sim}+(-b{\sim})+d+b{\sim}+d{\sim}<b+d+c{\sim}+(-d{\sim})+b{\sim}+d{\sim}$ by adding $b{\sim}$ and $d{\sim}$ to both sides.
$b+a{\sim}+d+d{\sim}<b+d+c{\sim}+b{\sim}$ by the inverse property of addition.
$(b+d)+(a{\sim}+d{\sim})<(b+d)+(c{\sim}+b{\sim})$
$a{\sim}+d{\sim}<c{\sim}+b{\sim}$

(Prof. Simmons, 3-30-18: The logic is correct. A couple of points about the write-up: Try to use a bit more in the way of complete sentences and white space to help the reader parse the argument. For instance, it's confusing to write '$X < Y= X'<Y'$' when $X=X'$ and $Y=Y'$; instead, write '$X < Y \Longrightarrow X'<Y'$' or '$X < Y$ implies $X'<Y'$'.

LaTeX tip: use \widetilde{a} for $\widetilde{a}$. )

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