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## 1.4 Exercise 1

Omigod, this was a kilometric problem, and kind of boring too.  I guess once in a lifetime every mathematician (or schoolboy) should go ahead and get his hands dirty proving identities only using axioms.  Here goes, although I'll complete this problem five letters at a time (until I reach the end), over the span of a few days.

"Prove the following laws of algebra for $\mathbb{R}$, using only the following axioms (I)-(V):

Algebraic Properties of the Reals

I. $(x + y) + z = x + (y+z)$, $(x \cdot y) \cdot z = x \cdot (y \cdot z)$ for all $x, y, z$ in $\mathbb{R}$.

II. $x + y = y + x$,  $x \cdot y = y \cdot x$ for all $x, y$ in $\mathbb{R}$.

III. There exists a unique element of $\mathbb{R}$ called zero, denoted by 0, such that $x+0=x$ for all $x \in \mathbb{R}$. There exists a unique element of $\mathbb{R}$ called one, different from 0 and denoted by 1, such that $x \cdot 1 = x$ for all $x \in \mathbb{R}$.

IV. For each $x$ in $\mathbb{R}$, there exists a unique $y$ in $\mathbb{R}$ such that $x+y=0$.  For each $x$ in $\mathbb{R}$ different from 0, there exists a unique $y$ in $\mathbb{R}$ such that $x \cdot y = 1$.

V. $x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ for all $x, y, z \in \mathbb{R}$.

-----

(a) If $x+y = x$, then $y = 0$

(b) $0 \cdot x = 0$ [Hint: Compute $(x+0)\cdot x$]

(c) $-0 = 0$

(d) $-(-x) = x$

(e) $x(-y) = -(xy) = (-x)y$

(f) $(-1)x = -x$

(g) $x(y-z) = xy - xz$

(h) $-(x+y) = -x -y; -(x-y) = -x + y$

(i) If $x \neq 0$ and $x \cdot y = x$, then $y = 1$

(j) $x/x = 1$ if $x \neq 0$

(k) $x/1 = x$

(l) $x \neq 0$ and $y \neq 0$, then $xy \neq 0$

(m) $(1/y)(1/z) = 1/(yz)$ if $y, z \neq 0$

(n) $(x/y)(w/z) = (xw)/(yz)$ if $y, z \neq 0$

(o) $(x/y) +(w/z) = (xz + wy)/(yz)$ if $y, z \neq 0$

(p) $x \neq 0 \Rightarrow 1/x \neq 0$

(q) $1/(w/z) = z/w$ if $w, z \neq 0$

(r) $(x/y)/(w/z) = (xz)/(yw)$ if $y, w, z \neq 0$

(s) $(ax)/y = a(x/y)$ if $y \neq 0$

(t) $(-x)/y = x/(-y) = -(x/y)$ if $y \neq 0$"

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 34.)

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SOLUTION

(a)

(b)

(c)

(d)

(e)

First we show the first equality.

Now the second equality.

Finally, by transitivity of equality, since $x(-y) = -(xy)$ and $-(xy) = (-x)y$, it follows that $x(-y) = (-x)y$.

(f)

(g)

(h)

First we show the first equation.

Now the second.

(i)

(j)

(k)

(l)

Suppose otherwise, that $x y = 0$.  Then:

contradicts the fact that $y$ is nonzero.  Thus, $x y \neq 0$ holds.

(m)

(n)

With $z \neq 0, y \neq 0$,

(o)

With $y \neq 0, z \neq 0$,

(p)

Suppose otherwise, and $\frac{1}{x} = 0$.  But then:

(q)

(r)

(s)

(t)

This is true by Part (e), using compact notation.

\subsubsection[Exercise 1]{Prove the following laws of algebra'' for $\mathbb{R}$, using only the following axioms (I)-(V): \newline
\emph{Algebraic Properties of the Reals} \newline
I. $(x + y) + z = x + (y+z)$, \newline
$(x \cdot y) \cdot z = x \cdot (y \cdot z)$ for all $x, y, z$ in $\mathbb{R}$. \newline
II. $x+y = y+x$ \newline
$x \cdot y = y \cdot x$ for all $x, y$ in $\mathbb{R}$.  \newline
III. There exists a unique element of $\mathbb{R}$ called \emph{zero}, denoted by 0, such that $x+0=x$ for all $x \in \mathbb{R}$. \newline
There exists a unique element of $\mathbb{R}$ called \emph{one}, different from 0 and denoted by 1, such that $x \cdot 1 = x$ for all $x \in \mathbb{R}$. \newline
IV. For each $x$ in $\mathbb{R}$, there exists a unique $y$ in $\mathbb{R}$ such that $x+y=0$. \newline For each $x$ in $\mathbb{R}$ different from 0, there exists a unique $y$ in $\mathbb{R}$ such that $x \cdot y = 1$. \newline
V. $x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ for all $x, y, z \in \mathbb{R}$. \newline \newline
(a) If $x+y = x$, then $y = 0$ \newline
(b) $0 \cdot x = 0$ [Hint: Compute $(x+0)\cdot x$] \newline
(c) $-0 = 0$ \newline
(d) $-(-x) = x$ \newline
(e) $x(-y) = -(xy) = (-x)y$ \newline
(f) $(-1)x = -x$ \newline
(g) $x(y-z) = xy - xz$ \newline
(h) $-(x+y) = -x -y; -(x-y) = -x + y$ \newline
(i) If $x \neq 0$ and $x \cdot y = x$, then $y = 1$ \newline
(j) $x/x = 1$ if $x \neq 0$ \newline
(k) $x/1 = x$ \newline
(l) $x \neq 0$ and $x \cdot y = x$, then $y = 1$ \newline
(m) $(1/y)(1/z) = 1/(yz)$ if $y, z \neq 0$ \newline
(n) $(x/y)(w/z) = (xw)/(yz)$ if $y, z \neq 0$ \newline
(o) $(x/y) +(w/z) = (xz + wy)/(yz)$ if $y, z \neq 0$ \newline
(p) $x \neq 0 \Rightarrow 1/x \neq 0$ \newline
(q) $1/(w/z) = z/w$ if $w, z \neq 0$ \newline
(r) $(x/y)/(w/z) = (xz)/(yw)$ if $y, w, z \neq 0$ \newline
(s) $(ax)/y = a(x/y)$ if $y \neq 0$ \newline
(t) $(-x)/y = x/(-y) = -(x/y)$ if $y \neq 0$}
(a)
\newline \newline
\begin{eqnarray*}
-x + x + y & = & -x + x \verb|     |\textrm{Additive Prop. of Eq.} \\
0 + y & = & 0 \verb|     |\textrm{Axiom IV} \\
y & = & 0 \verb|     |\textrm{III}
\end{eqnarray*}  \newline \newline
(b)
\newline \newline
\begin{eqnarray*}
(x+0)x & = & x \cdot x + x \cdot 0 \verb|     |\textrm{Hint, V} \\
x \cdot x & = & x \cdot x + x \cdot 0 \verb|     |\textrm{III} \\
(-x \cdot x) + x \cdot x & = & (-x \cdot x) + x \cdot x + x \cdot 0  \verb|     |\textrm{Additive Prop. of Eq.} \\
0 & = & 0 + x \cdot 0 \verb|     |\textrm{IV} \\
0 & = & x \cdot 0 \verb|     |\textrm{III} \\
x \cdot 0 & = & 0 \verb|     | \textrm{Symmetric Prop. of Eq.}
\end{eqnarray*}  \newline \newline
(c)
\newline \newline
\begin{eqnarray*}
0 + (-0) & = & 0 \verb|     |\textrm{IV} \\
(-0) + 0 & = & 0 \verb|     |\textrm{II}  \\
-0 & = & 0 \verb|     |\textrm{III}
\end{eqnarray*}  \newline \newline
(d)
\newline \newline
\begin{eqnarray*}
x + (-x) & = & 0 \verb|     |\textrm{IV} \\
x + (-x) + -(-x) & = & 0 + -(-x) \verb|     |\textrm{Additive Prop. of Eq.}  \\
x + 0 & = & -(-x) \verb|     |\textrm{IV, III} \\
x & = & -(-x) \verb|     |\textrm{III} \\
-(-x) & = & x \verb|     |\textrm{Symmetric Prop. of Eq.}
\end{eqnarray*}  \newline \newline
(e)
\newline \newline
First we show the first equality.
\begin{eqnarray*}
x \cdot 0 & = & 0  \verb|     |\textrm{Part (b)} \\
x (y + -y) & = & 0 \verb|     |\textrm{IV}  \\
xy + x(-y) & = & 0 \verb|     |\textrm{V} \\
-(xy) + xy + x(-y) & = & -(xy) + 0 \verb|     |\textrm{Additive Prop. of Eq.} \\
0 + x(-y) & = & -(xy) \verb|     |\textrm{IV, III} \\
x(-y) & = & -(xy) \verb|     |\textrm{II, III}
\end{eqnarray*}
Now the second equality.
\begin{eqnarray*}
y \cdot 0 & = & 0  \verb|     |\textrm{Part (b)} \\
y (x + -x) & = & 0 \verb|     |\textrm{IV}  \\
yx + y(-x) & = & 0 \verb|     |\textrm{V} \\
xy + (-x)y & = & 0 \verb|     |\textrm{II} \\
-(xy) + xy + (-x)y & = & -(xy) + 0 \verb|     |\textrm{Additive Prop. of Eq.} \\
0 + (-x)y & = & -(xy) \verb|     |\textrm{IV, III} \\
(-x)y & = & -(xy) \verb|     |\textrm{II, III} \\
-(xy) & = & (-x)y \verb|     |\textrm{Symmetric Prop. of Eq.}
\end{eqnarray*}
Finally, by transitivity of equality, since $x(-y) = -(xy)$ and $-(xy) = (-x)y$, it follows that $x(-y) = (-x)y$

## 1.3 Exercise 15

Yes, after months, I'm back ;-).  This was an interesting problem to me because it explores a bit more deeply the concept of the LUBP.

"Assume that the real line has the least upper bound property.

(a) Show that the sets $[0,1] = \{x \ \vert \ 0 \leq x \leq 1\}$ and $[0,1) = \{x \ \vert \ 0 \leq x < 1\}$ have the least upper bound property.

(b) Does $[0, 1] \times [0, 1]$ in the dictionary order have the least upper bound property?  What about $[0,1] \times [0, 1)$?  What about $[0,1) \times [0,1]$?"

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)

-------

SOLUTION

(a)

Pick any nonempty $A_0 \subset [0,1]$, bounded above in $[0,1]$.  Note that $A_0 \subset [0,1] \subset \mathbb{R}$, and so $A_0 \subset \mathbb{R}$, by transitivity of inclusion (a partial order axiom).  Since $\mathbb{R}$ has the LUBP (by assumption), the set of upper bounds of $A_0$ (in $\mathbb{R}$), say $[b, \infty)$, has a least upper bound, namely $b$.   Restricted to $A = [0,1]$, the upper bounds are $[b, \infty) \cap [0, 1] = [b, 1]$, and $b$ is still the LUB in $A$.  It follows that all nonempty subsets that are bounded above in $A$ have a least upper bound in $A$, and $A$ has the LUBP.  Following the exact same recipe works with all nonempty subsets of $A' = [0,1)$ that are bounded above.  As a point of clarification, notice that the set $A_0' = A' = [0, 1)$ is not bounded above in $A'$, and so it is not an impediment to the base set $A'$ having the LUBP.

(b)

Yes,  $[0, 1] \times [0, 1]$ in the dictionary order have the least upper bound property.  Pick any nonempty subset of $A = [0, 1] \times [0, 1]$ that is bounded above, as the interval $(x_1 \times y_1, x_2 \times y_2)$, $[x_1 \times y_1, x_2 \times y_2)$, $(x_1 \times y_1, x_2 \times y_2]$, xor $[x_1 \times y_1, x_2 \times y_2]$ with $x, y \in [0,1]$ of course.  For all these, the set of upper bounds are $[x_2 \times y_2, 1 \times 1]$, and the LUB is $x_2 \times y_2$.

No, $[0,1] \times [0, 1)$ does not have the LUBP (in the dictionary order).  We need only show a counterexample: take the interval  $(x_1\times y_1, x_1 \times 1)$ (nonempty, bounded above) with $x \in [0, 1]$ and $y \in [0, 1)$.  The set of upper bounds is $(x_1 \times 0, 1 \times 1)$, and this clearly has no smallest element.  In other words, by picking the smallest upper bound one could think of, say $x_2 \times 0$, the idea is that one can always come up with a (strictly) smaller one, say $\frac{x_2}{2} \times 0$.

Yes, $[0,1) \times [0,1]$ has the LUBP (in the dictionary order).  Pick any nonempty subset of $A = [0, 1) \times [0, 1]$ that is bounded above, as the interval $(x_1 \times y_1, x_2 \times y_2)$, $[x_1 \times y_1, x_2 \times y_2)$, $(x_1 \times y_1, x_2 \times y_2]$, xor $[x_1 \times y_1, x_2 \times y_2]$ with $x \in [0, 1)$ and $y \in [0, 1]$.  For all these, $x_2 \times y_2$ is the LUB, since it's included in the upper bound set and it is its least element.

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## Busy People

People.... mYsTiFy. Some I don't get.

They act like we've never met.

They've plans, their afternoons are set.

They work, in their sleep they fidget.

Heart disease? They'll go ahead and fret.

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"Hello! Sorry, I've got to jet!"

Or, they'll stare through their socket,

Eyes angry, eyes wet,

their pens on their desk going tet-tet-tet.

Get out, like a rocket!

They've a gun in their pocket!

It doesn't seem they will let,

nor go fishing, cast a net,

chillax, get a pet

and take it to the vet.

Shet.

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