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1.4 Exercise 1

June 19th, 2009 No comments

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)

1.4.1.a

(b)

1.4.1.b

(c)

1.4.1.c

(d)

1.4.1.d

(e)

First we show the first equality.

1.4.1.e.I

Now the second equality.

1.4.1.e.II

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

(f)

1.4.1.f

(g)

1.4.1.g

(h)

First we show the first equation.

1.4.1.h.I

Now the second.

1.4.1.h.II

(i)

1.4.1.i

(j)

1.4.1.j

(k)

1.4.1.k

(l)

Suppose otherwise, that  x y = 0 .  Then:

1.4.1.l

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

(m)

1.4.1.m

(n)

With  z \neq 0, y \neq 0 ,

1.4.1.n

(o)

With  y \neq 0, z \neq 0 ,

1.4.1.o

(p)

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

1.4.1.p

and this contradicts Axiom III.

(q)

1.4.1.q

(r)

1.4.1.r

(s)

1.4.1.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

June 18th, 2009 No comments

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.)

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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.

Busy People

June 17th, 2009 No comments

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.

It's about their office and their mile-high docket.

"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.

 

Categories: Miscellaneous, Poetry