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)

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
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