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 , using only the following axioms (I)-(V):

*Algebraic Properties of the Reals*

I. , for all in .

II. , for all in .

III. There exists a unique element of called *zero*, denoted by 0, such that for all . There exists a unique element of called *one*, different from 0 and denoted by 1, such that for all .

IV. For each in , there exists a unique in such that . For each in different from 0, there exists a unique in such that .

V. for all .

-----

(a) If , then

(b) [Hint: Compute ]

(c)

(d)

(e)

(f)

(g)

(h)

(i) If and , then

(j) if

(k)

(l) and , then

(m) if

(n) if

(o) if

(p)

(q) if

(r) if

(s) if

(t) if "

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

-------

SOLUTION

(a)

(b)

(c)

(d)

(e)

First we show the first equality.

Now the second equality.

Finally, by transitivity of equality, since and , it follows that .

(f)

(g)

(h)

First we show the first equation.

Now the second.

(i)

(j)

(k)

(l)

Suppose otherwise, that . Then:

contradicts the fact that is nonzero. Thus, holds.

(m)

(n)

With ,

(o)

With ,

(p)

Suppose otherwise, and . But then:

and this contradicts Axiom III.

(q)

(r)

(s)

(t)

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

\subsubsection[Exercise 1]{Prove the following ``laws of algebra'' for

, using only the following axioms (I)-(V): \newline

\emph{Algebraic Properties of the Reals} \newline

I.

, \newline

II.

\newline

III. There exists a unique element of

called \emph{zero}, denoted by 0, such that

for all

. \newline

There exists a unique element of

called \emph{one}, different from 0 and denoted by 1, such that

for all

. \newline

IV. For each

in

, there exists a unique

in

such that

. \newline For each

in

different from 0, there exists a unique

in

such that

. \newline

V.

for all

. \newline \newline

(a) If

, then

\newline

(b)

[Hint: Compute

] \newline

(c)

\newline

(d)

\newline

(e)

\newline

(f)

\newline

(g)

\newline

(h)

\newline

(i) If

and

, then

\newline

(j)

if

\newline

(k)

\newline

(m)

if

\newline

(n)

if

\newline

(o)

if

\newline

(p)

\newline

(q)

if

\newline

(r)

if

\newline

(s)

if

\newline

(t)

if

}

(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

and

, it follows that