the proposition in question. To achieve this, the steps in a proof must follow
logically from previous steps or be justified by some other agreedupon set of
facts. In addition to being valid, these steps must also fit coherently together to
form a cogent argument. Mathematics has a specialized vocabulary, to be sure,
1.2.
Some Preliminaries
9
but that does not exempt a good proof from being written in grammatically
correct English.
The one proof we have seen at this point (to Theorem 1.1.1) uses an indirect
strategy called
proof by contradiction
. This powerful technique will be employed
a number of times in our upcoming work. Nevertheless, most proofs are direct.
(It also bears mentioning that using an indirect proof when a direct proof is
available is generally considered bad manners.)
A direct proof begins from
some valid statement, most often taken from the theorem’s hypothesis, and
then proceeds through rigorously logical deductions to a demonstration of the
theorem’s conclusion.
As we saw in Theorem 1.1.1, an indirect proof always
begins by negating what it is we would like to prove. This is not always as easy
to do as it may sound. The argument then proceeds until (hopefully) a logical
contradiction with some other accepted fact is uncovered.
Many times, this
accepted fact is part of the hypothesis of the theorem. When the contradiction is
with the theorem’s hypothesis, we technically have what is called a
contrapositive
proof.
The next proposition illustrates a number of the issues just discussed and
introduces a few more.
Theorem 1.2.6.
Two real numbers
a
and
b
are equal if and only if for every
real number
ϵ
>
0
it follows that

a
−
b

<
ϵ
.
Proof.
There are two key phrases in the statement of this proposition that
warrant special attention.
One is “for every,” which will be addressed in a
moment. The other is “if and only if.” To say “if and only if” in mathematics
is an economical way of stating that the proposition is true in two directions.
In the forward direction, we must prove the statement:
(
⇒
)
If
a
=
b
, then for every real number
ϵ
>
0
it follows that

a
−
b

<
ϵ
.
We must also prove the converse statement:
(
⇐
)
If for every real number
ϵ
>
0
it follows that

a
−
b

<
ϵ
, then we must
have
a
=
b
.
For the proof of the first statement, there is really not much to say. If
a
=
b
,
then

a
−
b

= 0, and so certainly

a
−
b

<
ϵ
no matter what
ϵ
>
0 is chosen.
For the second statement, we give a proof by contradiction. The conclusion
of the proposition in this direction states that
a
=
b
, so we assume that
a
̸
=
b
.
Heading o
ff
in search of a contradiction brings us to a consideration of the phrase
“for every
ϵ
>
0.” Some equivalent ways to state the hypothesis would be to
say that “for all possible choices of
ϵ
>
0” or “no matter how
ϵ
>
0 is selected,
it is always the case that

a
−
b

<
ϵ
.” But assuming
a
̸
=
b
(as we are doing at
the moment), the choice of
ϵ
0
=

a
−
b

>
0
poses a serious problem.
We are assuming that

a
−
b

<
ϵ
is true for
every
ϵ
>
0, so this must certainly be true of the particular
ϵ
0
just defined. However,
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 Spring '18
 Real Numbers, G.H. Hardy