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representation of real numbers∗
perucho†
2013-03-22 3:24:19
0.1
Introduction
It is well-known that there are several methods to introduce the real numbers.
We shall follow an inductive method which is instructive as well as elementary.
Apart from that such treatment is modern, interesting and is obtained through
two theorems and a lemma, which are relatively easy to understand. So that our
aim will be to prove the above propositions. Our starting point is the following
theorem.
Theorem 1. Let {ai } be a sequence of positive integers such that ai > 1, for
all i ≥ 1. Then any real number ρ is uniquely expressible by
ρ = b0 +
∞
X
i=1
bi
Qi
j=1
aj
,
(1)
where the bi are integers satisfying the inequalities 0 ≤ bi ≤ ai − 1 for all i ≥ 1,
and bi < ai − 1 for infinitely many i.
∞
Proof. Let {bi }∞
0 be a sequence of integers and {ρi }1 a sequence of real numbers
defined by the equations
ρ1 = ρ − b0 ,
b0 = [ρ],
(2)
and for all i ≥ 1
bi = [ai ρi ],
ρi+1 = ai ρi − bi ,
(3)
denoting [·] the integral part function. Clearly ρi+1 is the fractional part of ai ρi ,
therefore for all i ≥ 1 we have,
0 ≤ ρi < 1.
(4)
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1
Next we multiply (4) by ai whence 0 ≤ ai ρi < ai , but bi = [ai ρi ], so that
0 ≤ bi ≤ ai − 1,
an inequality required by the theorem.
Now, by (3) and (4), and applying induction on ρi , we can establish that
ρ = b0 + ρ1 = b0 +
b1
ρ2
b1
b2
ρ3
+
= b0 +
+
+
a1
a1
a1
a1 a2
a1 a2
= · · · = b0 +
n
X
ρn+1
+ Qn
.
j=1 aj
j=1 aj
bi
Qi
i=1
Now we define
d n = b0 +
n
X
i=1
bi
Qi
j=1
aj
(5)
,
thus from (4), (5) and by the hypothesis ai ≥ 2, we arrive to
1
ρn+1
< n,
0 ≤ ρ − d n = Qn
2
a
j
j=1
because the fractional part ρn+1 < 1. Then if we let n grows beyond of any
bound, ρ − dn will be so close to zero as we want and such a observation implies
the representation (1).
We still need to prove the another inequality of the theorem, i.e. bi < ai − 1 for
infinitely many i, but also the uniqueness of representation (1). To do that we
need to make use of the identity
∞
X
an+i − 1
= 1.
Qi
j=1 an+j
i=1
(6)
(It is legitimate to consider this identity as a lemma, as we need it to prove this
theorem as well as the next one).
We shall prove later this identity. Let us prove the inequality bi < ai − 1 by
tertio excluso; thus suppose that there is a fixed integer n such that bi = ai − 1
for all i > n. From (1) and (6) we get
ρ = b0 +
n
X
i=1
= b0 +
= b0 +
bi
Qi
j=1 aj
n
X
i=1
n
X
i=1
bi
Qi
j=1
aj
bi
Qi
j=1
aj
+
∞
X
ai − 1
Qi
j=1 aj
i=n+1
∞
X
an+i − 1
Qi
a
j=1 j i=1
j=1 an+j
+ Qn
+ Qn
1
1
j=1
2
aj
,
and comparing this with (5) one realizes that ρn+1 = 1, contradicting (4).
Finally we must prove the uniqueness of the representation (1). So that we
suppose
∞
X
ci
.
ρ = c0 +
Qi
j=1 aj
i=1
Here the integers ci satisfy the same conditions as do the bi . It is necessary
(and sufficient!) to show that ci = bi for every i. The condition ci < ai − 1, for
infinitely many i, altogether with the identity (6) imply that
∞
X
i=1
ci
Qi
j=1
aj
< 1,
so we see that c0 is the integral part of ρ, i.e. c0 = [ρ], but also from (2) b0 = [ρ],
therefore c0 = b0 . Next we shall again use tertio excluso. On this way let us
suppose that for some n ≥ 1 the pair bn and cn are unequal. There is no loss of
generality in assuming that n is the smallest integer with this property (which is
justified by a simple inductive argument), and that bn > cn , so that bn − cn ≥ 1.
Thus we have
∞
∞
X
X
bi
ci
=
.
Qi
Qi
a
j=1 j
j=1 aj
i=n
i=n
(There is no contradiction at all in this equality, as it is easily seen in the next
below equation).
It is obvious that these series are absolutely convergent, so we may rearrange
terms to obtain
∞
X
c i − bi
bn − c n
1
= Qn
≥ Qn
.
Qi
a
j=1 j
j=1 aj
j=1 aj
i=n+1
But then recalling that ci < ai − 1, we see tat ci − bi < ai − 1. From this fact
and (6), we can write
∞
X
∞
X
ci − bi
ai − 1
<
Qi
Qi
j=1 aj
j=1 aj
i=n+1
i=n+1
∞
X
an+i − 1
1
= Qn
.
Qi
j=1 aj i=1
j=1 aj
j=1 an+j
= Qn
1
This is a contradiction with respect to the inequality found before, thus the
proof of this theorem is complete.
0.2
Some implications
First we remarked that Theorem 1 is a generalization of the standard decimal
expansion for a real number ρ. This may be seen by taking all the integers
3
ai = 10. Thus, if ρ > 0, (1) gives the decimal representation
ρ = b0 +
∞
X
bi
= b0 .b1 b2 · · · .
i
10
i=1
(7)
Second, if ρ < 0, we must write its decimal representation on the form (7)and
then changing all signs. Third, any real ρ could have an ambiguous decimal
representation, e.g. ρ = 1.5699 · · · , having an infinite successive sequence of 9’s,
which also involves a geometric series in 10−i in turns implying (at the limit)
that also ρ = 1.57. For that reason, (7) represents that number with an infinite
succession of 0,s, that is, ρ = 1.57 = 1.5700 · · · . The reason for this resides in
that an infinite succession of 9’s is ruled out by the condition of the theorem
that bi < ai − 1 for infinitely many i, a condition that in the present example
takes the form bi < 9 for infinitely many i.
Now we prove (6).
Lemma 1. For the integers sequence {ai }, where ai > 1 for every i ≥ 1, we
have
∞
X
an+i − 1
= 1.
Qi
j=1 an+j
i=1
Proof. Let us take the partial sum
Sm =
m
m
m
X
X
X
an+i − 1
an+i
1
=
−
Qi
Qi
Qi
j=1 an+j
j=1 an+j
j=1 an+j
i=1
i=1
i=1
m
=
=1+
m
X
an+i
Qi
j=1 an+j
i=2
=1+
m
X
=1+
i=2
j=1 an+j
m
X
an+i
Qi
j=1 an+j
m
X
=1+
i=2
an+i
Qi
i=2
m
X
m
X
an+i
1
an+1 X
+
−
Qi
Qi
an+1 i=2 j=1 an+j
j=1 an+j
i=1
Qi
m+1
X
i=2
m+1
X
i=2
an+i
−
i=2
Qi
an+i
j=1
−
−
an+j
j=1 an+j
m
X
−
i=2
Qi
1
Qi
j=1 an+j−1
an+i
Qi
an+i j=1 an+j−1
−
n+m+1
Qm+1
(n + m + 1) j=1 an+j−1
an+i
j=1
an+j
1
,
j=1 an+j
− Qm
Q
since when in the
operator an+j−1 7→ an+j , then the index value of j, at its
upper limit, m + 1 7→ m, but its lower limit does not. Thus, central sums cancel
and the last term vanishes because, by hypothesis, we have
lim Sm =
m→∞
∞
X
an+i − 1
1
,
= 1 − lim Qm
Qi
m→∞
j=1 an+j
j=1 an+j
i=1
and the lemma is proved.
4
Theorem 1 also represents an irrational number whenever we add a couple
of additional conditions. Thus we have the following important theorem.
Theorem 2. Let us consider the same integers sequence {ai } described in the
preceding theorem, and that the integers bi satisfying the inequalities of that
result. In addition, let us assume that infinite integers bi are positive, and that
each prime number divides infinitely many ai . Then ρ is irrational.
Proof. We contradict the thesis by supposing ρ = p/q is rational (p, q, coprime).
By the last hypothesis in the preceding theorem,
Qn we can choose an integer n
sufficiently large in order to q be a divisor of j=1 aj . Now we may use (1)
replacing the LHS by our rational number assumption, next multiplying both
side by the latter product, and rearranging terms we get (we do the partition
sum i = 1, . . . , n; n + 1, . . . ∞)
Qn
Qn
n
∞
X
X
b0 j=1 aj
bn+i
j=1 aj (p − b0 q)
−
=
.
(8)
Qi
Qi
q
j=1 aj
j=1 an+j
i=1
i=1
By hypothesis, the LHS of (8) is obviously an integer. However, we have proved
already that the last inequality of theorem 1 requires that bn+i < an+i − 1 for
infinitely many i, so that, from the RHS of (8) and the lemma we see that
∞
X
i=1
bn+i
Qi
j=1
an+j
<
∞
X
an+i − 1
= 1,
Qi
j=1 an+j
i=1
a clear contradiction, proving the theorem.
0.3
Example
e is irrational.
All we know there are different ways to prove the irrationality of e. In particular,
it results illustrative if we adapt our theorems and its hypotheses (all of which
are true in this case) to this problem. From Taylor’s expansion
∞
X
1
e=
.
i!
i=0
Let us use (1), by setting ρ = e, b0 = 2, bi−1 = 1 for all i ≥ 2, and aj = j + 1,
0 ≤ j ≤ i − 1. Thus,
∞
X
1
e=2+
.
Qi
j=1 j
i=1
So far our discussion on real numbers. An interesting approach on transcendental numbers as well as and extensive bibliography on real numbers are given,
for instance, in [?].
5
References
[1] I. Niven, IRRATIONAL NUMBERS, Ch. VII, pp. 83-88; also p. 157, The
Mathematical Association of America, 2005.
6