Joseph Liouville’s construction of a transcendental number
Topmatter:
Some ideas pulled from exercises 6.6 – 6.8 in, Galois Theory (second edition) by Ian Stewart,
page 69. Some from elsewhere.
Here I present a (probably modified) version of Louiville’s work that
explicitly constructed a transcendental number, though some work done
after his work of 1844. Cantor’s later work constructed none, but had
important ideas of cardinality.
Joseph Liouville
There are countless web references to transcendental numbers, a Wikipedia
My beaten-up text
page. There are biographies of Joseph Liouville, discussions of
transcendental numbers, and even a page on the “fifteen most famous
transcendental numbers.” But I could not find a good construction, though this page
has some discussion on a higher level than this one, as does this link.
Disclaimer:
Slowly typeset by Dean Moore, October – December, 2010, Boulder, Colorado, USA, tweaks afterward.
In March 2009 I worked the proof for pure entertainment, later typeset to slowly piece through, to
understand what I did, and fill in gaps. And my field was analysis, not algebra – number theory, where
my knowledge is at best fuzzy. Why in graduate school at CU – Boulder a mathematician studying
analysis bought a book on Galois Theory is a lost mystery, though years later worked through it.
I aimed at a simplistic explanation with many references (probably too many) that could be understood by
an undergrad in math with some understanding of real analysis, number theory, rings and fields, a little set
theory. Familiarity with elementary algebra as exponents is assumed. Some understanding of limits of
sequences and infinite sums is a must. Understanding of some mathematical notation is assumed; one
reference is here. If by some weirdness someone finds it and finds it useful, great. Send me an e-mail,
dean at deanlm dot com, or if I made some typo, flub or poor logic.
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Background:
Rational numbers, the ancient Greeks:
Ancient Greek mathematicians first assumed all numbers rational, i.e., ratios of integers
𝑝
as 𝑞 , with 𝑝, 𝑞 𝜖 ℤ, 𝑞 ≠ 0. Easy examples are
5
𝑦0 = ,
7
13
32 ∙ 7 ∙ 127 ∙ 7853
𝑦1 = 13 =
, or 𝑦3 = 3.14159265 =
1
28 ∙ 57
The Greeks knew of the square root of two, √2 from obvious considerations from the
Pythagorean Theorem (diagram, right: the hypotenuse of a right triangle of both short
legs of length 1, as 12 + 12 = 2), but initially assumed it rational. Eventually a
Greek proved the square root of two was irrational, i.e., cannot be expressed as a
fraction of two integers. This may be proved many ways.
So for 𝑝, 𝑞 𝜖 ℤ, 𝑞 ≠ 0 to write
𝑝
= √2
𝑞
is impossible. One may get arbitrarily “close” (i.e., limit), but equality cannot happen.
Notably for the polynomial 𝑓(𝑥) = 𝑥 2 − 2, the number 𝑥1 = √2
satisfies 𝑓(𝑥1 ) = 0. So √2 is the root of a polynomial in rational (here
integer) coefficients.
So we had an easy way to include numbers like √2 in our number system,
and in terms of simple familiar things: roots of polynomials in integer
coefficients, ideas studied since at least the ancient Egyptians and
Babylonians.
Algebraic numbers:
Thus algebraic numbers: all numbers that are roots of some polynomial
Babylonian mathematics on √𝟐. Image credit
here.
𝑛
𝑓(𝑥) = ∑ 𝑎𝑖 𝑥 𝑖 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + … + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛 𝑥 𝑛 ,
𝑖=0
𝑛 𝜖 ℕ, 𝑎𝑖 𝜖 ℚ ∀ 𝑖 𝜖 {0, … 𝑛}
in rational coefficients (equivalently, integer: one may multiply by a least common multiple to
make all coefficients integers, since 𝑓(𝑥) is set equal zero). That these form an algebraically
closed field is proven elsewhere. This result is in various books; some discussion here. As an
example, a number as
(3 +
7
(52
−
1
24
11
3 5
717 )
52
73
+ 1)
−1
is algebraic, a root of some integer-coefficient polynomial. From Galois Theory, roots of
polynomials of degree five and higher cannot in general be expressed as
combinations of radicals of integers / rationals, as in general such
polynomials are “not solvable.”
However, no one proved all numbers are algebraic. The algebraics turn out
an ℵ0 field, actually small in a set-theoretic sense, the first infinity, where the
real numbers (or the complexes) are 2ℵ0 (sometimes identified with ℵ1, but
this set theory and not currently decidable, and gets afield for this discussion)
fields, the second infinity.
The transcendentals:
But cardinality waited for Cantor, whose work postdated Liouville.
Algebraic numbers led to the idea of the transcendental number: a number
that is not the root of any polynomial in rational (equivalent, integer)
coefficients or algebraic coefficients (as the algebraics are algebraically
closed, this is equivalent to integer coefficients). However, no one explicitly constructed one,
though in 1677 the mathematician James Gregory attempted to prove π
Georg Cantor
transcendental (this history here and here). The term transcendental goes back to at
least Leibniz in 1673, if more modern formulations seem to trace to Euler in 1748.
Transcendentals were first proven to exist in 1844 by the French mathematician Joseph
Liouville, though he did not then construct an explicit decimal number but a
continued fraction. The first decimal proven transcendental was the Liouville
constant which was proven transcendental in 1850, not 1844 as stated in some
web references. It is a “Liouville number,” a bit odd, and the first “naturally”
occurring transcendentals were later proven e (Hermite, 1873) and π
(Lindemann, 1882), neither of which are Liouville numbers.
As an aside, when π was proven transcendental it proved that by the methods
of ancient Greek geometers it is impossible to square the circle. This
answered a question had been open for thousands of years.
After Liouville, Cantor constructed no transcendental numbers, but showed
they had to exist, and his work had important implications of cardinality, that
they comprise a “big” 2ℵ0 set. It’s going afield, but Cantor showed the algebraics
are a countable set, but his “diagonal argument” showed the real line uncountable,
Joseph Liouville
hence transcendentals must be uncountable, that is, of cardinality 2ℵ0 .
As a foretaste, let 𝑥0 be an irrational algebraic, a root of a polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] of degree 𝑛.
Liouville took 𝑛, and bounds of the derivative of 𝑓 “near” 𝑥0 , and set a “speed limit” on how fast
a sequence of rationals can limit on an irrational algebraic number, then showing only a finite set
of rationals could beat a stronger speed limit. He constructed sequence of rationals limiting to an
irrational number. He then proved an infinite subset of the sequence violated the strong speed
limit, so the limit could not be algebraic, hence, was transcendental.
Construction of a transcendental number:
To do in detail the construction is long and tedious with a few proofs. First we define a few
things that for the remainder of this discussion will not vary.
Definition 1
We define and fix a polynomial, its degree, and an irrational algebraic number that is a root of it:
Suppose 𝑥0 𝜖 ℝ ∽ ℚ, and, for an 𝑛𝑡ℎ − degree polynomial not identical zero (i.e., 𝑛 > 0),
𝑓(𝑥) 𝜖 ℤ[𝑥] (notably Lemma 2 and all else below depends on the polynomial having coefficients
in ℤ, equivalent to ℚ; we henceforth assume this, so our polynomial is not assumed monic),
𝑓(𝑥) = ∑𝑛𝑖=0 𝑎𝑖 𝑥 𝑖 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + … + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛 𝑥 𝑛 ,
𝑛 𝜖 ℕ, 𝑎𝑖 𝜖 ℤ ∀ 𝑖 𝜖 {0, … 𝑛}
we have 𝑓(𝑥0 ) = 0, that is, 𝑥0 is an
irrational algebraic number. The number
𝑥0 is fixed for the remainder of this
discussion.
Note 𝑓 is fixed, will be referred to without
redefinition. The degree of 𝑓,
𝑛 = deg(𝑓) comes up again, is fixed,
important later on. Note 𝑛 ≥ 2, or 𝑥0 is
rational, the single zero of a polynomial
like 𝑓(𝑥) = 2𝑥 − 9. This is of small
importance later on. Note next lemma
depends on 𝑓 being a polynomial with
coefficients in ℤ.
Graph of the fifth-order polynomial 𝒇(𝒙) = 𝒙𝟓 − 𝒙 − 𝟏. The lone real root at 𝒙 ≈ 𝟏. 𝟏𝟔𝟕𝟑
is a real algebraic; the other four roots are complex algebraics. This polynomial cannot be
factored over ℤ, and is not “solvable” (mentioned here), so no roots may be obtained via
algebra and extraction of roots on integers/rationals.
Lemma 2.
For a polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] of degree 𝑛, we have the following result:
𝑝
𝑝
1
If 𝑝, 𝑞 𝜖 ℤ, 𝑞 ≠ 0 (assume 𝑞 ≥ 1, safe) and 𝑓 (𝑞 ) ≠ 0, |𝑓 (𝑞 )| ≥ 𝑞𝑛 is always true.
Proof.
𝑝
𝑝
As 𝑓 (𝑞 ) ≠ 0 assume 𝑓 (𝑞 ) = 𝑐 for 𝑐 𝜖 ℚ ∽ {0}; note 𝑐 is easily rational as 𝑓 is a polynomial
over ℤ. So
𝑛
𝑝
𝑝 𝑖
𝑓 ( ) = ∑ 𝑎𝑖 ( ) = 𝑐 𝜖 ℚ ∽ {0}
𝑞
𝑞
𝑖=0
Taking absolute values of both sides we have:
𝑝 𝑖
𝑝
|𝑓 (𝑞 )| = |∑𝑛𝑖=0 𝑎𝑖 (𝑞 ) | = |𝑐|.
Now multiply both sides by 𝑞 𝑛 (again, assuming 𝑞 ≥ 1):
𝑛
𝑛
𝑖=0
𝑖=0
𝑝
𝑝 𝑖
𝑛 |𝑐|
𝑛
𝑛
𝑞
= 𝑞 |𝑓 ( )| = 𝑞 |∑ 𝑎𝑖 ( ) | = |∑ 𝑎𝑖 𝑝𝑖 ∙ 𝑞 𝑛−𝑖 |
𝑞
𝑞
Note in rightmost sum that 𝑛 − 𝑖 ≥ 0 is always true, so all terms of the right expression are
integers and 𝑞 𝑛 |𝑐| is some natural number. So 𝑞 𝑛 |𝑐| 𝜖 ℕ = {1, 2, 3, … }, is at least one.
Thus,
𝑝
𝑞 𝑛 |𝑓 ( )| ≥ 1
𝑞
𝑝
1
Dividing by 𝑞 𝑛 we get |𝑓(𝑞 )| ≥ 𝑞𝑛, our first lemma. □
Further discussion 3
As an example, 𝑥 = √2 satisfies 𝑓(𝑥) = 𝑥 2 − 2. While √2 = 1.41421 … is irrational, its digits
never ending with no known pattern, we can approximate it with
𝑝1
𝑞1
=
141421
10000
=
7 ∙ 89 ∙ 227
10000
(coprime form, and I used an online number factorer). So 𝑞1 = 10000. But
𝑝1
𝑝1 2
|𝑓 ( )| = |( ) − 2| ≅ 1.00759 ∙ 10−5
𝑞1
𝑞1
1
1
Note here 𝑛 = 2 (the degree of the polynomial 𝑓(𝑥) = 𝑥 2 − 2), and 𝑞2 = 100002 = 10−10 , so that
1
𝑝
1
|𝑓 (𝑞 )| ≥ 𝑞𝑛 seems obvious, by about five orders of magnitude.
Convention 4
𝑝
For 𝑥0 an irrational root of the polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥], suppose 𝑞 is “close” to 𝑥0 in the sense
𝑝
𝑝
of 𝑥0 − 1 < 𝑞 < 𝑥0 + 1, and that 𝑞 is closer to 𝑥0 than any to other root of 𝑓. In a so-so graph:
𝑝
𝑞
𝑓(𝑥)
𝑥0 − 1
𝑥0 + 1
𝑥0
By calculus results, on a bounded interval the derivative 𝑓 ′ of 𝑓 is bounded, thus
∃ 𝑀, 0 < 𝑀 < ∞ such that |𝑓 ′ (𝑦)| ≤ 𝑀 ∀ 𝑦 𝜖 (𝑥0 − 1, 𝑥0 + 1). We will be looking at rational
𝑝 ∞
sequences {𝑞𝑖 }
𝑖
𝑖=1
𝑝
with lim (𝑞𝑖 ) = 𝑥0 , so after a finite point any such sequence is in the
𝑖→∞
𝑖
interval (𝑥0 − 1, 𝑥0 + 1). As the rationals are dense in the real line, such a sequence exists.
Our next proposition establishes a “speed limit.” Note it doesn’t depends on 𝑥0 being irrational,
𝑝
merely that 𝑞 ≠ 𝑥0 , but this is always true for irrational𝑥0 .
Proposition 5.
𝑝
We maintain that for any 𝑞 as in Convention 4 above, for 𝑥0 our earlier-defined irrational root of
the 𝑛𝑡ℎ -degree polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥], that
𝑝
1
| − 𝑥0 | ≥
𝑞
𝑀 ∙ 𝑞𝑛
Proof.
By the Mean Value Theorem, between
𝑝
𝑝
𝑞
𝑞
𝑝
𝑞
and 𝑥0 the derivative 𝑓′ of 𝑓 assumes a “mean value”:
∃ 𝑐 𝜖 (min{ , 𝑥0 }, max{ , 𝑥0 }) such that
𝑝
𝑓 (𝑞 ) − 𝑓(𝑥0 )
𝑝
𝑞 − 𝑥0
𝑝
𝑝
= 𝑓′(𝑐)
As 𝑞 is closer to 𝑥0 than to any other root of 𝑓, 𝑓 (𝑞 ) ≠ 0 and Lemma 2 holds. As 𝑓(𝑥0 ) = 0, we
may restate this,
𝑝
𝑓 (𝑞 ) − 𝑓(𝑥0 )
𝑝
𝑞 − 𝑥0
𝑝
𝑓 (𝑞 ) − 0
= 𝑝
=
−
𝑥
0
𝑞
𝑝
𝑓 (𝑞 )
= 𝑓 ′ (𝑐)
𝑝
𝑞 − 𝑥0
Taking absolute values, from Convention 4 this gives
𝑝
𝑓 (𝑞 )
|𝑝
| = | 𝑓′(𝑐)| ≤ 𝑀
−
𝑥
0
𝑞
But from Lemma 2 we know that
𝑝
1
|𝑓 ( )| ≥ 𝑛 .
𝑞
𝑞
Putting this together with |𝑓′(𝑐)| ≤ 𝑀,
𝑝
1
1
1
𝑀 ≥ |𝑓 ′ (𝑐)| = |𝑓 ( )| ∙ 𝑝
≥ 𝑛 ∙ 𝑝
𝑞
|𝑞 − 𝑥0 | 𝑞 |(𝑞 − 𝑥0 )|
We drop the middle terms and rearrange to derive
𝑝
1
| − 𝑥0 | ≥
𝑞
𝑀𝑞 𝑛
This verifies our inequality. □
Comment 6
𝑝 ∞
Again, we will be looking at rational sequences {𝑞𝑖 }
𝑖
𝑖=1
𝑝
with 𝑞𝑖 → 𝑥0 , so after a finite point
𝑖
Convention 4 will be true. For an algebraic 𝑥0 for each “large” 𝑖 we have
𝑝𝑖
1
| − 𝑥0 | ≥
𝑞𝑖
𝑀𝑞𝑖𝑛
𝑝 ∞
So a rational sequence {𝑞𝑖 }
𝑖
𝑖=1
can’t converge to the irrational algebraic 𝑥0 “too fast,” where
again, 𝑛 = deg(𝑓), a constant fixed above, the degree of an rational-coefficient polynomial (a
minimal polynomial is best, gives the best bounds) of which 𝑥0 is a root.
𝑝
The last showed |𝑞 − 𝑥0 | has to be bigger than something. The next proposition shows that for
𝑝
only a finite set of rationals can |𝑞 − 𝑥0 | be smaller than a stronger bound, and is important in
our final theorem that proves the existence of a non-algebraic number.
We note the Thue–Siegel–Roth theorem has stronger bounds than the next. Its proof is not easy,
and led to a Fields Medal. What’s truly important is that 𝑘 > 𝑛 = deg(𝑓).
Proposition 7.
Fix 𝐾 > 0 and 𝑘 > 𝑛 = deg(𝑓). For some index set 𝐼 there exists an at most finite set 𝑆 of
unique rational numbers,
𝑆={
𝑝𝑖
| 𝑖 𝜖 𝐼, 𝑝𝑖 , 𝑞𝑖 𝜖 ℤ, 𝑞𝑖 ≠ 0}
𝑞𝑖
such that
|
𝑝𝑖
𝐾
− 𝑥0 | < 𝑘
𝑞𝑖
𝑞𝑖
That is, the index set 𝐼 is finite, hence 𝑆 is a finite set. We assume {𝑞𝑖 ≥ 1, ∀ 𝑖 𝜖 𝐼}.
Proof.
We do a proof by contradiction. Assume for an infinite index set (as the rational numbers form
an ℵ0 set, we choose 𝐼 = ℕ) there exists a set
𝑆={
with each
𝑝𝑖
𝑞𝑖
𝑝𝑖
𝑞𝑖
| 𝑖 𝜖 ℕ, 𝑝𝑖 , 𝑞𝑖 𝜖 ℤ, 𝑞𝑖 ≥ 1}
is unique within 𝑆, such that for all 𝑖 𝜖 𝐼 and for our fixed 𝐾 > 0 and 𝑘 > 𝑛 that
|
𝑝𝑖
𝐾
− 𝑥0 | < 𝑘
𝑞𝑖
𝑞𝑖
Now we show {𝑞𝑖 | 𝑖 𝜖 ℕ} is an unbounded set.
If the denominators {𝑞𝑖 | 𝑖 𝜖 ℕ} are restricted to a bounded set (i.e., live within a finite interval, so
there can only be a finite number of 𝑞𝑖 ) we get a contradiction: as the set 𝑆 is infinite and each
is unique within the set 𝑆, we have 𝑝𝑖 → ∞ forced. As the 𝑞𝑖 are bounded and 𝑝𝑖 → ∞, after
some finite point the inequality
𝑝
𝐾
| 𝑖 − 𝑥0 | < 𝑘
𝑞𝑖
𝑞𝑖
𝑝
must be violated: as 𝑝𝑖 → ∞, the term |𝑞𝑖 − 𝑥0 | → ∞, but
𝑖
𝐾
𝑞𝑖𝑘
is bounded.
𝑝𝑖
𝑞𝑖
Thus 𝑞𝑖 → ∞ is forced. As 𝑘 − 𝑛 ≥ 1, we have 𝑞𝑖𝑘−𝑛 → ∞. By the above inequality we
𝑝 ∞
𝑝
have lim (𝑞𝑖 ) = 𝑥0 , thus may drop a finite set of {𝑞𝑖 }
𝑖→∞
𝑖
𝑖
, apply Proposition 5 and on a “small”
𝑖=1
interval bound the derivative 𝑓 ′ of 𝑓 by some constant 𝑀.
Now we derive our contradiction. Remembering 𝑥0 is an irrational root of our degree 𝑛
polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] and applying Proposition 5 we get:
𝑝𝑖
𝐾
1
>
|
−
𝑥
|
≥
0
𝑞𝑖
𝑀 ∙ 𝑞𝑖𝑛
𝑞𝑖𝑘
Tossing the middle and re-arranging, the following is always true:
𝑀 ∙ 𝐾 > 𝑞𝑖𝑘−𝑛
As 𝑞𝑖 → ∞, for some 𝑖0 𝜖 ℕ we have 𝑞𝑖 > 𝐾 ∙ 𝑀′ whenever 𝑖 ≥ 𝑖0 . As 𝑘 − 𝑛 ≥ 1, for all 𝑖 ≥ 𝑖0 we
have forced
𝑀 ∙ 𝐾 < 𝑞𝑖𝑘−𝑛
violating the last inequality. Thus our initial assumption was wrong, verifying the proposition. □
Discussion 8
Putting together Proposition 4 and Proposition 6 with 𝐾 = 1, let 𝑥0 be an algebraic number, 𝑀 as
before, and let 𝑓(𝑥) 𝜖 ℤ[𝑥] satisfy 𝑓(𝑥0 ) = 0. Fix 𝑘 = 𝑛 + 1 > 𝑛 = deg(𝑓).
𝑝 ∞
In the next equation for any rational sequence {𝑞𝑖 }
𝑖
𝑖=1
𝑝
𝑝
with 𝑞𝑖 → 𝑥0 , whenever 𝑞𝑖 is “close” to 𝑥0 ,
𝑖
𝑖
i.e., after some finite point, we have
1
𝑝𝑖
𝑛 ≤ | − 𝑥0 | (always true)
𝑀 ∙ 𝑞𝑖
𝑞𝑖
For the next we have a best a finite set of 𝑞𝑖 ’s:
𝑝𝑖
1
𝑝𝑖
| − 𝑥0 | < 𝑛+1 (a finite set of )
𝑞𝑖
𝑞𝑖
𝑞𝑖
Now we create a transcendental number, i.e., a number 𝑥1 such that 𝑥1 is not the root of any
polynomial in 𝑓(𝑥) 𝜖 ℤ[𝑥].
Definition 9
We define 𝑥1 in a special way, and for now assume it an algebraic number, the root of a
polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] of some degree 𝑛. Let
∞
𝑥1 = ∑ 10−𝑛!
𝑛=1
using factorial function, as in 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120. The number 𝑥1 is easily an infinite
𝑝 ∞
sum of rational numbers, and at each point 𝑖 we define an infinite sequence of rationals {𝑞𝑖 }
𝑖
as
𝑖=1
𝑖
𝑝𝑖
1
1
1
1
1
= ∑ 10−𝑛! = 1! + 2! + 3! + ⋯ + (𝑖−1)! +
=
𝑞𝑖
10
10
10
10𝑖!
10
𝑛=1
1
1
1
1
+
+
+ ⋯ 𝑖!
10 100 1000000
10
It is easy to assign the denominator at each finite point 𝑖 as 𝑞𝑖 = 10𝑖! . Note by standard calculus
𝑝 ∞
tests like the ratio test the sequence {𝑞𝑖 }
𝑖
𝑖=1
is convergent to 𝑥1,
𝑝
i.e., lim (𝑞𝑖 ) = 𝑥1 .
𝑖→∞
𝑖
Note 𝑥1 is called the Liouville Constant, and 𝑥1 =
0.110001000000000000000001000 … with the 1’s getting further and
further apart. As digits of 𝑥1 never repeat, it is easily irrational.
Theorem 10.
There does not exist a polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] of which 𝑥1 is a root.
Proof.
The geometric constant π is
transcendental, but this was proven
in 1882, over forty years after
Liouville’s proof. This meant that
squaring the circle is impossible,
solving a problem that had been open
for thousands of years.
Assume the contrary, i.e., that the number 𝑥1 is a root of some polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥], and 𝑓(𝑥)
is of degree 𝑛 𝜖 ℕ; the degree 𝑛 is important, and we will refer to it.
First we draw a bound. Note that for any natural numbers 𝑗 > 𝑖 ≥ 1, the base-ten expansion of the
fraction
1
10𝑖!
has a 1-digit at least one to the left of the leftmost 1-digit of the base-ten expansion of the fraction
1
(𝑖+𝑗)!
10
To summarize this as a formula, whenever 𝑗 > 𝑖 ≥ 1 are natural numbers,
1
10(𝑖+𝑗)!
<
1
.
10𝑖!
Go back to our definition of 𝑥1 . Note that at any finite point 𝑖 that
𝑖
∞
∞
𝑛=1
𝑛=1
𝑛=𝑖+1
𝑝𝑖
| − 𝑥1 | = |∑ 10−𝑛! − ∑ 10−𝑛! | = ∑ 10−𝑛! =
𝑞𝑖
1
10(𝑖+1)!
+
1
10(𝑖+2)!
+⋯
In the next, on the left when we tack 1-digits further right it is irrelevant, thus from our bound we derive:
∞
∑ 10−𝑛! =
𝑛=𝑖+1
1
10(𝑖+1)!
+
1
10(𝑖+2)!
+⋯ <
1
10𝑖!
So we draw an important bound:
∞
𝑝𝑖
10
1
| − 𝑥1 | = ∑ 10−𝑛! < (𝑖+1)! = 𝑖!
𝑞𝑖
10
10
𝑛=𝑖+1
We need to do a little algebra. Note that any polynomial 𝑓(𝑥) 𝜖 ℤ[𝑥] whose root is an irrational
algebraic is of at least degree 2 (this was talked about, bottom of Definition 1), so 𝑛 = deg(𝑓) ≥
2. For fixed 𝑘 > 𝑛, let 𝑖 ≥ 𝑘 > 𝑛, so 𝑖 > 2. As 𝑖! ≥ 3! = 6, the next is easy:
𝑖+1>𝑘+
1
𝑖!
Re-arranging,
(𝑖 + 1)𝑖! > 𝑘 ∙ 𝑖! + 1
Now writing (𝑖 + 1)𝑖! = (𝑖 + 1)! and re-arranging a little more,
(𝑖 + 1)! − 1 > 𝑘 ∙ 𝑖!
So the next is easy, rules of exponents:
1
10(𝑖+1)!−1
𝑝 ∞
Fix 𝑘 = 𝑛 + 1. Note {𝑞𝑖 }
𝑖
<
1
10𝑘∙𝑖!
is an infinite sequence of rational numbers, and we now use our
𝑖=𝑘
algebra result. For each 𝑖 ≥ 𝑘 > 𝑛:
𝑝𝑖
10
1
1
1 𝑘
1
| − 𝑥1 | < (𝑖+1)! = (𝑖+1)!−1 < 𝑘∙𝑖! = ( 𝑖! ) = 𝑘
𝑞𝑖
𝑞𝑖
10
10
10
10
𝑝 ∞
But 𝑘 > 𝑛 = deg(𝑓), and {𝑞𝑖 }
𝑖
is an infinite set, a direct contradiction to Proposition 7 and
𝑖=𝑘
impossible for 𝑥1 an algebraic number. Thus 𝑥1 is not an algebraic number. □
Final conclusion 11.
−𝑛!
By Theorem 10 the number 𝑥1 = ∑∞
cannot be the root of any polynomial over ℤ
𝑛=1 10
(equivalent, ℚ), hence is not algebraic, that is to say, 𝑥1 is a transcendental number. □