Irrational Numbers & Arguments: Theorems & Applications

Irrational Numbers and Arguments
CANBANKAN
Contents
1
Introduction
1.1 Form of this content . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Soft Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
II
II
II
2 Dirichlet and Pigeonhole
2.1 Dirichlet’s Approximation Theorem . . . . . . . . . . . . . . . . . . . . .
2.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
3 Beatty Sequences
4
4 Density
4.1 Functional Equations . . . .
4.1.1 First configuration .
4.1.2 Second Configuration
4.1.3 Example Problem . .
4.2 Basis . . . . . . . . . . . . .
4.3 Linear Dependence over Z .
4.4 Exercises . . . . . . . . . . .
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5 Detailed Solutions to Selected Problems
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1 Introduction
Thanks to Aritra Mondal for supplying the TeX template, and Aritra Mondal and k12byda5h for
proofreading this article and providing me with valuable suggestions.
1.1 Form of this content
The handout discusses various methods to deal with arguments involving irrational numbers,
fractional parts, and functional equations.
1.2 Soft Prerequisites
The reader is expected to know what functional equations are, and have the mathematical
maturity to handle (mod t) where t = 1 or is not necessarily an integer.
II
2 Dirichlet and Pigeonhole
2.1 Dirichlet’s Approximation Theorem
Let ||x|| denote the distance between x and the closest integer to x. More rigorously, ||x|| = {x}
if {x} < 12 and 1 − {x} otherwise.
Theorem 2.1 — For any integer N > 1 and real α there exists an integer 1 ≤ q ≤ N such
that
||qα|| ≤
1
N +1
Proof. Consider {α}, {2α}, · · · , {N α}. Sort them into 0 < a1 < a2 < · · · < aN < 1. Let
d1 = a1 , dk = ak − ak−1 for 2 ≤ k ≤ N and dN +1 = 1 − aN .
Since
N
+1
X
di = 1 → E[di ] =
i=1
There exists k such that dk <
1
N +1
1
. Let ak = {xα} and ak−1 = {yα} then
N +1
|| (|x − y|) α|| ≤
1
N +1
2.2 Applications
One surprising application of it is being able to prove existence of a, b such that a2 + b2 = p.
There are other ways to prove this, such as using a Gauss sum to generate an element z ∈ Z[i]
√
such that |z| = p, but it works here.
Theorem 2.2 — Let p ≡ 1(mod 4) be a prime. There exists a, b ∈ N such that a2 + b2 = p
Proof. Let x2 ≡ −1(mod p) where x is an integer, which exists as p ≡ 1(mod 4).
1
√
By dirichlet approximation theorem on N = ⌊ p⌋ and α = xp , there exists q < N with
1
||qα|| < √
p
I claim a = q, b = p||qα|| works. Since b ≡ xa or b ≡ −xa(mod p), p | a2 + b2 . Also, a, b ∈ N
√
√
and a, b < p. Therefore, 0 < a2 + b2 < 2 p2 = 2p, so a2 + b2 = p, as desired.
Exercise 2.1. Let p ≡ 1(mod 6). Prove there exists 0 < a, b < p such that
p5 |(a + b)p − ap − bp
We will now present a more complex example that uses the same principle with a lot of
motivation:
Example 2.1. (Vietnam 2021 TST 6) Let n ≥ 3 be a positive integer and p be a prime
number such that p > 6n−1 − 2n + 1. Let S be the set of n positive integers with different
residues modulo p. Show that there exists a positive integer c such that there are exactly
two ordered triples (x, y, z) with distinct elements, such that x + y − z − c is divisible by
p.
Okay, the condition that there are exactly two ordered tuples suggest only (x, y, z) and (y, x, z)
satisfy the condition for c. If the domain is not in Zp but in Z or R, then the problem is
easy: sort the elements of S as s1 < s2 < · · · < sn , then (s1 , s2 , sn ) works for size reasons, for
x + y − z ≥ s1 + s2 − sn with equality only holding at (x, y, z) ∈ {(s1 , s2 , sn ), (s2 , s1 , sn )}
If the si ’s are within some interval of length p3 , we’re done by the same argument. Luckily, we
are still done if R(asj + b, p) is within an interval of length p3
We can shift b, so we can have
asj + b
p
<
1
6
and be done. This should ring some bells about dirichet’s approximation theorem and its proof
principles.
We can first settle the case where p > 6n with a simple pigeonhole argument that satisfies
asj
p
<
1
6
Consider blocks (x1 , x2 , · · · , xn ) where xj ∈ {0, 1, 2, 3, 4, 5}.
2
Define
as1
as2
asn
f (a) =
6
, 6
,··· , 6
p
p
p
There exists 1 ≤ a1 , a2 ≤ 6n + 1 st f (a1 ) = f (a2 ). Then |a1 − a2 | satisfies the criteria.
Now, we invoke the other degree of freedom b that finishes the same argument for p > 6n−1 . In
(Z6 )n , we put (c, · · · , xn + c) in the same group for fixed x2 , · · · , xn and c varies in Z6 . Repeat
the same pigeonhole principle argument suggests there is a point in f (a1 ) = (c, x2 +c, · · · , xn +c)
and f (a2 ) = (d, x2 + d, · · · , xn + d).
WLOG a1 > a2 , then a1 − a2 satisfies f (a1 − a2 ) has all entries in {d − c, d − c − 1} mod 6,
(a −a )s
d−c+1
which means { 1 p 2 j } is contained in d−c−1
6 ,
6
Finally, we optimize the argument a bit to make it work for p > 5 · 6n−2 + 1. As mentioned
before, only (x2 − x1 , x3 − x1 , · · · , xn − x1 ) matters. By pigeonhole principle, there exists
j ∈ {0, · · · , 5} such that at least 5 · 6n−3 + 1 satisfy x2 − x1 ≡ j(mod 6). Repeating the same
argument, there exists j2 , · · · , jn−1 and at least 6 a’s (call a1 , · · · , a6 ) such that they all have
xt − x1 = j t .
Define xi,n st f (ai ) = (ci , j2 + ci , · · · , jn−1 + c, xi,n + ci ). We have
(ai − aj )sk
ci − cj − 1 ci − cj + 1
∈
,
p
6
6
for all 1 ≤ k ≤ n − 1. Therefore, we only need to worry about sn working.
For 1 ≤ i ≤ 6, let g(i) ∈ [0, 1) such that
g(i) ≡
ai sn ci
− (mod 1)
p
6
Suppose g(i1 ) < g(i2 ) < · · · < g(i6 ) < 1 + g(i1 ). Say t satisfies g(it ) − g(it−1 ) < 16 , for the case
where 1 + g(i1 ) − g(i6 ) < 16 is similar. Then
(ait − ait−1 )sn cit − cit−1
ait−1 sn cit−1
ait sn cit
1
−
=
−
−
−
<
p
6
p
6
p
6
6
Therefore, picking a = ait − ait−1 satisfies
cit − cit−1 − 1 cit − cit−1 + 1
asj
∈
,
(mod 1)
p
6
6
For all 1 ≤ k ≤ n. The conclusion follows.
3
3 Beatty Sequences
The Beatty sequence for an irrational α is defined as ⌊α⌋, ⌊2α⌋, · · · ,
Theorem 3.1 — The Beatty sequence for α and for β partition N only if
1
α
+
1
β
= 1.
This is not too hard to prove. Hint: Consider the number of terms less than N in the Beatty
sequence for α.
Problem 1. (Serbia MO 22/6) Let p and q be different primes, and α ∈ (0, 3) a real number.
Prove that in sequence
⌊α⌋, ⌊2α⌋, ⌊3α⌋ . . .
there is a number less than 2pq and divisible by p or q.
Problem 2. Let n be a positive integer. Prove that
n
X
(−1)⌊k(
√
2−1)⌋
≥ 0.
k=1
Problem 3. Do there exist two positive reals α, β such that each positive integer appears exactly
once in the following sequence?
2020, [α], [β], 4040, [2α], [2β], 6060, [3α], [3β], · · ·
If so, determine all such pairs; if not, prove that it is impossible.
4
4 Density
Call a set S ⊂ R dense if the following is true:
For any x ∈ R and ϵ ∈ R+ , there exists s ∈ S such that |s − x| < ϵ.
We can therefore use the definition of density to stuff something from a dense set between
two numbers with possibly small but fixed difference to get a contradiction. It has many
applications. To begin with, it is a great tool for solving functional equations.
4.1 Functional Equations
4.1.1 First configuration
Theorem 4.1 — If f, g, h : R+ → R+ such that
f (x + g(y)) = f (x) + h(y)
then
g(y)
doesn’t depend on y.
h(y)
Proof. Suppose
g(y)
g(z)
<
h(y)
h(z)
The idea is note that f (x + mg(z) − ng(y)) = f (x) + mh(z) − nh(y) if x + mg(z) − ng(y) > 0.
Since Q is dense, there exists integers m, n such that
g(y)
m
h(y)
<
<
g(z)
n
h(z)
Notice that mg(z) − ng(y) > 0, mh(z) − nh(y) < 0,
Therefore, for any x, we can select k ∈ N such that
f (x) + k(mh(z) − nh(y)) < 0
Thus, f (x + mkg(z) − nkg(y)) < 0 but x + mkg(z) − nkg(y) > 0
5
4.1.2 Second Configuration
Theorem 4.2 — Let c > 0 be a constant. If f, g : R+ → R+ such that
f (x + g(y)) = f (x) + cg(y)
for all x, y ∈ R+ , then either f (x) ≥ cx for all x, or Im(g) = dZ for some d ∈ R
Sketch. The key idea is that if f (x) < cx for some x, then if a1 t1 + · · · + ak tk > 0 where ai ∈ Z
and t1 , · · · , tk ∈ Im(g) then a1 t1 + · · · + ak tk > p for some constant p depending on cx − f (x),
which proves the theorem. The details are left as an exercise for the reader.
Remark. There does exist a function such that
f (x + g(y)) = f (x) + cg(y)∀x, y ∈ R+
f is nonlinear and Im(g) ̸⊂ dZ for any real d.
For example, if Im(g) = {1,
√ e}, then let S be {a + be|a, b ∈ Z}, there exists a function
where c = 1, f (1) = 10, f ( 2) = π.
If we ignore the size restraint that f ’s range is the set of positive reals, the only restraint
on f is that f (x) − f (y) = c(x − y) for all x − y ∈ S.
4.1.3 Example Problem
Problem 4. Solve over R+ :
f (f (x)f (f (x)) + y) = f (y) + xf (x)
Basically, this problem gets trivialized by the two above configurations. I will leave this as an
exercise to the readers.
Example 4.1. Find all functions f from the interval (1, ∞) to (1, ∞) with the following
property: if x, y ∈ (1, ∞) and x2 ≤ y ≤ x3 , then
(f (x))2 ≤ f (y) ≤ (f (x))3 .
The answer is f (x) ≡ xc for some c ∈ R>0 , which clearly work. All logs are done in the
same base (for example, e) Proof. Let g(log x) = log(f (x)). Then g is from R+ → R+ . Let
a = log x, b = log y. If 2a ≤ b ≤ 3a then 2g(a) < g(b) < 3g(a)
6
Let h(log x) = log g(x). Then h is from R → R. Let c = log a, d = log b, α = log 2, β = log 3,
then if d − c ∈ [α, β] then h(d) − h(c) ∈ [α, β].
This implies for positive integer t, u, h(x + tα) − h(x) ≥ tα and h(x + uβ) − h(x) ≤ uβ If t, u
are large,
h(x + tα + ϵ) − h(x) ≥ tα, h(x + uβ − ϵ) − h(x) ≤ uβ
still holds.
I claim h(x) − h(y) = x − y by proving h(x) − h(y) ≥ x − y, and h(x) − h(y) ≤ x − y is similar.
This proves h(x) − x is constant, so h(x) = x + C, g(x) = eh(log x) = elog x+C = cx where c = eC ,
and f (x) = eg(log x) = ec log x = xc
Select t, u large such that h(x) − h(y) < tα − uβ < x − y, which exist because
ϵ1 , ϵ2 > 0 such that x − tα − ϵ1 = y − uβ + ϵ2 = z. Then
β
α
∈
/ Q. Select
h(z) = h(x − t(α + ϵ1 /t)) ≤ h(x) − tα < h(y) − uβ ≤ h(y − u(β − ϵ/u)) = h(z)
Which works if t, u are so large that α +
ϵ1
t ,β
−
ϵ2
u
∈ [α, β]
Here, the key idea is to stuff tα − uβ into [h(x) − h(y), x − y] to get the desired contradiction.
Problem 5. Find all f : R+ → R+ such that
f (x + y) = f (x) + f (y)
Problem 6. Find all f : R+ → R+ satisfying the following condition: for any three distinct real
numbers a, b, c, a triangle can be formed with side lengths a, b, c, if and only if a triangle can
be formed with side lengths f (a), f (b), f (c)
4.2 Basis
Sometimes, the set of solutions to an FE satisfy the following:
If f work, g work, then f ± g works.
In this case, it is very helpful to think of f, g as part of a ”basis”.
Here is an example problem.
Example 4.2. Solve over R+ :
xf (x) + yf (y) = (x + y)f
7
x2 + y 2
x+y
Before we start with the proof, we guess the solution set: f (x) = ax + b. We know it has two
degrees of freedom: a, b. Therefore, if we know f (x1 ), f (x2 ) and if f (x) = ax + b is the only
solution, the function is unique. Therefore, we fix two degrees of freedom, which in turn fixes
a, b, and then consider a set S = {x|f (x) = ax + b}. We prove it’s dense and in turn to prove
S = R+
Solution: I claim the only functions that work are f (x) = ax + b where a,b are nonnegative
with at least one of them positive. They clearly work.
Let f (1) = t + u and f (1.01) = 1.01t + u. Let S = {x : f (x) = xt + u}
Main Claim: S is dense.
The motivation for this claim is that if this is true, then we can use bounding to prove that S
is actually the entire domain.
Proof. We first we prove it for [1,1.01]
If l < r, l, r ∈ S, then
l2 + r 2
∈S
l+r
Furthermore, if k, m satisfy
l2 +k2
l+k
= r and
r2 +m2
r+m
= l, they are also in S.
Suppose there exist m in [l, r] such that we cannot get any number x with |m − x| < ϵ for some
ϵ
Then we divide and conquer: in this interval, this difference is not too small compared to simple
binary search, and as it gets smaller it gets closer to binary search. The details will be given to
the reader as an exercise.
With this in mind, we can reach arbitrarily large values (and fill them in). Reaching arbitrarily
small values require more analysis: if x0 = 1.01, x1 = 1 we can repeatedly do the following:
Take
xi+1
xi
∈ [0.99, 1] through density in [xi , 1.1]
Where x2i+1 + t2 = (xi+1 + t)x for a carefully chosen t ∈ S that should be smaller than cxi but
greater than xi for a c independent of i that follows from density.
With this in mind, if f (k) > tk + u, we can select
k 2 + ϵ2
∈S
k+ϵ
such that
kf (k) > (k + ϵ)f
k 2 + ϵ2
k+ϵ
by expanding the right hand side and picking epsilon smaller than the difference (which exists
by density).
8
To be more clear, if kf (k) = k (tk + u + δ) and we can pick ϵ small enough such that
2
k + ϵ2
(k + ϵ)(t
+ u) − k(tk + u) < kδ
k+ϵ
This suggests f (ϵ) < 0, the desired contradiction.
Problem 7. Find all functions f : R → R such that for all real numbers a, b, and c:
(i) If a + b + c ≥ 0 then f (a3 ) + f (b3 ) + f (c3 ) ≥ 3f (abc).
(ii) If a + b + c ≤ 0 then f (a3 ) + f (b3 ) + f (c3 ) ≤ 3f (abc).
Problem 8. Determine all functions f : (0, ∞) → R satisfying
y
1
x+
f (y) = f (xy) + f
x
x
for all x, y > 0.
Problem 9. Find all functions f : Z → R such that
f (x + y) + f (1) + f (xy) = f (x) + f (y) + f (1 + xy)
4.3 Linear Dependence over Z
The dense set is typically of the form {mα + nβ|m, n ∈ Z}. It is dense if
be very useful.
α
β
∈
/ Q and that can
Theorem 4.3 (Weak Kronecker) — If α, β, 1 are linearly independent over Z, then
({nα}, {nβ}) over [0, 1)2 is dense. In other words, for any [l1 , r1 ], [l2 , r2 ] ⊆ [0, 1), there
exists an integer n such that {na} ∈ [l1 , r1 ], {nb} ∈ [l2 , r2 ], which readily implies the
density (by shifting the rectangle boundaries for ({na}, {nb}) to make them disjoint but
arbitrarily close)
Proof. To prove this claim, we will first prove a lemma.
Lemma. For any ϵ > 0, there exists n such that ||nα||, ||nβ|| < ϵ
9
1
Proof. Since ϵ > 0, there exists an integer m such that ϵ > m
. Divide [0, 1) × [0, 1) into
b b+1
a a+1
×
,
,
m m
m m
where 0 ≤ a, b < m. Place points ({nα}, {nβ}). There are m2 regions, so if there are sufficiently
many points, two points will be in the same region, call ({xα}, {xβ}) and ({yα}, {yβ}). Then
x − y works.
Pick
ϵ<
r1 − l1 r2 − l2
,
,n
1000
1000
such that ||nα||, ||nβ|| < ϵ.
WLOG, {nα}, {nβ} < ϵ, for the other cases are similar. Let x = {nα} and y = {nβ}.
1
Pick k = ⌈ A+l
x ⌉, where A is an integer to be determined later. Then A + l1 < kx < A + l1 + x,
so {kx} ∈ (l1 , l1 + x). Also,
A
y l1 y
y l1 y
+
< ky < A +
+y
x
x
x
x
Since xy ∈
/ Q, it follows that {A xy } is dense in (0, 1), so we can choose k such that {ky} ∈ [l2 , r2 ].
nk satisfies the original criteria.
Here are a few exercises:
4.4 Exercises
Problem 10 (China TST 2022 Test 3 Problem 2). Two positive real numbers α, β satisfies that for
any positive integers k1 , k2 , it holds that ⌊k1 α⌋ ̸= ⌊k2 β⌋, where ⌊x⌋ denotes the largest integer
less than or equal to x. Prove that there exist positive integers m1 , m2 such that
m1 m2
+
=1
α
β
Problem 11 (China TST 2017 Test 2 Problem 6). Let M be a subset of R such that the following
conditions are satisfied:
• For any x ∈ M, n ∈ Z, one has that x + n ∈ M.
• For any x ∈ M , one has that −x ∈ M .
• Both M and R\M contain an interval of length larger than 0.
For any real x, let M (x) = {n ∈ Z+ |nx ∈ M }. Show that if α, β are reals such that
M (α) = M (β), then we must have one of α + β and α − β to be rational.
10
Problem 12. Let a, b, c, d be positive irrational numbers such that
⌊an⌋ + ⌊bn⌋ = ⌊cn⌋ + ⌊dn⌋
For all n ∈ N. Prove that one of the following holds:
• a ≡ c(mod 1) and b ≡ d(mod 1)
• a ≡ d(mod 1) and b ≡ c(mod 1)
• a + b ∈ N and c + d ∈ N
5 Detailed Solutions to Selected Problems
Problem 1. Let p and q be different primes, and a ∈ (0, 3) a real number. Prove that in
sequence
[a] , [2a] , [3a] . . .
exists number less than 2pq, divisible by p or q.
Solution 1. The case where p = 2 implies the sequence will contain all odd numbers up to 4q,
so assume p, q ≥ 3.
n o
Note p|[ta] ↔ ta
< p1
p
ta
Let tp be the minimal number such that { (t+1)a
p } > { p }, and let
(t + 1)a
ta
Sp = t|
>
and (t + 1)a < 2pq
p
p
The key is to bound
contradiction.
t
pa
or
t+1
p a
closely, and then use a similar bound for q to derive a
Case 1: Not all elements in Sp are divisible by tp .
Claim: 2t ∈
/ Sp .
Proof: Let m be the minimal integer such that mt ∈
/ S, (m − 1)t ∈ S → mt + 1 ∈ S.
n o
Let ϵ = 1 − ta
p . From
(m − 1)t ∈ S → 1 − (m − 1)ϵ = { (m−1)ta
}>1−
p
mt ∈
/S→ϵ>
This implies
a−1
p .
This implies ϵ <
a−1
p(m−1)
a
pm .
m
m−1
>
a
a−1 ,
so a > m, which means m = 2 is forced, proving our claim.
11
a
Let J = (t+1)a
because 2t + 1 ∈ S → 2J < ap . Therefore,
− 1. Then, note p1 < J < 2p
p
a
a
1
a
a
0 < 3−a
p < 3J − p < 2p , which implies p < 3J − p < 2p if (3(t + 1) − 1)a < 2pq. Repeating the
a
same argument yields p1 < (2n + 1)J − n ap < 2p
if ((2n + 1)(t + 1) − n)a < 2pq
2−a
We get 2t+1
a
−
1
∈
,
0
if n(2t + 1) + (t + 1) < 2pq
2p
a . Since n(2t + 1) < 2p, we can get
(4n+2)p
n≥q−1
Case 2: all elements in Sp are a multiple of tp .
Let
ϵ=1−
ta
p
Note { mta
p } = 1 − mϵ. Pick the largest m such that
(mt + 1)a < 2pq
Since ta < p it follows that m ≥ 2q − 1. We also have
{
(mt + 1)a
1
}>
p
p
which means
1 − (2q − 1)ϵ < 1 −
so ϵ <
a−1
p
a−1
p(2q−1)
This means
1>
t
a−1
·a>1−
p
p(2q − 1)
Now we combine the same bound for q.
a falls into case 1 for p, q: for some t, u,
2t + 1 2u + 1
a−2
−
a<
2p
2q
(4q − 2)p
but LHS ≥
a
2pq ,
impossible.
a falls into case 1 for p, case 2 for q: analogously we get
a<3
a falls into case 2 for p, q:
a
pq
<
a
2pq
<
a−2
(4q−2)p
+
a−1
p(2q−1) ,
impossible if
a−1
p(2q−1) .
All three cases are impossible. The conclusion follows.
Problem 10. Two positive real numbers α, β satisfies that for any positive integers k1 , k2 ,
it holds that ⌊k1 α⌋ ̸= ⌊k2 β⌋, where ⌊x⌋ denotes the largest integer less than or equal to x.
Prove that there exist positive integers m1 , m2 such that mα1 + mβ2 = 1. Solution 10. Assume
α, β ∈
/ Q, for the other case is obvious.
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Note the number of k1 satisfying ⌊k1 α⌋ < X is precisely ⌊ X
α ⌋.
Let a = α1 and b = β1 . From ⌊k1 α⌋ =
̸ ⌊k2 β⌋ we can see if ⌊Xa⌋ > ⌊(X − 1)a⌋ then ⌊Xb⌋ =
⌊(X − 1)b⌋, for otherwise there exists a solution (k1 , k2 ) to ⌊k1 α⌋ = ⌊k2 β⌋ = X
In other words, if {xa} > 1 − a then {xb} < 1 − b.
Claim: There exists integers m, n, k such that ma + nb = k.
Proof of Claim: We first prove there exists n such that ||na||, ||nb|| < ϵ where ||na|| =
min{1 − {na}, {na}}.
We can easily find ||na|| < ϵ3 , and we note ||na||, ||2na||, · · · , ||⌈ 1ϵ ⌉na|| are all smaller than ϵ.
We can prove with pigeonhole principle at least one of ||tnb|| where 1 ≤ t ≤ ⌈ 1ϵ ⌉ works.
Suppose {na}, {nb} < ϵ. Let x = {na}, y = {nb}. Assume for the sake of contradiction
not rational.
x
y
is
If there exists integers M, A, B such that M x ∈ (A, A + a) and M y ∈ (B, B + b), then I am
done, because {M na} = {(M n − 1)a} + 1 and {M nb} = {(M n − 1)b} + 1.
A+a
A+a
We take M = ⌊ A+a
x ⌋. This guarantees x( x − 1) = A + a − x ≤ M x ≤ x x = A + a. Since
a > x, M x ∈ (A, A + a)
We have M y ∈ ((A + a) xy − y, (A + a) xy ). We need to show for some A, there exists an integer
B such that (A + a) xy − y ≥ B and (A + a) xy ≤ B + b. If xy is irrational, there exists A such
that y ≤ {(A + a) xy } ≤ b. Therefore, xy must be rational.
Now, let na = C + x, nb = D + y then x = qy for some q ∈ Q. Therefore,
(na − C) = q(nb − D)
Let q =
u
v
na−C
nb−D
= q →
where u, v ∈ Z then v(na − C) = u(nb − D) → (vn)a − (un)b = vC − uD
Back to the problem if m, n are of opposite signs, say ma ≡ nb(mod 1), then pick ||tb|| = ϵ.
ab
If {tb} = ϵ < mn
then {tmb} = mϵ < ab
n < b and {tma} = {tnb} <
{tma} < {(tm − 1)a} and {tmb} < {(tm − 1)b}, contradiction.
ab
m
< a. It follows that
ab
ab
If {tb} = 1−ϵ > 1− mn
then {tmb} = 1−mϵ > 1− ab
n > 1−b and {tma} = {tnb} > 1− m > 1−a.
It follows that {tma} > {(tm + 1)a} and {tmb} > {(tm + 1)b}, contradiction.
Therefore, m, n are of the same sign. WLOG m, n, k are all positive and k is minimal.
Assume for the sake of contradiction k ≥ 2. Thus, either ma ≥ 1 or nb ≥ 1. WLOG nb ≥ 1, so
it suffices to take {Xa} < a and {Xb} < n1 < b
WLOG gcd(k, m, n) = 1. Let {Xa} = Z + ϵ, then I want Xk − Zm ≡ 1(mod n), which wins
Xk−(Z+ϵ)m
m
because { X(k−ma)
} = { Xk
}
n
n − Xa n } = {
n
But this is doable by density; (X, Xa(mod n)) are dense in Zn × [0, n) so we can clearly take
Xk ≡ Zm + 1(mod n) and {Xa} < ϵ for any ϵ > 0
(another way to think about it is that { Xa
n } is dense in [0, 1) if we only consider X ≡ c(mod n)
for some fixed c)
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