Lecture 16
Inequalities about irrational numbers
𝑚
16.1. Problem. Prove that if the positive integers 𝑚, 𝑛 satisfy the inequality √3 ‒ 𝑛 > 0 then
𝑚
√3−1
√3 ‒ 𝑛 ≥ 𝑚𝑛 .
𝑚
Solution: Since √3 ‒ 𝑛 > 0 and 𝑚, 𝑛 are positive integers, 3𝑛2 > 𝑚2 . So 3𝑛2 ≥ 𝑚2 + 1. But the
equality 3𝑛2 = 𝑚2 + 1 is impossible because 𝑚2 + 1 is not divisible by 3. Therefore 3𝑛2 ≥ 𝑚2 +
2, from which we obtain successively
3𝑛2 −𝑚2 ≥ 2,
(√3𝑛 − 𝑚)(√3𝑛 + 𝑚) ≥ 2,
2
√3𝑛 − 𝑚 ≥ √3𝑛+𝑚.
We will now consider two cases.
𝑚
1st case. Suppose that 𝑛 ≥ 1. In this case
2
√3𝑛+𝑚
≥
√3−1
. Let us prove this. It can be simplified to
𝑚
2𝑚 ≥ (√3 − 1)(√3𝑛 + 𝑚). By expanding we obtain 2𝑚 ≥ 3𝑛 − √3𝑛 + √3𝑚 − 𝑚.
Rewrite the last inequality as 3𝑚 − √3𝑚 ≥ 3𝑛 − √3𝑛, which is true because 𝑚 ≥ 𝑛. So, we
proved that in this case √3𝑛 − 𝑚 ≥
2
√3𝑛+𝑚
≥
√3−1
. By dividing both sides to 𝑛 we obtain the
𝑚
required inequality.
𝑚
𝑚
√3−1
2nd case. Suppose now that 𝑛 < 1. Then √3 ‒ 𝑛 > √3 − 1 ≥ 𝑚𝑛 .
Note: The problem was inspired by the problem 5.11 in Russian book [1]. It was proposed in
𝑚
1
Romanian Mathematical Olympiad in 1978. In the problem it was asked to prove √7 ‒ 𝑛 > 𝑚𝑛 if
𝑚
it is known that √7 ‒ 𝑛 > 0 and 𝑚, 𝑛 are positive integers. We have also used ideas from the
𝑚
1
solution of a similar problem in [2] which asks to prove the inequality │√2 ‒ 𝑛 │> 3𝑛2 for all
integers 𝑚, 𝑛 (𝑛 ≠ 0).
𝑚
𝑚
1
16.2. Problem. Prove that for any rational number 𝑛 the inequality│√5 ‒ 𝑛 │> 5𝑛2 holds true.
Solution: It is known that
1 9 11
2.2 < √5 < 2.25 = 2 = <
4 4
4
1st case:
𝑚
11
9
𝑚
11 9
2
1
1
If 𝑛 ≥ 4 then noting the inequality ‒ √5 > ‒ 4 we obtain 𝑛 − √5 > 4 ‒ 4 = 4 = 2 > 5𝑛2 for
𝑛 = 1,2,3 …
2nd case:
11
𝑚
If 4 > 𝑛 > √5 then
𝑚2
𝑚2
𝑚2
1
− 5 ≥ 𝑛2 . Indeed, since 𝑛2 > 5, 𝑚2 > 5𝑛2 ; 𝑚2 − 5𝑛2 > 0; 𝑚2 −
𝑛2
1
𝑚
𝑚
1
5𝑛2 ≥ 1; 𝑛2 − 5 ≥ 𝑛2 . If we factorize we obtain ( 𝑛 − √5) ( 𝑛 + √5) ≥ 𝑛2 .
𝑚
11
9
+ √5 <
+ = 5
𝑛
4
4
𝑚
1
1
1
− √5 ≥ 2 ∙ 𝑚
> 2
𝑛
𝑛
5𝑛
𝑛 + √5
3rd case:
𝑚2
𝑚
𝑚2
1
If √5 > 𝑛 >0 then 5 − 𝑛2 ≥ 𝑛2 . Indeed, since 5 > 𝑛2 , 5𝑛2 > 𝑚2 ; 5𝑛2 − 𝑚2 > 0; 5𝑛2 − 𝑚2 ≥ 1;
𝑚2
1
𝑚
𝑚
1
5 − 𝑛2 ≥ 𝑛2 . Again, if we factorize, then we obtain (√5 − 𝑛 ) (√5 + 𝑛 ) ≥ 𝑛2 .
𝑚
𝑚
1
√5 + 𝑛 > √5 + √5 = 2√5; 2√5 < 5; √5 − 𝑛 ≥
1
𝑚
𝑛2 (√5+ )
𝑛
> 5𝑛2.
4th case:
𝑚
𝑚
𝑚
1
If 𝑛 ≤ 0 then │√5 − 𝑛 │ = √5 − 𝑛 > √5 > 1 > 5𝑛2 for 𝑛 = 1,2,3 ….
Note: The problem was inspired by the problem 17.3 in [2] and explains to a certain degree why
it is difficult to approximate the golden ratio
1+√5
2
by rationals.
𝑚
𝑚
1
16.3. Exercise. Prove that for any rational number 𝑛 the inequality│√2021 ‒ 𝑛 │> 2020𝑛2 holds
true.
𝑚
𝑚
1
16.4. Exercise. Prove that for any rational number 𝑛 the inequality│√3 ‒ 𝑛 │> 4𝑛2 holds true.
𝑚
𝑚
1
16.5. Exercise. Prove that for any rational number 𝑛 the inequality│√7 ‒ 𝑛 │> 6𝑛2 holds true.
16.6. Problem. Which of the numbers is greater?
√2
a) √3
√3
√2
or √2 , b) √2
or √3.
Solution.
a) Raise both numbers to the second power. Let us show that 3√2 > 2√3 . By raising both
√6
sides to the power √6 we obtain (3√2 )
obvious because 9√3 > 8√3 > 8√2 .
> (2√3 )
√6
or (32 )√3 > (23 )√2. But this is
3
b) Raise both numbers to the second power. First note that √2 < 2. We can prove this by
3
9
raising both sides to the second power. Indeed, 2 < 4. Then 2√2 < 22 . On the other hand,
3
we know that 22 < 3. Indeed, if we raise both sides to the second power then we get
23 < 32 . This is true 8 < 9.
References.
[1] Зарубежные математические олимпиады. Под ред. Сергеева И.Н. М.: Наука. Гл. ред.
физ.-мат. лит., 1987. — (Б-ка мат. кружка). — 416 с.
[2] Hans Rademacher, Otto Toeplitz, Von Zahlen und Figuren: Proben mathematischen Denkens
für Liebhaber der Mathematik, Springer Berlin Heidelberg, 2000 - 174 Seiten. (Russian
translation: ONTI, Moscow, 1935)
16.7. Prove that there are irrational numbers 𝑎 > 1, 𝑏 > 1 such that lim |𝑎𝑛 − 𝑏 𝑚 | = +∞,
𝑛→∞
𝑚→∞
where 𝑚, 𝑛 ∈ 𝛮.
Solution. Let 𝑏 > 1 be an irrational algebraic number and 𝑑 > 1 be an irrational algebraic
number of order 𝑘 > 1. It is known that ([1], p. 270) 𝑎 = 𝑏 𝑑 is an irrational number. By
𝑚
𝑐
Liouville’s theorem ([1], p. 258) there is 𝑐 > 0, such that |𝑑 − 𝑛 | ≥ 𝑛𝑘 for all rational numbers
𝑚
𝑛
. Let us prove that 𝑎 = 𝑏 𝑑 and 𝑏 satisfy the conditions:
lim |𝑎𝑛 − 𝑏 𝑚 | = lim |𝑏 𝑑𝑛 − 𝑏 𝑚 |.
𝑛→∞
𝑚→∞
If 𝑛 ≥ 𝑚 then lim |𝑏 𝑑𝑛 − 𝑏 𝑚 | ≥ lim |𝑏 𝑑𝑚 − 𝑏
𝑛→∞
𝑚→∞
𝑚→∞
So, let 𝑛 < 𝑚.
𝑛→∞
𝑚→∞
𝑚|
= +∞.
𝑚
𝑚
lim |𝑏 𝑑𝑛 − 𝑏 𝑚 | = lim 𝑏 𝑚 |(𝑏 𝑛 )𝑑− 𝑛 − 1| ≥ lim 𝑏 𝑛 |(𝑏 𝑛 )𝑑− 𝑛 − 1| ≥
𝑛→∞
𝑚→∞
𝑛→∞
𝑚→∞
±
𝑛→∞
𝑚→∞
𝑐
±
|𝑏 𝑥 𝑘−1 − 1|
𝑐
≥ lim 𝑏 𝑛 |(𝑏 𝑛 ) 𝑛𝑘 − 1| = lim
𝑛→∞
𝑥→+∞
𝑥∈𝑅
𝑏 −𝑥
=
±
′
𝑐
|𝑏 𝑥 𝑘−1 − 1|
= lim
(𝑏 −𝑥 )′𝑥
𝑥→+∞
𝑥∈𝑅
𝑐(𝑘 − 1)𝑏
𝑥
= lim
𝑥→+∞
𝑥𝑘
𝑐
𝑥± 𝑘−1
𝑥
= +∞.
𝑥∈𝑅
√2
It is possible to give a simpler solution using these specific numbers: a=√2 , b=√2, 𝑑 = √2,
1
√2
k=2, с=1/3. Note that |𝑎𝑛 − 𝑏 𝑚 | = |2 2 𝑛 − 22𝑚 |. If 𝑛 > 𝑚 then we obtain
1
√2
1
√2
1
√2−1
𝑚
|2 2 𝑛 − 22𝑚 | > |2 2 𝑚 − 22𝑚 | = 22𝑚 (2 2
− 1).
Both factors in the last expression go to infinity for 𝑚 → ∞. If 𝑛 ≤ 𝑚 the we can use (see
𝑚
1
above) the inequality |√2 − 𝑛 | > 3𝑛2 . Then
√2
1
1
𝑛 ± 1
1
𝑛
𝑚
|2 2 𝑛 − 22𝑚 | = 22𝑚 ⋅ |(22 )(√2− 𝑛 ) − 1| >
1
1
> 22𝑚 ⋅ |(22 ) 3𝑛2 − 1| > 22𝑚 ⋅ |2±6𝑛 − 1| ≥
1
±
1
1
1
1
6𝑛 − 1
2
≥ 22𝑛 ⋅ |2±6𝑛 − 1| = ±22𝑛 ⋅ (2±6𝑛 − 1) = ±
1
2−2𝑛
The last expression can be evaluated using L’Hospital’s rule:
1
1
𝑥 1
1
±
±
±
6𝑥 ⋅ (∓
2
) ln 2
6𝑥
2 6𝑥
2
−1
2
2
6𝑥
± lim
= ± lim
= lim
= +∞.
1
1
𝑥→+∞
𝑥→+∞
𝑥→+∞ 3𝑥 2
1
− 𝑥
− 𝑥
2
2
2
2 (− 2) ln 2
√2
1
So, in the case 𝑛 ≤ 𝑚 for 𝑛 and 𝑚 ≥ 𝑛 going to infinity, |2 2 𝑛 − 22𝑚 | → +∞.
Note that irrationality of the number 2√2 is part of Hilbert’s seventh problem which was solved
by A. Gelfond.
References.
1. Бухштаб А.А. «Теория чисел». УЧПЕДГИЗ. М. 1960.
2. «Квант» №12, 1973