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Some Inequalities on T3 Tree

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DOI: 10.4236/apm.2018.88043    266 Downloads   515 Views   Citations
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ABSTRACT

The article proves several inequalities derived from nodal multiplication on T3 tree. The proved inequalities are helpful to estimate certain quantities related with the T3 tree as well as examples of proving an inequality embedded with the floor functions.

1. Introduction

The T3 tree, which first appeared in [1] and was formerly introduced in [2] , is a perfect complete binary tree that is considered to be a new tool to study integers. The tree can reveal many new properties of integers such as the symmetric properties discovered in [3] and [4] , the genetic property found in [5] , and other properties introduced in [6] and [7] . The tree also shows its big potentiality in factorization of big semiprimes, as seen in [8] and [9] . A recent study found several inequalities related with estimation of multiplication on the tree. This article introduces the main results.

2. Preliminaries

This section lists for later sections the necessary preliminaries, which include definitions, notations and lemmas.

2.1. Definitions and Notations

Symbol T3 is the T3 tree that was introduced in [1] and [2] and symbol N ( k , j ) is by default the node at position j on level k of T3, where k 0 and 0 j 2 k 1 . Number of the level by default begins at zero and index of the position also by default begins at zero. Symbol x is the floor function, an integer function of real number x that satisfies inequality x 1 < x x , or equivalently x x < x + 1 . Symbol A B means conclusion B can be derived from condition A.

For convenience in deduction of a formula, comments are inserted by symbols that express their related mathematical foundations. For example, the following deduction

A = B ( L ) = C ( P ) D

means that, lemma (L) supports the step from B to C, and proposition (P) supports the step from C to D.

2.2. Lemmas

Lemma 1. (See in [1] ) T3 tree has the following fundamental properties.

(P1). Every node is an odd integer and every odd integer bigger than 1 must be on the T3 tree. Odd integer N with N > 1 lies on level log 2 N 1 .

(P2). On level k with k = 0 , 1 , , there are 2 k nodes starting by 2 k + 1 + 1 and ending by 2 k + 2 1 , namely, N ( k , j ) [ 2 k + 1 + 1 , 2 k + 2 1 ] with j = 0 , 1 , , 2 k 1 .

(P3). N ( k , j ) is calculated by

N ( k , j ) = 2 k + 1 + 1 + 2 j , j = 0 , 1 , , 2 k 1

(P4). Multiplication of arbitrary two nodes of T3, say N ( m , α ) and N ( n , β ) , is a third node of T3. Let J = 2 m ( 1 + 2 β ) + 2 n ( 1 + 2 α ) + 2 α β + α + β ; the multiplication N ( m , α ) × N ( n , β ) is given by

N ( m , α ) × N ( n , β ) = 2 m + n + 2 + 1 + 2 J

If J < 2 m + n + 1 , then N ( m , α ) × N ( n , β ) = N ( m + n + 1 , J ) lies on level m + n + 1 of T3; whereas, if J 2 m + n + 1 , N ( m , α ) × N ( n , β ) = N ( m + n + 2 , χ ) with χ = J 2 m + n + 1 lies on level m + n + 2 of T3.

Lemma 2. (See in [10] ) Let α and x be a positive real numbers; then it holds

α x 1 < α x < α ( x + 1 )

Particularly, if α is a positive integer, say α = n , then it yields

n x n x n ( x + 1 ) 1

3. Main Results with Proofs

Proposition 1. For positive integer k and real number x > 0 , it holds

0 2 k x 2 k x { 1 2 k , 0 k log 2 x x , k > log 2 x (1)

Proof. It can see by Lemma 2 that,

2 k x 2 k 2 k x 2 k = x

and

2 k x 2 k x ( 2 k x 2 k + 1 2 k ) x = 1 2 k

Meanwhile, when 2 k > x , or k > log 2 x log 2 x 0 , x 2 k = 0 ; thus

2 k x 2 k x = x .

Consequently (1) holds.

Proposition 2. Let N ( m , α ) and N ( n , β ) be nodes of T3 with 0 m n ; let

J = N ( m , α ) × N ( n , β ) 1 2 2 m + n + 1 (2)

then when J < 2 m + n + 1

2 N ( m , α ) × N ( n , β ) 1 2 m + n + 1 3

N ( m , α ) × N ( n , β ) 1 2 m + n + 2 = 1

N ( m , α ) × N ( n , β ) 1 2 m + n + 2 + σ ( σ 0 ) = 0

and when J 2 m + n + 1

2 N ( m , α ) × N ( n , β ) 1 2 m + n + 2 3

N ( m , α ) × N ( n , β ) 1 2 m + n + 3 = 1

N ( m , α ) × N ( n , β ) 1 2 m + n + 3 + σ ( σ 0 ) = 0

Proof. By Lemma 1 (P4), it knows, when J < 2 m + n + 1 , N ( m , α ) × N ( n , β ) lies on level m + n + 1 of T3 and thus 2 m + n + 2 + 1 N ( m , α ) × N ( n , β ) 2 m + n + 3 1 ; hence it holds

2 = 2 m + n + 2 2 m + n + 1 N ( m , α ) × N ( n , β ) 1 2 m + n + 1 2 m + n + 3 2 2 m + n + 1 < 4

and

1 = 2 m + n + 2 2 m + n + 2 N ( m , α ) × N ( n , β ) 1 2 m + n + 2 2 m + n + 3 2 2 m + n + 2 = 2 1 2 m + n + 1 < 2

Thus

2 N ( m , α ) × N ( n , β ) 1 2 m + n + 1 3

and

N ( m , α ) × N ( n , β ) 1 2 m + n + 2 = 1

and thus

N ( m , α ) × N ( n , β ) 1 2 m + n + 2 + σ ( σ 1 ) = 0

Similarly, when J 2 m + n + 1 , N ( m , α ) × N ( n , β ) lies on level m + n + 2 of T3 and 2 m + n + 3 + 1 N ( m , α ) × N ( n , β ) 2 m + n + 4 1 and it holds

2 = 2 m + n + 3 2 m + n + 2 < N ( m , α ) × N ( n , β ) 1 2 m + n + 2 2 m + n + 4 2 2 m + n + 2 = 4 1 2 m + n + 1 < 4 2 N ( m , α ) × N ( n , β ) 1 2 m + n + 2 3

and

1 = 2 m + n + 3 2 m + n + 3 < N ( m , α ) × N ( n , β ) 1 2 m + n + 3 2 m + n + 4 2 2 m + n + 3 = 2 1 2 m + n + 2 < 2 N ( m , α ) × N ( n , β ) 1 2 m + n + 3 = 1 N ( m , α ) × N ( n , β ) 1 2 m + n + 3 + σ ( σ 1 ) = 0

Proposition 3. Let N ( m , α ) be a node of T3 and n be an integer with 0 m n ; then it holds

1 < N ( m , α ) 1 2 n + 1 1 < 1 (3)

0 < N ( m , α ) + 1 2 n + 1 2 (4)

Thus for arbitrary integer σ 0

1 2 σ < N ( m , α ) 1 2 n + 1 + σ 1 2 σ < 1 2 σ (5)

0 < N ( m , α ) + 1 2 n + 1 + σ 2 1 σ (6)

Proof. Considering that 2 m + 1 + 1 N ( m , α ) 2 m + 2 1 holds for arbitrary m 0 , it yields

1 + 1 2 n m = 2 m + 1 2 n + 1 1 N ( m , α ) 1 2 n + 1 1 2 m + 2 2 2 n + 1 1 = 1 2 n m 1 1 2 n 1 (7)

and

0 < 1 2 n m + 1 2 n = 2 m + 1 + 2 2 n + 1 N ( m , α ) + 1 2 n + 1 2 m + 2 2 n + 1 = 2 2 n m 2 (8)

Consider in (7)

1 2 n m 1 1 2 n 1 = { 1 1 2 n < 1 , n = m 1 2 n < 0 , n = m + 1 1 2 n m 1 1 2 n 1 < 0 , n > m + 1

and

1 2 n m 1 = { 0 , n = m 1 + 1 2 n m > 1 , n > m

it knows (3) and (4) hold and consequently (5) and (6) hold.

Proposition 4. Let N ( m , α ) and N ( n , β ) be nodes of T3 with 0 m n ; then it holds

N ( m , α ) + N ( m , α ) 1 2 n + 1 N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) N ( m , α ) + 1 2 n + 1 (9)

and thus for arbitrary integer σ 0 it holds

N ( m , α ) 2 σ + N ( m , α ) 1 2 n + 1 + σ N ( m , α ) × N ( n , β ) 1 2 n + 1 + σ N ( m , α ) 2 σ 1 N ( m , α ) + 1 2 n + 1 + σ (10)

Consequently, it yields

N ( m , α ) N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) 1 (11)

N ( m , α ) 1 2 1 N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 1 (12)

and

N ( m , α ) 2 2 2 < N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 2 (13)

N ( m , α ) 1 2 2 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 (13*)

Proof. The condition that N ( n , β ) is a node of T3 leads to

2 n + 1 + 1 N ( n , β ) 2 n + 2 1

Then direct calculation shows

N ( m , α ) × N ( n , β ) 1 2 n + 1 N ( m , α ) N ( m , α ) 1 2 n + 1 = N ( m , α ) × N ( n , β ) 1 2 n + 1 N ( m , α ) N ( m , α ) + 1 2 n + 1 = N ( m , α ) × ( N ( n , β ) ( 2 n + 1 + 1 ) ) 2 n + 1 0

and

N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) + N ( m , α ) + 1 2 n + 1 = N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) + N ( m , α ) + 1 2 n + 1 = N ( m , α ) × ( N ( n , β ) ( 2 n + 2 1 ) ) 2 n + 1 0

Hence (9) holds.

Multiplying each item in (9) by 1 2 σ for integer σ 1 immediately yields (10).

By definition of the floor function, it holds

N ( m , α ) × N ( n , β ) 1 2 n + 1 1 < N ( m , α ) × N ( n , β ) 1 2 n + 1 N ( m , α ) × N ( n , β ) 1 2 n + 1

By the Inequalities (3), (4) and (9) it yields

N ( m , α ) + N ( m , α ) 1 2 n + 1 1 _ _ N ( m , α ) × N ( n , β ) 1 2 n + 1 1 < N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) 1 N ( m , α ) 1 < N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) 1 N ( m , α ) N ( m , α ) × N ( n , β ) 1 2 n + 1 2 N ( m , α ) 1

which says (11) holds.

Likewise, by definition of the floor function and referring to the Inequalities (5), (6) and (10), it yields

N ( m , α ) × N ( n , β ) 1 2 n + 2 1 < N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 2 + N ( m , α ) 1 2 n + 2 1 N ( m , α ) × N ( n , β ) 1 2 n + 2 1 < N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) N ( m , α ) + 1 2 n + 2 N ( m , α ) 2 + N ( m , α ) 1 2 n + 2 1 < N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 1 N ( m , α ) 1 2 + N ( m , α ) 1 2 n + 2 1 2 _ _ < N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 1 N ( m , α ) 1 2 1 2 < N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 1 N ( m , α ) 1 2 1 N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) 1 (14)

which is the (12).

Similarly, the Inequalities (10) and the definition of the floor function lead to

N ( m , α ) 2 2 + N ( m , α ) 1 2 n + 3 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 2 N ( m , α ) + 1 2 n + 3

and

N ( m , α ) × N ( n , β ) 1 2 n + 3 1 < N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) × N ( n , β ) 1 2 n + 3

Then referring to the Inequalities (5) and (6), it immediately results in

N ( m , α ) 2 2 + N ( m , α ) 1 2 n + 3 1 < N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 2 N ( m , α ) + 1 2 n + 3 N ( m , α ) 2 2 + N ( m , α ) 1 2 n + 3 1 2 2 _ _ 3 4 < N ( m , α ) × N ( n , β ) 1 2 n + 3 < N ( m , α ) 2 N ( m , α ) 2 2 1 4 3 4 < N ( m , α ) × N ( n , β ) 1 2 n + 3 < N ( m , α ) 1 2 + 1 2 N ( m , α ) 2 2 1 < N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 2

N ( m , α ) 2 2 < 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 N ( m , α ) 1 2 3 2 < 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 N ( m , α ) 1 2 1 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 (15)

which is just the (13).

Proposition 5. Let N ( m , α ) and N ( n , β ) be nodes of T3 with 0 m n ; then it holds for integer 0 s m

N ( m , α ) 2 s + 2 + 1 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 2 N ( m , α ) 1 (16)

and

N ( m , α ) 1 2 2 s + 2 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1 (17)

Proof. By Lemma 2 and Proposition 1, it holds when 0 s m

2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 ( P 1 ) N ( m , α ) × N ( n , β ) 1 2 n + 1 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 1 ( L 2 ) N ( m , α ) × N ( n , β ) 1 2 n + 1 + 1 2 s + 2

and

2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 ( L 2 ) 2 s + 2 × N ( m , α ) × N ( n , β ) 1 2 n + s + 2 ( 2 × N ( m , α ) × N ( n , β ) 1 2 n + 2 + 1 2 ) = 2 × N ( m , α ) × N ( n , β ) 1 2 n + 1 N ( m , α ) × N ( n , β ) 1 2 n + 1 + 1 ( L 2 ) 2 N ( m , α ) × N ( n , β ) 1 2 n + 1 1 N ( m , α ) × N ( n , β ) 1 2 n + 1 + 1 = N ( m , α ) × N ( n , β ) 1 2 n + 1

That is

N ( m , α ) × N ( n , β ) 1 2 n + 1 + 1 2 s + 2 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 N ( m , α ) × N ( n , β ) 1 2 n + 1

By (11) it holds

N ( m , α ) + 1 2 s + 2 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 2 2 N ( m , α ) 1

which is just the (16).

Similarly it holds

N ( m , α ) × N ( n , β ) 1 2 n + 2 + 1 2 s + 2 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) × N ( n , β ) 1 2 n + 2

and by (12) it yields

N ( m , α ) 1 2 2 s + 2 2 s + 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 + s 2 N ( m , α ) × N ( n , β ) 1 2 n + 3 N ( m , α ) 1

4. Conclusion

The T3 tree is emerging its value in studying integers. A lot of equations and inequalities will be research objectives. Since most of the inequalities on the T3 tree are in the form of floor functions, their proofs are often skillful. The inequalities proved in this article are not only quite useful for knowing the T3 tree, but also excellent samples for proving inequalities with the floor functions. Hope it helpful to the readers of interests.

Acknowledgements

The research work is supported by the State Key Laboratory of Mathematical Engineering and Advanced Computing under Open Project Program No. 2017A01, Department of Guangdong Science and Technology under project 2015A010104011, Foshan Bureau of Science and Technology under projects 2016AG100311, Project gg040981 from Foshan University. The authors sincerely present thanks to them all.

Conflicts of Interest

The authors declare no conflicts of interest.

Cite this paper

Wang, X. (2018) Some Inequalities on T3 Tree. Advances in Pure Mathematics, 8, 711-719. doi: 10.4236/apm.2018.88043.

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