Journal of Applied Mathematics and Physics
Vol.07 No.08(2019), Article ID:94419,11 pages
10.4236/jamp.2019.78121

The Aleksandrov Problem in Non-Archimedean 2-Fuzzy 2-Normed Spaces

Meimei Song, Haixia Jin*

Science of College, Tianjin University of Technology, Tianjin, China

Copyright © 2019 by author(s) and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: July 19, 2019; Accepted: August 16, 2019; Published: August 19, 2019

ABSTRACT

We introduce the definition of non-Archimedean 2-fuzzy 2-normed spaces and the concept of isometry which is appropriate to represent the notion of area preserving mapping in the spaces above. And then we can get isometry when a mapping satisfies AOPP and (*) (in article) by applying the Benz’s theorem about the Aleksandrov problem in non-Archimedean 2-fuzzy 2-normed spaces.

Keywords:

Non-Archimedean 2-Fuzzy 2-Normed Space, Isometry, Benz’s Theorem

1. Introduction

Let X , Y be two metric spaces. For a mapping f : X Y , for all x 1 , x 2 X , if f satisfies,

d Y ( f ( x 1 ) , f ( x 2 ) ) = d X ( x 1 , x 2 )

where d X ( , ) , d Y ( , ) denote the metrics in the spaces X , Y , then f is called an isometry. It means that for some fixed number p > 0 , assume that f preserves distance p; i.e., for all x 1 , x 2 in X, if d X ( x 1 , x 2 ) = p , we can get d Y ( f ( x 1 ) , f ( x 2 ) ) = p . Then we say p is a conservative distance for the mapping f. Whether there exists a single conservative distance for some f such that f is an isometry from X to Y, is the basic issue of conservative distances. It is called the Aleksandrov problem.

Theorem 1.1. ( [1] ) Let X , Y be two real normed linear spaces (or NLS) with dim X > 1 , dim Y > 1 and Y is strictly convex, assume that a fixed real number p > 0 and that a fixed integer N > 1 . Finally, if f : X Y is a mapping satisfies

1) x 1 x 2 = p f ( x 1 ) f ( x 2 ) p

2) x 1 x 2 = N p f ( x 1 ) f ( x 2 ) N p

for all x 1 , x 2 X . Then f is an affine isometry. we can call Benz’s theorem.

We can see some results about the Aleksandrov problem in different spaces in [2] - [10] . A natural question is that: Whether the Aleksandrov problem can be proved in non-Archimedean 2-fuzzy 2-normed spaces under some conditions. So in this article, we will give the definition of non-Archimedean 2-fuzzy 2-normed spaces according to [11] [12] [13] [14] , then by applying the Benz’s theorem to fix the value of p and N to solve problems.

If a function from a field K to [ 0, ) satisfies

(T1) | a | 0 , | a | = 0 a = 0 ;

(T2) | a b | = | a | | b | ;

(T3) | a + b | max { | a | , | b | } .

for all a , b K , then the field K is called a non-Archimedean field.

We can know | 1 | = | 1 | = 1 , | a | 1 for all a N from the above definition. An example of a non-Archimedean valuation (or NAV) is the function | | taking | 0 | = 0 and others into 1.

In 1897, Hence in [15] found that p-adic numbers play a vital role in the complex analysis, the norm derived from p-adic numbers is the non-Archimedean norm, the analysis of the non-Archimedean has important applications in physics.

Definition 1.2. Let X be a vector space and dim X 2 . A function , : X [ 0, ) is called non-Archimedean 2-norm, if and only if it satisfies

(T1) x 1 , x 2 0 , x 1 , x 2 = 0 iff x 1 , x 2 are linearly dependent;

(T2) x 1 , x 2 = x 2 , x 1 ;

(T3) r x 1 , x 2 = | r | x 1 , x 2 ;

(T4) x 1 + x 2 , y max { x 1 , y , x 2 , y }

for all x 1 , x 2 , y X , r K . Then ( X , , ) is called non-Archimedean 2-normed space over the field K.

Definition 1.3. An NAV | | in a linear space X over a field K. A function F : X × [ 0,1 ] is said to be a non-Archimedean fuzzy norm on X, if and only if for all x , x 1 , x 2 X and s , t ,

(F1) F ( x , s ) = 0 with s 0 ,

(F2) F ( x , s ) = 1 iff x = 0 for all s > 0 ,

(F3) F ( c x , s ) = F ( x , s | c | ) , for c 0 and c K ,

(F4) F ( x 1 + x 2 , s + t ) min { F ( x 1 , s ) , F ( x 2 , t ) } ,

(F5) F ( x , ) is a nondecreasing function of s R and lim s F ( x , s ) = 1 .

Then ( X , F ) is known as a non-Archimedean fuzzy normed space (or F-NANS).

Theorem 1.4. Let ( X , F ) be an F-NANS. Assume the condition that:

(F6) F ( x , s ) > 0 for all s > 0 x = 0 .

Define x α = inf { s : F ( x , s ) α } , α ( 0 , 1 ) . We call these α-norms on X or the fuzzy norm on X.

Proof: 1) Let x α = 0 , it implies that inf { s : F ( x , s ) α } = 0 , then for all s R , s > 0 , F ( x , s ) α > 0 , so x = 0 ;

Conversely, assume that x = 0 , by (F2), F ( x , s ) = 1 for all s > 0 , then inf { s : F ( x , s ) α } = 0 for all α ( 0,1 ) , so x α = 0 .

2) By (F3), if c 0 , then

c x α = inf { s : F ( c x , s ) α } = inf { s : F ( x , s | c | ) α }

Let t = s | c | , then

c x α = inf { | c | t : F ( x , t ) α } = | c | inf { t : F ( x , t ) α } = | c | x α

If c = 0 , then

c x α = 0 = c x α

3) We have

max { x α , y α } = max { inf { s , F ( x , s ) α } , inf { t , F ( x , t ) α } } = inf { max { s , t } , F ( x , s ) α , F ( x , t ) α } inf { s + t , F ( x + y , s + t ) F ( x + y , max { s , t } ) min { F ( x , s ) α , F ( x , t ) α } α } inf { r , F ( x + y , r ) α } = x + y α

Example 1.5. Let ( X , ) be a non-Archimedean normed space. Define

F ( x , s ) = ( s s + x , s > 0 , 0 , s 0.

for all x X , Then ( X , F ) is a F-NANS.

Definition 1.6. Let Z be any non-empty set and ( Z ) be the set of all fuzzy sets on Z. For Z 1 , Z 2 ( Z ) and λ K , define

Z 1 + Z 2 = { ( z 1 + z 2 , μ 1 μ 2 ) | ( z 1 , μ 1 ) Z 1 , ( z 2 , μ 2 ) Z 2 }

and

λ Z 1 = { ( λ z 1 , μ 1 ) | ( z 1 , μ 1 ) Z 1 }

Definition 1.7. A non-Archimedean fuzzy linear space X ^ = X × ( 0 , 1 ] over the field K, we define the addition and scalar multiplication operation of X as following: ( x 1 , μ 1 ) + ( x 2 , μ 2 ) = ( x 1 + x 2 , μ 1 μ 2 ) , λ ( x 1 , μ 1 ) = ( λ x 1 , μ 1 ) , if for every ( x 1 , μ 1 ) X , we have a related non-negative real numebr, ( x 1 , μ 1 ) is the fuzzy norm of ( x 1 , μ 1 ) in such that

(T1) ( x 1 , μ 1 ) = 0 x 1 = 0 , μ 1 ( 0 , 1 ] ;

(T2) λ ( x 1 , μ 1 ) = | λ | ( x 1 , μ 1 ) ;

(T3) ( x 1 , μ 1 ) + ( x 2 , μ 2 ) max { ( x 1 , μ 1 μ 2 ) , ( x 2 , μ 1 μ 2 ) } ;

(T4) ( x 1 , t μ t ) = t ( x 1 , μ t ) for all μ t ( 0,1 ] .

for every ( x 1 , μ 1 ) , ( x 2 , μ 2 ) X , λ K , then we say that X is an F-NANS.

Definition 1.8. Let X be a non-empty non-Archimedean field set, ( X ) be the set of all fuzzy sets on X. If f 1 ( X ) , then f 1 = { ( x 1 , μ 1 ) : x 1 X , μ 1 ( 0 , 1 ] } . Clearly, | f 1 ( x 1 ) | 1 , so f 1 is a bounded function. Let K , then ( X ) is a non-Archimedean linear space over the field K and the addition, scalar multiplication are defined as follows

f 1 + f 2 = { ( x 1 , μ 1 ) + ( x 2 , μ 2 ) } = { ( x 1 + x 2 , μ 1 μ 2 ) | ( x 1 , μ 1 ) f 1 , ( x 2 , μ 2 ) f 2 }

and

λ f 1 = { ( λ x 1 , μ 1 ) | ( x 1 , μ 1 ) f 1 }

If for every f ( X ) , there is a related non-negative real number f called the norm of f in such that for all f 1 = ( x 1 , μ 1 ) , f 2 = ( x 2 , μ 2 ) (X)

(T1) f = 0 iff f = 0 . For

f = { ( x 1 , μ 1 ) } = 0

x 1 = 0 , μ 1 ( 0 , 1 ]

f = 0.

(T2) λ f = | λ | f , λ K . For

λ f = { λ ( x 1 , μ 1 ) } = { | λ | ( x 1 , μ 1 ) } = | λ | f

(T3) f 1 + f 2 max { f 1 , f 2 } . For

f 1 + f 2 = { ( x 1 , μ 1 ) + ( x 2 , μ 2 ) } = { ( x 1 + x 2 ) , ( μ 1 μ 2 ) } max { ( x 1 , μ 1 μ 2 ) , ( x 2 , μ 1 μ 2 ) } max { f 1 , f 2 }

Then the linear space ( X ) is a non-Archimedean normed space.

Definition 1.9. ( [4] ) A 2-fuzzy set on X is a fuzzy set on ( X ) .

Definition 1.10. A NAV | , | in a linear space ( X ) over a field K. If a function F : ( X ) 2 × [ 0,1 ] is a non-Archimedean 2-fuzzy 2-norm on X (or a fuzzy 2-norm on ( X ) ), iff for all f 1 , f 2 , f 3 ( X ) , s , t ,

(F1) F ( f 1 , f 2 , s ) = 0 for s 0 ;

(F2) F ( f 1 , f 2 , s ) = 1 iff f 1 , f 2 are linearly dependent for all s > 0 ;

(F3) F ( f 1 , f 2 , s ) = N ( f 2 , f 1 , s ) ;

(F4) F ( c f 1 , f 2 , s ) = N ( f 1 , f 2 , s | c | ) , for c 0 and c K ;

(F5) F ( f 1 , f 2 + f 3 , s + t ) min { F ( f 1 , f 2 , s ) , F ( f 1 , f 3 , t ) } ;

(F6) F ( f 1 , f 2 , ) is a nondecreasing function of R and lim s F ( f 1 , f 2 , s ) = 1 ;

Then ( ( X ) , F ) is called a non-Archimedean fuzzy 2-normed space (or FNA-2) or ( X , F ) is a non-Archimedean 2-fuzzy 2-normed space.

Theorem 1.11. Let ( ( X ) , F ) be an FNA-2. Suppose the condition that:

(F7) N ( f 1 , f 2 , s ) > 0 for all s > 0 f 1 and f 2 are linearly dependent.

Define f 1 , f 2 α = inf { t : N ( f 1 , f 2 , s ) α , α ( 0 , 1 ) } . We call these α-2-norms on ( X ) or the 2-fuzzy 2-norm on X.

Proof: It is similar to the proof of Theorem 1.4.

2. Main Result

From now on, if we have no other explanation, let dim ( X ) 2 , dim ( Y ) 2 . = f h , g h α , = ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β

Definition 2.1. Let ( X ) , ( Y ) be two FNA-2 and a mapping ψ : ( X ) ( Y ) . If for all f , g , h ( X ) and α , β ( 0,1 ) , we have

ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β = f h , g h α ( ∇ )

then ψ is called 2-isometry.

Definition 2.2. For a mapping ψ : ( X ) ( Y ) and f , g , h (X)

1) If = 1 , then = 1 , we say ψ satisfies the area one preserving property (AOPP).

2) If = n , then = n , we say ψ satisfies the area n for each n (AnPP).

Definition 2.3. We say a mapping ψ : ( X ) ( Y ) preserves collinear, if f , g , h mutually disjoint elements of ( X ) , then exist some real number t we have

ψ ( g ) ψ ( h ) = t ( ψ ( f ) ψ (h))

Next, we denote ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β f h , g h α ( ) .

Lemma 2.4. Let ( X ) and ( Y ) be two FNA-2. If 1 , a mapping ψ : ( X ) ( Y ) satisfies ( ) and AOPP, then we can get ( ) where 1 .

Proof: 1) Firstly, we prove that f preserves collinear. We assume that = 0 , according to ( ) , we get

ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β = 0

then ψ ( f ) ψ ( h ) and ψ ( g ) ψ ( h ) are linearly dependent. So we obtain that ψ preserves collinear.

2) Secondly, we prove that when 1 , we can get ( ) .

If

<

Let ω = h + f h f h , g h α , then ω h , g h α = 1 , so

ψ ( ω ) ψ ( h ) , ψ ( g ) ψ ( h ) β = 1 (Δ)

Since

ω f , g h α = f h f h , g h ( f h ) , g h = 1

according to ( ) , we have

ψ ( ω ) ψ ( f ) , ψ ( g ) ψ ( h ) β ω f , g h α = 1

Since f preserves collinear, so there exists a real number s such that

ψ ( ω ) ψ ( h ) = s ( ψ ( f ) ψ (h))

and

ψ ( ω ) ψ ( f ) = ( s 1 ) ( ψ ( f ) ψ (h))

So, we get

ψ ( ω ) ψ ( h ) , ψ ( g ) ψ ( h ) β = | s | | s 1 | + = ψ ( ω ) ψ ( f ) , ψ ( g ) ψ ( h ) β + < 1 + = 1

This contradicts with Δ .

Lemma 2.5. Let ( X ) and ( Y ) be two FNA-2. If a mapping ψ : ( X ) ( Y ) satisfies AOPP and preserves collinear, then

1) ψ is an injective;

2) if ϕ ( f ) = ψ ( f ) ψ ( 0 ) , then ϕ ( f + g ) = ϕ ( f ) + ϕ ( g ) and ϕ ( λ f ) = λ ϕ ( f ) with 0 < λ < 1 .

Proof: 1) We prove ψ is injective. Let f , g ( X ) , since dim ( X ) 2 , there exists an element h ( X ) such that f h , g h are linearly independent. Hence 0 .

Let γ = h + g h f h , g h α , then f h , γ h α = 1 , and ψ satisfies AOPP, so

ψ ( f ) ψ ( h ) , ψ ( γ ) ψ ( h ) β = 1

we can see ψ ( h ) ψ ( f ) . So the mapping ψ is injective.

2) Let f , g , h mutually disjoint elements of ( X ) and f = g + h 2 , so f h = g f ( ) . Since ψ is injective and preserves collinear, there exist s 0 such that

ψ ( g ) ψ ( f ) = s ( ψ ( h ) ψ (f))

Since dim ( X ) 2 , there exist an element f 1 ( X ) such that g f , f 1 f α 0 . Let η = f + f 1 f g f , f 1 f α , then g f , η f α = 1 and

ψ ( g ) ψ ( f ) , ψ ( η ) ψ ( f ) β = 1.

So,

ψ ( h ) ψ ( f ) , ψ ( η ) ψ ( f ) β = | 1 s | .

Since ( ) , we get h f , η f α = 1 and

ψ ( h ) ψ ( f ) , ψ ( η ) ψ ( f ) β = 1.

According to the mapping ψ is injective, so s = 1 , and

ψ ( g + h 2 ) = ψ ( g ) + ψ ( h ) 2

Let ϕ ( f ) = ψ ( f ) ψ ( 0 ) , so we have

ϕ ( g + h 2 ) = ϕ ( g ) + ϕ ( h ) 2

Therefore

ϕ ( f 2 ) = ϕ ( f + 0 2 ) = ϕ ( f ) 2

and

ϕ ( f + g ) = ϕ ( 2 f + 2 g 2 ) = ϕ ( 2 f ) 2 + ϕ ( 2 g ) 2 = ϕ ( f ) + ϕ (g)

So ϕ is additive.

From the lemma 2.4, we know that if 1 , then ϕ satisfies 2-isometry.

0 = λ f , f α = ψ ( λ f ) ψ ( 0 ) , ψ ( f ) ψ ( 0 ) β = ϕ ( λ f ) , ϕ ( f ) β

so ϕ ( λ f ) and ϕ ( f ) is linearly dependent i.e. ϕ ( λ f ) = s ϕ ( f ) .

Next we assume f , g α = λ ,

1 λ f , g α = f λ 0 , g 0 α = 1

and

1 = ϕ ( f λ ) ϕ ( 0 ) , ϕ ( g ) ϕ ( 0 ) β = ϕ ( f λ ) , ϕ ( g ) β = 1 | s | ϕ ( f ) , ϕ ( g ) β = 1 | s | f , g α

Thus λ = | s | , if s = λ , then ϕ ( λ f ) = λ ϕ ( f ) , but

| λ 1 | f , g α = λ f f , g 0 α = ϕ ( λ f ) ϕ ( f ) , ϕ ( g ) ϕ ( 0 ) β = λ ϕ ( f ) ϕ ( f ) , ϕ ( g ) ϕ ( 0 ) β = ( λ + 1 ) ϕ ( f ) , ϕ ( g ) β = ( λ + 1 ) f , g α

so | λ 1 | = ( λ + 1 ) . It contradicts with 0 < λ < 1 . Thus ϕ ( λ f ) = λ ϕ ( f ) .

Lemma 2.6. Let ( X ) and ( Y ) be FNA-2. If 1 , a mapping ψ : ( X ) ( Y ) satisfies ( ) and AOPP, then we can get for all f , g , h ( X ) , we can get ( ) .

Proof: From lemma 2.4, we know ψ preserves collinear.

For any f , g , h ( X ) , there exist two numbers m , n such that m n .

So,

ψ ( f m ) ψ ( h m ) , ψ ( g ) ψ ( h ) β f h m , g h α 1 n

and

( f m ) ϕ ( h m ) , ϕ ( g ) ϕ ( h ) β 1 n

By lemma 2.5, we have

1 m ( ϕ ( f ) ϕ ( h ) ) , ϕ ( g ) ϕ ( h ) β 1 n

ϕ ( f ) ϕ ( h ) , ϕ ( g ) ϕ ( h ) β m n

Thus

ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β m n

Lemma 2.7. Let ( X ) and ( Y ) be two FNA-2. If a mapping ψ : ( X ) ( Y ) satisfies AOPP and ( ) for all f , g , h ( X ) with 1 , then ψ satisfies AnPP.

Proof: Let f , g , h ( X ) and n N . Let

= n , g i = h + i n ( g h )

and

f h , g i + 1 g i α = 1 , i = 0 , 1 , , n 1.

So,

ψ ( f ) ψ ( h ) , ψ ( g i + 1 ) ψ ( g i ) β = 1 , i = 0 , 1 , , n 1.

We know ψ preserves collinear. So there exist a number t such that

ψ ( g 2 ) ψ ( g 1 ) = t ( ψ ( g 1 ) ψ ( g 0 ) )

Therefore

Then we have t = ± 1 . By lemma 2.5, t = 1 , so

ψ ( g 2 ) ψ ( g 1 ) = ψ ( g 1 ) ψ ( g 0 ) .

In the same way, we can get

ψ ( g i + 1 ) ψ ( g i ) = ψ ( g i ) ψ ( g i 1 ) , i = 0 , 1 , , n 1.

Hence

ψ ( g ) ψ ( h ) = ψ ( g n ) ψ ( g 0 ) = ψ ( g n ) ψ ( g n 1 ) + ψ ( g n 1 ) ψ ( g n 2 ) + + ψ ( g 1 ) ψ ( g 0 ) = n ( ψ ( g 1 ) ψ ( g 0 ) )

Therefore

= ψ ( f ) ψ ( h ) , n ( ψ ( g 1 ) ψ ( g 0 ) ) β = n ψ ( f ) ψ ( h ) , ψ ( g 1 ) ψ ( g 0 ) β = n

Theorem 2.8. Let ( X ) and ( Y ) be two FNA-2. If a mapping ψ : ( X ) ( Y ) satisfies AOPP and ( ) for all f , g , h ( X ) with 1 , then ψ is 2-isometry.

Proof: Since lemma 2.4, we just need to prove that ( ) with > 1 .

We can assume that when > 1 , for all f , g , h ( X ) , we have < n 0 + 1 . and there exist a number n 0 such that

Let τ = f + n 0 + 1 f h , g h α ( f h ) , then

τ f , g h α = n 0 + 1

and

τ h , g h α = n 0 + 1

Since ψ preserves collinear, there exist a number c such that

ψ ( τ ) ψ ( f ) = c ( ψ ( h ) ψ (f))

Since 2),

n 0 + 1 = ψ ( τ ) ψ ( f ) , ψ ( g ) ψ ( h ) β = | c | | c 1 | + = ψ ( τ ) ψ ( h ) , ψ ( g ) ψ ( h ) β + < n 0 + 1 + = n 0 + 1

which is contradiction, so

n 0 + 1

Therefore, we get ( ) with > 1 . Hence

ψ ( f ) ψ ( h ) , ψ ( g ) ψ ( h ) β = f h , g h α

for all f , g , h ( X ) .

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Song, M.M. and Jin, H.X. (2019) The Aleksandrov Problem in Non-Archimedean 2-Fuzzy 2-Normed Spaces. Journal of Applied Mathematics and Physics, 7, 1775-1785. https://doi.org/10.4236/jamp.2019.78121

References

  1. 1. Benz, W. and Berens, H. (1987) A Contribution to a Theorem of Ulam and Mazur. Aequationes Mathematicae, 34, 61-63. https://doi.org/10.1007/BF01840123

  2. 2. Bag, T. and Samanta, S.K. (2003) Finite Dimensional Fuzzy Normed Linear Spaces. The Journal of Fuzzy Mathematics, 11, 687-705.

  3. 3. Chu, H.Y., Park, C.G. and Park, W.G. (2004) The Aleksandrow Problem in Linear 2-Normed Spaces. Journal of Mathematical Analysis and Applications, 289, 666-672. https://doi.org/10.1016/j.jmaa.2003.09.009

  4. 4. Somasundaram, R.M. and Beaula, T. (2009) Some Aspects of 2-Fuzzy 2-Normed Linear Spaces. Bulletin of the Malaysian Mathematical Sciences Society, 32, 211-221.

  5. 5. Zheng, F.H. and Ren, W.Y. (2014) The Aleksandrow Problem in Quasi Convex Normed Linear Space. Acta Scientiarum Natueralium University Nankaiensis, No. 3, 49-56.

  6. 6. Huang, X.J. and Tan, D.N. (2017) Mapping of Conservative Distances in p-Normed Spaces ( ). Bulletin of the Australian Mathematical Society, 95, 291-298. https://doi.org/10.1017/S0004972716000927

  7. 7. Ma, Y.M. (2000) The Aleksandrov Problem for Unit Distance Preserving Mapping. Acta Mathematica Science, 20B, 359-364. https://doi.org/10.1016/S0252-9602(17)30642-2

  8. 8. Wang, D.P., Liu, Y.B. and Song, M.M. (2012) The Aleksandrov Problem on Non-Archimedean Normed Spaces. Arab Journal of Mathematical Science, 18, 135-140. https://doi.org/10.1016/j.ajmsc.2011.10.002

  9. 9. Ma, Y.M. (2016) The Aleksandrov-Benz-Rassias Problem on Linear n-Normed Spaces. Monatshefte für Mathematik, 180, 305-316. https://doi.org/10.1007/s00605-015-0786-8

  10. 10. Huang, X.J. and Tan, D.N. (2018) Mappings of Preserves n-Distance One in N-Normed Spaces. Aequations Mathematica, 92, 401-413. https://doi.org/10.1007/s00010-018-0539-6

  11. 11. Xu, T.Z. (2013) On the Mazur-Ulanm Theorem in Non-Archimedean Fuzzy n-Normed Spaces. ISRN Mathematical Analysis, 67, 1-7. https://doi.org/10.1155/2013/814067

  12. 12. Chang, L.F. and Song, M.M. (2014) On the Mazur-Ulam Theorem in Non-Archimedean Fuzzy 2-Normed Spaces. Mathematica Applicata, 27, 355-359.

  13. 13. Alaca (2010) New Perspective to the Mazur-Ulam Problem in 2-Fuzzy 2-Normed Linear Spaces. Iranian Journal of Fuzzy Systems, 7, 109-119.

  14. 14. Park, C. and Alaca, C. (2013) Mazur-Ulam Theorem under Weaker Conditions in the Framework of 2-Fuzzy 2-Normed Linear Spaces. Journal of Inequalities and Applications, 2018, 78. https://doi.org/10.1186/1029-242X-2013-78

  15. 15. Hensel, K. (1897) über eine neue Begründung der Theorie der algebraischen Zahlen. Jahresbericht der Deutschen Mathematiker-Vereinigung, 6, 83-88.