ON THE RATIO OF DIRECTED LENGTHS IN THE TAXICAB PLANE AND RELATED PROPERTIES

Münevver Özcan and Rüstem Kaya

Abstract. In this work, it is shown that a point of division divides a related line segment in the same ratio both in the taxicab and Euclidean planes. Consequently, the coordinates of the division point can be determined by the same formula as in the Euclidean plane. In the latter parts of the paper, taxicab analogues of Ceva’s and Menelaus’ Theorems and the theorem of directed lines are given.

1. Introduction. A family of “metrics”, including the taxicab metric, have been published by H. Minkowski [9] at the beginning of the last century. Later, taxicab plane geometry was introduced in [8] and developed in [5] using the taxicab metric in the coordinate plane by

d_{T} (P_{1}, P_{2}) = |x_{1} − x_{2}| + |y_{1} − y_{2}|

instead of the Euclidean metric

d_{E}(P_{1}, P_{2}) = \sqrt{(x_{1} − x_{2})^{2} + (y_{1} − y_{2})^{2}}

where P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}).

A few problems related to the taxicab geometry have been studied and improved by some authors, see [1, 2, 3, 4, 7, 10, 11, 12, 13, 14]. The taxicab geometry was constructed by simply replacing the Euclidean distance function d_{E} by the taxicab distance function d_{T}. Therefore it seems interesting to study the taxicab analogues of the topics which include the concept of distance in Euclidean geometry. These topics are division point, directed lengths, ratio of directed lengths, Menelaus’ Theorem, Ceva’s Theorem, and the theorem of directed lines.

2. Directed Taxicab Length and Division Point. Let X and Y be any two points on a directed straight line l. We define directed taxicab length of the line segment XY as follows:

d_{T} [XY] = \{ \begin{matrix}d_{T} (X, Y), & if XY and l have the same direction \\ −d_{T} (X, Y), & if XY and l have opposite direction. \end{matrix}

Thus, d_{T} [XY] = −d_{T} [Y X]. Clearly, directed length in the Euclidean plane can be defined in a similar way. That is

d_{E} [XY] = \{ \begin{matrix}d_{E}(X, Y), & if XY and l have the same direction \\ −d_{E}(X, Y), & if XY and l have opposite direction. \end{matrix}

If A, B, C are points on a same directed line and C is between points A and B, we denote this by ACB. If ACB, then C divides the line segment AB internally and the ratio of the directed taxicab lengths is a positive real number, that is d_{T} [AC] /d_{T} [CB] = λ > 0. If ABC or CAB then C divides the line segment AB externally, and d_{T} [AC] /d_{T} [CB] = λ < 0, that is, the line segments AC and CB have opposite directions. In both cases C is called the division point which divides the line segment AB in ratio λ. Clearly, C ≠ B. C = A ⇔ λ = 0 and (C is at infinity ⇔ λ = −1).

Let C and C′ be two points such that C divides a given line segment AB internally and C′ divides AB externally in the same proportion though with opposite signs. Thus, the ratio of the directed lengths, d_{T} [AC] /d_{T} [CB] = − d_{T} [AC′] /d_{T} [C′B] is the same positive number λ.

\underline{Theorem 1}. Let P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) be any two distinct points in the analytical plane. If Q = (x, y) is a point on the line passing through P_{1} and P_{2}, then

d_{T} [P_{1}Q] /d_{T} [QP_{2}] = d_{E} [P_{1}Q] /d_{E} [QP_{2}] .

That is, the ratios of the Euclidean and taxicab directed lengths are the same.

\underline{Proof}. If Q = P_{1} then both ratios are equal to 0. If Q is at infinity then both ratios are equal to −1. Therefore without loss of generality, let P_{1} ≠ Q ≠ P_{2}. It is enough to show that

\frac{|x_{1} − x| + |y_{1} − y|}{|x − x_{2}| + |y − y_{2}|} = \begin{matrix}√ & \overline{(x_{1} − x)^{2} + (y_{1} − y)^{2}} \\ √ & \overline{(x − x_{2})^{2} + (y − y_{2})^{2}} \end{matrix} .  (1)

Squaring both sides of Equation (1) one obtains

\frac{|x_{1} − x|^{2} + |y_{1} − y|^{2} + 2 |x_{1} − x| |y_{1} − y|}{|x − x_{2}|^{2} + |y − y_{2}|^{2} + 2 |x − x_{2}| |y − y_{2}|} = \frac{(x_{1} − x)^{2} + (y_{1} − y)^{2}}{(x − x_{2})^{2} + (y − y_{2})^{2}}

which is equivalent to

\begin{matrix}[(x − x_{2})^{2} + (y − y_{2})^{2}] & [|x_{1} − x|^{2} + |y_{1} − y|^{2} + 2 |x_{1} − x| |y_{1} − y|] \\ [(x_{1} − x)^{2} + (y_{1} − y)^{2}] & [|x − x_{2}|^{2} + |y − y_{2}|^{2} + 2 |x − x_{2}| |y − y_{2}|] \end{matrix} = 1.

Rearranging the last equality one gets

\frac{[(x_{1}−x)^{2}+(y_{1}−y)^{2}][(x−x_{2})^{2}+(y−y_{2})^{2}]+2|x_{1}−x||y_{1}−y|[(x−x_{2})^{2}+(y−y_{2})^{2}]}{[(x_{1}−x)^{2}+(y_{1}−y)^{2}][(x−x_{2})^{2}+(y−y_{2})^{2}]+2|x−x_{2}||y−y_{2}|[(x_{1}−x)^{2}+(y_{1}−y)^{2}]} = 1

which means that

\frac{2 |x_{1} − x| |y_{1} − y| [(x − x_{2})^{2} + (y − y_{2})^{2}]}{2 |x − x_{2}| |y − y_{2}| [(x_{1} − x)^{2} + (y_{1} − y)^{2}]} = 1

or simply

\begin{matrix}|x_{1} − x| & |y_{1} − y| \\ |x − x_{2}| & |y − y_{2}| \end{matrix} = \frac{(x_{1} − x)^{2} + (y_{1} − y)^{2}}{(x − x_{2})^{2} + (y − y_{2})^{2}} .  (2)

Examining the left side of Equation (2) one obtains

\begin{matrix}|x_{1} − x| & |y_{1} − y| \\ |x − x_{2}| & |y − y_{2}| \end{matrix} = \frac{(x_{1} − x) (y_{1} − y)}{(x − x_{2}) (y − y_{2})}  (3)

for all positions of Q on P_{1}P_{2}. Using Equation (3) in Equation (2) one obtains

\begin{matrix}(x_{1} − x) & (y_{1} − y) [(x − x_{2})^{2} + (y − y_{2})^{2}] \\ = (x − x_{2})(y − y_{2}) [(x_{1} − x)^{2} + (y_{1} − y)^{2}] \end{matrix}

which can be expressed as follows:

\begin{matrix}(x_{1} − x) & (x − x_{2}) [(x − x_{2})(y_{1} − y) + (x_{1} − x)(y − y_{2})] \\ = (y_{1} − y)(y − y_{2}) [(x − x_{2})(y_{1} − y) + (x_{1} − x)(y − y_{2})] . \end{matrix}

Rearranging this equality one gets

[(x − x_{2})(y_{1} − y) − (x_{1} − x)(y − y_{2})] [(x_{1} − x)(x − x_{2}) − (y_{1} − y)(y − y_{2})] = 0

(4)

If x_{1} = x_{2} then x = x_{1} = x_{2} and Equation (4) is obvious. If x_{1} ≠ x_{2} then

y = [(x_{2} − x)y_{1} − (x_{1} − x)y_{2}] / (x_{2} − x_{1})

since Q is on the line P_{1}P_{2}. Now, using this value of y in the first bracket of Equation (4) we get

(x − x_{2})(y_{1} − y) − (x_{1} − x)(y − y_{2})

= (x − x_{2})(y_{1} − \frac{(x_{2}−x)y_{1}−(x_{1}−x)y_{2}}{x_{2}−x_{1}} ) − (x_{1} − x)(\frac{(x_{2}−x)y_{1}−(x_{1}−x)y_{2}}{x_{2}−x_{1}} − y_{2})

= \frac{1}{x_{2}−x_{1}} [(x − x_{2})(xy_{1} − xy_{2} + x_{1}y_{2} − x_{1}y_{1}) − (x_{1} − x)(xy_{2} − xy_{1} + x_{2}y_{1} − x_{2}y_{2})]

= \frac{1}{x_{2}−x_{1}} [(x − x_{1})(x − x_{2})(y_{1} − y_{2}) − (x_{1} − x)(x − x_{2})(y_{2} − y_{1})] = 0

which shows that Equation (4) is satisfied.

The following corollary shows how one can find the coordinates of the division point which divides the line segment joining two given points in a given ratio, in the taxicab plane.

\underline{Corollary}. Let P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) be two distinct points in the taxicab plane. If Q = (x, y) divides the line segment P_{1}P_{2} in ratio λ then,

x = \frac{x_{1} + λx_{2}}{1 + λ} , y = \frac{y_{1} + λy_{2}}{1 + λ} ; λ ∈ ℝ, λ ≠ −1

as in the Euclidean plane.

\underline{Proof}. Although the Corollary follows from Theorem 1 we prefer to give a direct proof. The given formula is obvious when λ = 0 or λ = −1. If λ ≠ 0, −1 and Q divides the line segment P_{1}P_{2} in ratio λ, we have |d_{T} [P_{1}Q_{1}] /d_{T} [Q_{1}P_{2}]| = |λ|. That is,

\frac{|x_{1} − x| + |y_{1} − y|}{|x − x_{2}| + |y − y_{2}|} = |λ| .  (5)

Since P_{1} ≠ P_{2},

|λ| = |λ| (\frac{|x_{1} − x_{2}| + |y_{1} − y_{2}|}{|x_{1} − x_{2}| + |y_{1} − y_{2}|}) = \frac{|λx_{1} − λx_{2}| + |λy_{1} − λy_{2}|}{|x_{1} − x_{2}| + |y_{1} − y_{2}|} .

Adding x_{1} −x_{1} and y_{1} −y_{1} to the first and second summands in the numerator and similarly λx_{2} − λx_{2} and λy_{2} − λy_{2} in the denominator respectively, one obtains

|λ| = \frac{|λx_{1} + x_{1} − x_{1} − λx_{2}| + |λy_{1} + y_{1} − y_{1} − λy_{2}|}{|x_{1} + λx_{2} − λx_{2} − x_{2}| + |y_{1} + λy_{2} − λy_{2} − y_{2}|} .

Multiplying the numerator and the denominator of the right side of the last statement by 1/ |1 + λ|, one gets

|λ| = \frac{|\frac{λx_{1} + x_{1} − x_{1} − λx_{2}}{1 + λ} | + |\frac{λy_{1} + y_{1} − y_{1} − λy_{2}}{1 + λ} |}{|\frac{x_{1} + λx_{2} − λx_{2} − x_{2}}{1 + λ} | + |\frac{y_{1} + λy_{2} − λy_{2} − y_{2}}{1 + λ} |}

= \frac{|\frac{(1 + λ)x_{1}}{1 + λ} − \frac{x_{1} + λx_{2}}{1 + λ} | + |\frac{(1 + λ)y_{1}}{1 + λ} − \frac{y_{1} + λy_{2}}{1 + λ} |}{|\frac{x_{1} + λx_{2}}{1 + λ} − \frac{(1 + λ)x_{2}}{1 + λ} | + |\frac{y_{1} + λy_{2}}{1 + λ} − \frac{(1 + λ)y_{2}}{1 + λ} |}

= \frac{|x_{1−} \frac{x_{1} + λx_{2}}{1 + λ} | + |y_{1−} \frac{y_{1} + λy_{2}}{1 + λ} |}{|\frac{x_{1} + λx_{2}}{1 + λ} − x_{2}| + |\frac{y_{1} + λy_{2}}{1 + λ} − y_{2}|} .

Comparing this result with Equation (5) we obtain

x = \frac{x_{1} + λx_{2}}{1 + λ} and y = \frac{y_{1} + λy_{2}}{1 + λ} .

3. Theorems of Menelaus and Ceva in the Taxicab Plane. In this section, the taxicab analogues of the Theorems of Menelaus and Ceva are studied. In fact, the validity of these theorems is clear from the Theorem 1, but we prefer to state and give partial proofs for them.

\underline{Theorem 2}. (Menelaus’ Theorem.) Let \{P_{1}, P_{2}, P_{3}\} be a triangle and Q_{1}, Q_{2}, Q_{3} be on the lines that contain the sides P_{1}P_{2}, P_{2}P_{3}, P_{3}P_{1} respectively, in the taxicab plane. If Q_{1}, Q_{2}, Q_{3} are collinear, then

\frac{d_{T} [P_{1}Q_{1}]}{d_{T} [Q_{1}P_{2}]} · \frac{d_{T} [P_{2}Q_{2}]}{d_{T} [Q_{2}P_{3}]} · \frac{d_{T} [P_{3}Q_{3}]}{d_{T} [Q_{3}P_{1}]} = −1  (6)

where none of Q_{1}, Q_{2}, Q_{3} coincide with any of P_{1}, P_{2}, P_{3}.

\underline{Proof}. Several cases are possible, according to the positions of points P_{1}, P_{2}, P_{3} and Q_{1}, Q_{2}, Q_{3}. We give a proof of the theorem only in the following special case.

Let P_{i} = (x_{i}, y_{i}), i = 1, 2, 3 and x_{i} ≠ x_{i+1} and let Q_{1}, Q_{2}, Q_{3} be on a line l given by y = mx + k such that Q_{i} = l ∧ P_{i}P_{i+1 (mod 3)} and l is not parallel to the line P_{i}P_{i+1}, for i = 1, 2, 3 (Figure 1) . Clearly mx_{i} − y_{i} + k ≠ 0 since P_{i} ≠ Q_{j} for i, j = 1, 2, 3 and m ≠ (y_{i+1} − y_{i})(x_{i+1} − x_{i})^{−1}. The equation of the line P_{i}P_{i+1} is given by

y = (y_{i+1} − y_{i})(x_{i+1} − x_{i})^{−1}x − (x_{i}y_{i+1} − x_{i+1}y_{i})(x_{i+1} − x_{i})^{−1}.

It follows from a simple calculation that

Q_{i} = (\frac{x_{i}y_{i+1} − x_{i+1}y_{i} − kx_{i} + kx_{i+1}}{mx_{i} − mx_{i+1} − y_{i} + y_{i+1}} , \frac{mx_{i}y_{i+1} − mx_{i+1}y_{i} − ky_{i} + ky_{i+1}}{mx_{i} − mx_{i+1} − y_{i} + y_{i+1}} ).

Now let us find \frac{d_{T} [P_{i}Q_{i}]}{d_{T} [Q_{i}P_{i+1}]}.

Figure 1.

\frac{d_{T} [P_{1}Q_{1}]}{d_{T} [Q_{1}P_{2}]} = −\frac{d_{T} (P_{1}, Q_{1})}{d_{T} (Q_{1}, P_{2})}

= −\frac{|x_{1} − \frac{x_{1}y_{2}−x_{2}y_{1}−kx_{1}+kx_{2}}{mx_{1}−mx_{2}−y_{1}+y_{2}} | + |y_{1} − \frac{mx_{1}y_{2}−mx_{2}y_{1}−ky_{1}+ky_{2}}{mx_{1}−mx_{2}−y_{1}+y_{2}} |}{|\frac{x_{1}y_{2}−x_{2}y_{1}−kx_{1}+kx_{2}}{mx_{1}−mx_{2}−y_{1}+y_{2}} − x_{2}| + |\frac{mx_{1}y_{2}−mx_{2}y_{1}−ky_{1}+ky_{2}}{mx_{1}−mx_{2}−y_{1}+y_{2}} − y_{2}|}

= −\frac{|mx_{1}^{2}−mx_{1}x_{2}−x_{1}y_{1}+x_{2}y_{1}−kx_{2}+kx_{1}|+|mx_{1}y_{1}−mx_{1}y_{2}+y_{1}y_{2}−y_{1}^{2}−ky_{2}+ky_{1}|}{|mx_{2}^{2}−mx_{1}x_{2}−x_{2}y_{2}+x_{1}y_{2}−kx_{1}+kx_{2}|+|mx_{2}y_{2}−mx_{2}y_{1}+y_{1}y_{2}−y_{2}^{2}−ky_{1}+ky_{2}|}

= −\frac{|x_{1}(mx_{1}−y_{1}+k)−x_{2}(mx_{1}−y_{1}+k)|+|y_{1}(mx_{1}−y_{1}+k)−y_{2}(mx_{1}−y_{1}+k)|}{|x_{2}(mx_{2}−y_{2}+k)−x_{1}(mx_{2}−y_{2}+k)|+|y_{2}(mx_{2}−y_{2}+k)−y_{1}(mx_{2}−y_{2}+k)|}

= −\frac{(|x_{1} − x_{2}| + |y_{1} − y_{2}|) |mx_{1} − y_{1} + k|}{(|x_{1} − x_{2}| + |y_{1} − y_{2}|) |mx_{2} − y_{2} + k|}

= −\frac{|mx_{1} − y_{1} + k|}{|mx_{2} − y_{2} + k|}.

Similarly,

\frac{d_{T} [P_{2}Q_{2}]}{d_{T} [Q_{2}P_{3}]} = \frac{d_{T} (P_{2}, Q_{2})}{d_{T} (Q_{2}, P_{3})} = \frac{|mx_{2} − y_{2} + k|}{|mx_{3} − y_{3} + k|}

and

\frac{d_{T} [P_{3}Q_{3}]}{d_{T} [Q_{3}P_{1}]} = \frac{d_{T} (P_{3}, Q_{3})}{d_{T} (Q_{3}, P_{1})} = \frac{|mx_{3} − y_{3} + k|}{|mx_{1} − y_{1} + k|}

and consequently,

\frac{d_{T} [P_{i}Q_{i}]}{d_{T} [Q_{i}P_{i+1}]} = s\frac{|mx_{i} − y_{i} + k|}{|mx_{i+1} − y_{i+1} + k|} , s = \{ \begin{matrix}−1, & if i = 1 \\ 1, & if i = 2, 3. \end{matrix}

Now, it can be easily computed that

∏_{i=1}^{3}(d_{T} [P_{i}Q_{i}] /d_{T} [Q_{i}P_{i+1}]) = −1.

\underline{Theorem 3}. (Converse of Menelaus’ Theorem.) Let \{P_{1}, P_{2}, P_{3}\} be a triangle and Q_{1}, Q_{2}, Q_{3} be three points on the lines that contain the sides P_{1}P_{2}, P_{2}P_{3}, P_{3}P_{1}, respectively, in the taxicab plane. If

\frac{d_{T} [P_{1}Q_{1}]}{d_{T} [Q_{1}P_{2}]} · \frac{d_{T} [P_{2}Q_{2}]}{d_{T} [Q_{2}P_{3}]} · \frac{d_{T} [P_{3}Q_{3}]}{d_{T} [Q_{3}P_{1}]} = −1,

then Q_{1}, Q_{2}, Q_{3} are collinear. Note that none of Q_{1}, Q_{2}, Q_{3} are P_{1}, P_{2}, P_{3}.

\underline{Theorem 4}. (Ceva’s Theorem.) Let \{P_{1}, P_{2}, P_{3}\} be a triangle and lines l_{1}, l_{2}, l_{3} pass through the vertices P_{1}, P_{2}, P_{3}, respectively and intersect lines containing the opposite sides at points Q_{1}, Q_{2}, Q_{3}. The lines l_{1}, l_{2}, l_{3} are concurrent (or parallel) if and only if

\frac{d_{T} [P_{1}Q_{3}]}{d_{T} [Q_{3}P_{2}]} · \frac{d_{T} [P_{2}Q_{1}]}{d_{T} [Q_{1}P_{3}]} · \frac{d_{T} [P_{3}Q_{2}]}{d_{T} [Q_{2}P_{1}]} = 1.

Note that none of Q_{1}, Q_{2}, Q_{3} are P_{1}, P_{2}, P_{3}.

4. Theorems of Directed Lines (Strahlensätze). In general, it is well– known that the axiom of congruence and consequently properties of similarity for triangles are not valid in the taxicab plane. But, it follows from Theorem 1 that the following directed line theorem [6] is valid in it.

\underline{Theorem 5}. Let a pencil of lines be intersected by a family of parallel lines in the taxicab plane (see Figure 2).

(i) The ratios of the directed lengths of the corresponding segments on the lines belonging to the pencil are the same. For example,

\begin{aligned}d_{T} [SA] : d_{T} [SB] : d_{T} [SC] & = d_{T} [SA_{1}] : d_{T} [SB_{1}] : d_{T} [SC_{1}] \\ & = d_{T} [SA_{2}] : d_{T} [SB_{2}] : d_{T} [SC_{2}] \end{aligned}

or

d_{T} [SA_{1}] : d_{T} [SB_{1}] = d_{T} [A_{1}A_{2}] : d_{T} [B_{1}B_{2}] .

(ii) The ratios of the directed lengths of line segments on the parallel lines and corresponding segments on the lines belonging to the pencil, which are measured from the vertex, are the same. For example,

\begin{aligned}d_{T} [CB] : d_{T} [C_{1}B_{1}] : d_{T} [C_{2}B_{2}] & = d_{T} [SC] : d_{T} [SC_{1}] : d_{T} [SC_{2}] \\ & = d_{T} [SB] : d_{T} [SB_{1}] : d_{T} [SB_{2}] \end{aligned}

or

\begin{aligned}d_{T} [AB] : d_{T} [A_{1}B_{1}] : d_{T} [A_{2}B_{2}] & = d_{T} [SA] : d_{T} [SA_{1}] : d_{T} [SA_{2}] \\ & = d_{T} [SB] : d_{T} [SB_{1}] : d_{T} [SB_{2}] . \end{aligned}

(iii) The ratios of the lengths of the corresponding segments on the parallel lines are the same. That is,

d_{T} [AB] : d_{T} [BC] = d_{T} [A_{1}B_{1}] : d_{T} [B_{1}C_{1}] = d_{T} [A_{2}B_{2}] : d_{T} [B_{2}C_{2}] .

Notice that here a : b : c = a_{1} : b_{1} : c_{1} if and only if a/a_{1} = b/b_{1} = c/c_{1}.

Figure 2.

\underline{References}

1. Z. Akca and R. Kaya, “On the Taxicab Trigonometry,” Jour. of Inst. of Math. & Comp. Sci. (Math. Ser.), 10 (1997), 151–159.

2. C. Ekici, I. Kocayusufoglu and Z. Akca, “The Norm in Taxicab Geometry,” Tr. J. of Mathematics, 22 (1998), 295–307.

3. Y. P. Ho and Y. Liu, “Parabolas in Taxicab Geometry,” Missouri J. of Math., 8 (1996), 63–72.

4. R. Kaya, Z. Akca, I. Günaltitli and M. Özcan, “General Equation for Taxicab Conics and Their Classification,” Mitt. Math. Ges. Hamburg, 19 (2000), 135– 148.

5. E. F. Krause, Taxicab Geometry, Addison–Wesley, Menlo Park, California, 1975.

6. H. Kreul, K. Kulke, H. Pester, R. Schroedter, Mathematik Leicht Gemacht, Verlag Harri Deutsch–Thun–Frankfurt am Main, 1990.

7. R. Laatsch, “Pyramidal Sections in Taxicab Geometry,” Mathematics Magazine, 55 (1982), 205–212.

8. K. Menger, You Will Like Geometry, Guildbook of Illinois Institute of Technology Geometry Exhibit, Museum of Science & Industry, Chicago, Illinois, 1952.

9. H. Minkowski, Gesammelte Abhandlungen, Chelsea Publishing Co., New York, 1967.

10. B. E. Reynolds, “Taxicab Geometry,” Pi Mu Epsilon Journal, 7 (1980), 77–88.

11. D. J. Schattschneider, “The Taxicab Group,” Amer. Math. Monthly, 91 (1984), 423–428.

12. S. S. So and Z. S. Al–Maskari, “Two Simple Examples in Non–Euclidean Geometry,” Kansas Science Teacher (Journal of Mathematics and Science Teaching), 11 (1995), 14–18.

13. K. O. Sowell, “Taxicab Geometry – A New Slant,” Mathematics Magazine, 62 (1989), 238–248.

14. S. Tian, S. S. So and G. Chen, “Concerning Circles in Taxicab Geometry,” Int. J. Math. Educ. Sci. Technol., 28 (1997), 727–733.

Münevver Özcan Department of Mathematics Faculty of Science and Arts University of Osmangazi Eskişehir – TURKEY email: mozcan@ogu.edu.tr

Rüstem Kaya Department of Mathematics Faculty of Science and Arts University of Osmangazi Eskişehir – TURKEY email: rkaya@ogu.edu.tr