GROUP OF ISOMETRIES OF THE CC-PLANE

R. Kaya, Ö. Gelişgen, S. Ekmekçi, and A. Bayar

Abstract. In this work, it is shown that the group of isometries of the plane with respect to the Chinese Checkers metric is the semi-direct product of the Dihedral group D_{8} and T (2), where D_{8} is the (Euclidean) symmetry group of the regular octagon and T (2) is the group of all translations of the plane. Furthermore, some properties of the CC-plane are studied and the area formula for a triangle is given.

1. Introduction. One of the basic problems in geometric investigations for a given space S with a metric d is to describe the group G of isometries. If S is the Euclidean plane with the usual metric, then it is well-known that G consists of all translations, rotations, reflections and glide reflections of the plane. It is known that for the Euclidean plane, G = E(2) is the semi-direct product of its two subgroups O(2) (the orthogonal group) and T (2), where O(2) is the symmetry group of the unit circle and T (2) (the translation group) consists of all translations of the plane [2, 6, 8]. The group of isometries of the taxicab plane has been given in [7].

For the general problem stated above we use the analytical plane ℝ^{2} endowed with the Chinese Checker Metric d_{c} defined by

d_{c}(X, Y) = max \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\} + (\sqrt{2} − 1) min \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\} ,

where X = (x_{1}, y_{1}) and Y = (x_{2}, y_{2}). We use ℝ_{c}^{2} = (ℝ^{2}, d_{c}) for the Chinese Checkers plane (CC-plane). E. F. Krause [4] asked the question of how to develop a metric which would be similar to the movement made by playing Chinese Checkers. Later, the above metric was developed by G. Chen [1].

According to the definition of d_{c}-metric, the shortest path between the points A and B is the union of a vertical or a horizontal line segment and a line segment with the slope 1 or -1, as shown in Figure 1. Thus, the shortest distance d_{c} from A to B is the sum of the Euclidean lengths of these two line segments.

If the slope of the segment \overline{AB} is −1, 0, 1, ∞, then the d_{c}-distance is equal to the Euclidean distance between A and B. For the sake of brevity of notation we denote the slope of the vertical lines by ∞.

The unit circle in ℝ_{c}^{2} is the set of points (x, y) in the plane which satisfy the equation

max \{|x| , |y|\} + (\sqrt{2} − 1) min \{|x| , |y|\} = 1.

This is an octagon with vertices

\begin{matrix}A_{1} (1, 0) , A_{2} (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) , A_{3} (0, 1) , A_{4} (\frac{−1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) , \\ A_{5} (−1, 0) , A_{6} (\frac{−1}{\sqrt{2}}, \frac{−1}{\sqrt{2}}) , A_{7} (0, −1) , A_{8} (\frac{1}{\sqrt{2}}, \frac{−1}{\sqrt{2}}) \end{matrix}

as shown in Figure 2.

In the remaining part of this work, we will study the isometries of ℝ_{c}^{2}, determine its group of isometries, and give some properties of ℝ_{c}^{2}.

2. Isometries of the CC-Plane (ℝ_{c}^{2}). Since an isometry of a plane is defined to be a transformation which preserves the distances in the plane, an isometry of ℝ_{c}^{2} is therefore an isometry of the real plane with respect to the d_{c} metric.

\underline{Proposition 1}. Every Euclidean translation is an isometry of ℝ_{c}^{2}.

\underline{Proof}. Let T_{A}: ℝ_{c}^{2} → ℝ_{c}^{2} such that T_{A}(X) = A + X be the translation in the real plane ℝ^{2}, where A = (a_{1}, a_{2}) and X = (x_{1}, y_{1}) ∈ ℝ_{c}^{2}. For X = (x_{1}, y_{1}) and Y = (x_{2}, y_{2}) ∈ ℝ_{c}^{2}, we have

d_{c} (T_{A}(X), T_{A}(Y)) = max {|(a_{1} + x_{1}) − (a_{1} + x_{2})| , |(a_{2} + y_{1}) − (a_{2} + y_{2})|}

+ (\sqrt{2} − 1) min \{|(a_{1} + x_{1}) − (a_{1} + x_{2})| , |(a_{2}+y_{1}) − (a_{2} + y_{2})|\}

= max \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\} + (\sqrt{2} − 1) min \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\}

= d_{c} (X, Y) .

That is, T_{A} is an isometry.

Since CC-plane geometry is the study of Euclidean points, lines and angles in ℝ_{c}^{2}, we use the following definition and lemma to find the reflections.

\underline{Definition}. Let P and l be a point and a line in ℝ_{c}^{2}, and let Q denote the point on l such that P Q is perpendicular to l. If P′ is a point on the opposite side of the line l with respect to P such that d_{c} (P, Q) = d_{c} (P′, Q), then P′ is called the reflection of P.

Notice that it is enough to consider the lines passing through the origin as axes of reflections because of Proposition 1.

The following lemma will be useful in determining reflections in ℝ_{c}^{2}.

\underline{Lemma 2}. Let l be the line through the points A = (x_{1}, y_{1}) and B = (x_{2}, y_{2}) in the analytical plane and d_{E} denote the Euclidean metric. If l has slope m, then

d_{c}(A, B) = \frac{M}{\sqrt{m^{2} + 1}}d_{E}(A, B), where M = \{ \begin{matrix}1 + (\sqrt{2} − 1) |m| & if |m| ≤ 1 \\ |m| + \sqrt{2} − 1 & if |m| ≥ 1. \end{matrix}

\underline{Proof}. If l is parallel to the x-axis or y-axis, then m = 0 and M/\sqrt{m^{2} + 1} = 1 or m → ∞ and lim_{m→∞}(M/\sqrt{m^{2} + 1}) = 1. Then, d_{c}(A, B) = d_{E}(A, B) in both of the cases above. If l is not parallel to the x-axis and y-axis, then x_{1} ≠ x_{2} and y_{1} ≠ y_{2}, m = (y_{1} − y_{2})/(x_{1} − x_{2}), where m is the slope of l, and

d_{c}(A, B) = max \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\} + (\sqrt{2} − 1) min \{|x_{1} − x_{2}| , |y_{1} − y_{2}|\}

= \{ \begin{matrix}|x_{1} − x_{2}| (1 + (\sqrt{2} − 1) |m|) & if |m| ≤ 1 \\ |x_{1} − x_{2}| (|m| + \sqrt{2} − 1) & if |m| ≥ 1. \end{matrix}

Similarly,

d_{E}(A, B) = |x_{1} − x_{2}| \sqrt{1 + m^{2}} for all m ∈ ℝ

and consequently the given equality is valid.

The above proposition says that d_{c}-distance along any line is some positive constant multiple of Euclidean distance along the same line. Let ρ(m) denote this constant, that is, ρ(m) = M/\sqrt{m^{2} + 1}. For the graph of ρ see Figure 3.

Furthermore, one can immediately obtain the following corollaries.

\underline{Corollary 3}. If A, B, and X are any three collinear points in ℝ^{2}, then d_{E}(X, A) = d_{E}(X, B) if and only if d_{c}(X, A) = d_{c}(X, B).

\underline{Corollary 4}. If A, B, and X are any three distinct collinear points in the real plane, then

d_{c}(X, A)/d_{c}(X, B) = d_{E}(X, A)/d_{E}(X, B).

That is, the ratios of the Euclidean and d_{c}-distances along a line are the same.

Notice that the latter corollary gives us the validity of the Theorems of Menelaus and Ceva in ℝ_{c}^{2}. The following proposition determines the reflections which preserves distance in ℝ_{c}^{2}.

\underline{Proposition 5}. A reflection about the line y = mx in ℝ_{c}^{2} is an isometry if and only if

m ∈ \{0, ±1, ±(\sqrt{2} − 1), ±(\sqrt{2} + 1), ∞\} .

\underline{Proof}. Consider the Euclidean reflection φ about the line y = mx,

φ(P) = φ(x, y) = P′ = (x′, y′) = (\frac{(1 − m^{2})x + 2my}{1 + m^{2}} , \frac{2mx + (−1 + m^{2})y}{1 + m^{2}} ) .

If Q = \overline{P P′} ∩ \{(x, y) : y = mx\}, then d_{E}(P, Q) = d_{E}(P′, Q) implies d_{c}(P, Q) = d_{c}(P′, Q) by Corollary 3. That is, P′ is a d_{c}-reflection of P. Using Proposition 1, one can say that φ is an isometry of ℝ_{c}^{2} if and only if d_{c}(O, P) = d_{c}(O, P′). We claim that

d_{c}(O, P) = d_{c}(O, P′) if and only if m ∈ \{0, ±1, ±(\sqrt{2} − 1), ±(\sqrt{2} + 1), ∞\} .

If d_{c}(O, P) = d_{c}(O, P′), then ρ(m_{1}) = ρ(m_{2}) = k, where k is a constant and m_{1} and m_{2} are the slopes of the lines OP and OP′, respectively. Furthermore, if P is on the line y = mx, then P = P′, m_{1} = m_{2} = m, ρ(m) = k, and the derivative ρ′(m) = 0 for k ≠ 1. Thus,

(1 + m^{2})^{3/2}.ρ′(m) = \{ \begin{matrix}−1 − (\sqrt{2} − 1)m & if −∞ < m < −1 \\ −m − \sqrt{2} + 1 & if −1 < m < 0 \\ −m + \sqrt{2} − 1 & if 0 < m < 1 \\ 1 − (\sqrt{2} − 1)m & if 1 < m < ∞ \end{matrix}

and ρ′(m) = 0 gives us

m = −(\sqrt{2} + 1) = tg(5π/8) if − ∞ < m < −1
m = −(\sqrt{2} − 1) = tg(7π/8) if − 1 < m < 0
m = \sqrt{2} − 1 = tg(π/8)  if 0 < m < 1
m = \sqrt{2} + 1 = tg(3π/8)  if 1 < m < ∞.

In particular, if ρ(m) = k = 1, then m ∈ {0, ±1, ∞} as seen in Figure 3.

Conversely, if m_{1} and m_{2} are the slopes of the lines OP and OP′, respectively, where P′ = φ(P), then

\begin{aligned}d_{E}(O, P) & = d_{E}(O, P′) \\ d_{c}(O, P) & = ρ(m_{1})d_{E}(O, P) \\ d_{c}(O, P′) & = ρ(m_{2})d_{E}(O, P′) = ρ(m_{2})d_{E}(O, P). \end{aligned}

Now it is easy to check that ρ(m_{1}) = ρ(m_{2}) for all of the possible cases in the following table, which implies d_{c}(O, P) = d_{c}(O, P′).

\begin{matrix}m & 0 & \sqrt{2} − 1 & 1 & \sqrt{2} + 1 \\ m_{2} & −m_{1} & \frac{1−m_{1}}{1+m_{1}} & \frac{1}{m_{1}} & \frac{m_{1}+1}{m_{1}−1} \end{matrix}

\begin{matrix}m & ∞ & −\sqrt{2} − 1 & −1 & −\sqrt{2} + 1 \\ m_{2} & −m_{1} & \frac{1−m_{1}}{1+m_{1}} & \frac{1}{m_{1}} & \frac{m_{1}+1}{m_{1}−1} \end{matrix}

The above proposition shows that not all reflections preserve d_{c}- distances. The set of isometric reflections S_{c} consist of the eight Euclidean reflections about the lines in

\{x = 0, y = 0, y = ∓x, y = ∓ (\sqrt{2} − 1) x, y = ∓ (\sqrt{2} + 1) x\} .

\underline{Proposition 6}. There are only eight Euclidean rotations that preserve d_{c}-distances. In other words, the set of isometric rotations in ℝ_{c}^{2} is

R_{c} = \{r_{θ} | θ = k \frac{π}{4} , k = 0, 1, …, 7\} .

\underline{Proof}. In order to find the isometric rotations in ℝ_{c}^{2}, it is sufficient to determine the rotations which preserve the lengths of the sides of the d_{c}- unit circle. Consider the points A_{1} = (1, 0) and A_{2} = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) on the unit circle of ℝ_{c}^{2}. Rotating A_{1} and A_{2} by an angle θ, we get

r_{θ}(A_{1}) = (cos θ, sin θ)
r_{θ}(A_{2}) = (1/\sqrt{2})(cos θ − sin θ, sin θ + cos θ).

Clearly, d_{c}(A_{1}, A_{2}) = 2\sqrt{2} − 2. If r_{θ} preserves d_{c}-distance, we must look for θ which implies d_{c}(r_{θ}(A_{1}), r_{θ}(A_{2})) = 2\sqrt{2} − 2. Thus,

d_{c}(r_{θ}(A_{1}), r_{θ}(A_{2}))
= max \{|\frac{2−\sqrt{2}}{2} cos θ + \frac{\sqrt{2}}{2} sin θ| , |\frac{2−\sqrt{2}}{2} sin θ − \frac{\sqrt{2}}{2} cos θ|\}

+ (\sqrt{2} − 1) min \{|\frac{2−\sqrt{2}}{2} cos θ + \frac{\sqrt{2}}{2} sin θ| , |\frac{2−\sqrt{2}}{2} sin θ − \frac{\sqrt{2}}{2} cos θ|\}
= 2\sqrt{2} − 2.

Let α = \frac{2−\sqrt{2}}{2} cos θ + \frac{\sqrt{2}}{2} sin θ and β = \frac{2−\sqrt{2}}{2} sin θ − \frac{\sqrt{2}}{2} cos θ. Now, consider the following cases.

i) Let |α| ≥ |β|.

If α ≥ 0 and β ≤ 0, then cos θ + sin θ = \sqrt{2} which implies θ = π/4. If α ≥ 0 and β ≥ 0, then sin θ = 1 which implies θ = π/2. If α ≤ 0 and β ≥ 0, then cos θ + sin θ = \sqrt{2} which implies θ = 5π/4. If α ≤ 0 and β ≤ 0, then sin θ = −1 which implies θ = 3π/2.

ii) Let |α| < |β|.

Similar to case (i), we gets θ = 3π/4, θ = 0, θ = π, and θ = 7π/4.

From (i) and (ii), θ ∈ {0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4}. Finally, let m and m′ denote the slopes of the lines OX and OX′, respectively, where X′ = r_{θ}(X). As in Proposition 5, it can easily be check that ρ(m) = ρ(m′) for all of the possible cases in the following table, which implies d_{c}(O, X) = d_{c}(O, X′). That is, every r_{θ} ∈ R_{c} preserves all d_{c}-distances by Proposition 1.

\begin{matrix}θ & 0 & π/4 & π/2 & 3π/4 \\ m′ & m & \frac{1+m}{1−m} & −\frac{1}{m} & \frac{m−1}{m+1} \end{matrix}

\begin{matrix}θ & π & 5π/4 & 3π/2 & 7π/4 \\ m′ & m & \frac{1+m}{1−m} & −\frac{1}{m} & \frac{m−1}{m+1} \end{matrix}

Thus, we have the orthogonal group, consisting of eight reflections and eight rotations

O_{c}(2) = R_{c} ∪ S_{c},

which gives us the Dihedral group D_{8}, that is, the Euclidean symmetry group of the regular octagon. Now, let us show that all isometries of ℝ_{c}^{2} are in T (2).O_{c}(2).

\underline{Definition}. Let A = (a_{1}, a_{2}) and B = (b_{1}, b_{2}) be two points in ℝ_{c}^{2}. The minimum distance set of A and B is defined by

{X | d_{c}(A, X) + d_{c}(B, X) = d_{c}(A, B)}

and denoted by ÂB (Fig. 4).

Let m_{AB} denote the slope of the line through the points A and B. If m_{AB} = 0, ∓1, or the line is vertical, the set ÂB is the line segment joining A and B, that is, ÂB = \overline{AB}. We call ÂB the standard parallelogram with diagonal \overline{AB} .

\underline{Proposition 7}. Let ϕ: ℝ_{c}^{2} → ℝ_{c}^{2} be an isometry and let ÂB be the standard parallelogram. Then

ϕ(ÂB) = { \atop ϕ(A)ϕ(B)}.

\underline{Proof}. Let Y ∈ ϕ(ÂB). Then the following statements are equivalent.

Y ∈ ϕ(ÂB)

there exists an X ∈ ÂB such that Y = ϕ(X)

d_{c}(A, X) + d_{c}(X, B) = d_{c}(A, B)

d_{c}(ϕ(A), ϕ(X)) + d_{c}(ϕ(X), ϕ(B)) = d_{c}(ϕ(A), ϕ(B))

Y = ϕ(X) ∈ { \atop ϕ(A)ϕ(B)}.

\underline{Corollary 8}. Let ϕ: ℝ_{c}^{2} → ℝ_{c}^{2} be an isometry and let ÂB be the standard parallelogram. Then ϕ maps vertices to vertices and preserves the lengths of sides of ÂB.

\underline{Proposition 9}. Let f: ℝ_{c}^{2} → ℝ_{c}^{2} be an isometry such that f(O) = O. Then f ∈ R_{c} or f ∈ S_{c}.

\underline{Proof}. Let

A_{1} = (1, 0), A_{2} = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) , D = (1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

and consider the standard parallelogram ÔD.

It is clear from Figure 5 that f(A_{1}) ∈ \overline{A_{i}A_{i+1}}. Since f is an isometry by Proposition 6, f(A_{1}) and f(A_{2}) must be the vertices of the standard parallelogram { \atop Of(D)}. Also, when the slope of the parallel sides of a standard parallelogram is 0 or ∞, the slope of the other parallel sides is 1 or −1. Therefore, if f(A_{1}) ∈ \overline{A_{i}A}_{i+1}, then f(A_{1}) = A_{i} or f(A_{1}) = A_{i+1}. Similarly, f(A_{2}) = A_{i} or f(A_{2}) = A_{i+1}.

Case 1: If f(A_{1}) = A_{1}, then f(A_{2}) = A_{2} or f(A_{2}) = A_{8}.

Subcase 1.1: If f(A_{2}) = A_{2}, then f is a rotation with θ = 0.

Subcase 1.2: If f(A_{2}) = A_{8}, then f is a reflection about the line y = 0. Case 2: If f(A_{1}) = A_{2}, then f(A_{2}) = A_{1} or f(A_{2}) = A_{3}.

Subcase 2.1: If f(A_{2}) = A_{1}, then f is a reflection about the line y = (\sqrt{2} − 1)x.

Subcase 2.2: If f(A_{2}) = A_{3}, then f is a rotation with θ = π/4. Case 3: If f(A_{1}) = A_{3}, then f(A_{2}) = A_{2} or f(A_{2}) = A_{4}.

Subcase 3.1: If f(A_{2}) = A_{2}, then f is a reflection about the line y = x. Subcase 3.2: If f(A_{2}) = A_{4}, then f is a rotation with θ = π/2.

Case 4: If f(A_{1}) = A_{4}, then f(A_{2}) = A_{5} or f(A_{2}) = A_{3}.

Subcase 4.1: If f(A_{2}) = A_{5}, then f is a rotation with θ = 3π/4. Subcase 4.2: If f(A_{2}) = A_{3}, then f is a reflection about the line y = (\sqrt{2} + 1)x.

Case 5: If f(A_{1}) = A_{5}, then f(A_{2}) = A_{4} or f(A_{2}) = A_{6}.

Subcase 5.1: If f(A_{2}) = A_{4}, then f is a reflection about the line x = 0. Subcase 5.2: If f(A_{2}) = A_{6}, then f is a rotation with θ = π.

Case 6: If f(A_{1}) = A_{6}, then f(A_{2}) = A_{5} or f(A_{2}) = A_{7}.

Subcase 6.1: If f(A_{2}) = A_{5}, then f is a reflection about the line y=−(\sqrt{2}+1)x. Subcase 6.2: If f(A_{2}) = A_{7}, then f is a rotation with θ = 5π/4.

Case 7: If f(A_{1}) = A_{7}, then f(A_{2}) = A_{6} or f(A_{2}) = A_{8}.

Subcase 7.1: If f(A_{2}) = A_{6}, then f is a reflection about the line y = −x. Subcase 7.2: If f(A_{2}) = A_{8}, then f is a rotation with θ = 3π/2.

Case 8: If f(A_{1}) = A_{8}, then f(A_{2}) = A_{1} or f(A_{2}) = A_{7}.

Subcase 8.1: If f(A_{2}) = A_{1}, then f is a rotation with θ = 7π/4. Subcase 8.2: If f(A_{2}) = A_{7}, then f is a reflection about the line y = (1 − \sqrt{2})x.

\underline{Theorem 10}. Let f: ℝ_{c}^{2} → ℝ_{c}^{2} be an isometry. Then there exists a unique T_{A} ∈ T (2) and g ∈ O_{c}(2) such that f = T_{A} ∘ g.

\underline{Proof}. Let f(O) = A where A = (a_{1}, a_{2}). Define g = T_{−A} ∘ f. We know that g is an isometry and g(O) = O. Thus, g ∈ O_{c}(2) and f = T_{A} ∘ g by Proposition 9. The proof of uniqueness is trivial.

3. Area Formula For CC-Triangles. The area of a triangle in the Euclidean plane can be computed by the formula A = (1/2)bh, which is not, in general, valid in ℝ_{c}^{2}. Area formulas of a taxicab triangle are given in [3] and [5]. If one knows the d_{c}-lengths b_{c} and h_{c} of the base and the corresponding altitude, respectively, of a triangle in ℝ_{c}^{2}, how can its area be computed? The following theorem answers the question and gives the Euclidean area of a triangle in terms of d_{c}-distances.

\underline{Theorem 11}. Let b_{c} and h_{c} denote the d_{c}-lengths of a given side (base) and the corresponding altitude, respectively, of a triangle in ℝ_{c}^{2}. If the slope of base is m, then the area of the triangle can be computed by

A = \frac{1 + m^{2}}{2M^{2}} b_{c}h_{c}, where M = \{ \begin{matrix}1 + (\sqrt{2} − 1) |m| & if |m| ≤ 1 \\ |m| + \sqrt{2} − 1 & if |m| ≥ 1. \end{matrix}

\underline{Proof}. If b, h, and b_{c}, h_{c} are the Euclidean and d_{c}-lengths of the base and the corresponding altitude of a triangle, and if the slope of the base is m, then the slope of altitude is (−1/m) and

b = \frac{1}{ρ(m)}b_{c} = \frac{(1 + m^{2})^{1/2}}{M} b_{c}, h = \frac{1}{ρ(−m^{−1})}h_{c} = \frac{(1 + m^{2})^{1/2}}{M} h_{c}

by Lemma 2. Using these values of b and h in the area formula one obtains

S = \frac{1 + m^{2}}{2M^{2}} b_{c}h_{c}.

Clearly, the above result can be easily extended to the areas of polygons.

\underline{References}

1. G. Chen, Lines and Circles in Taxicab Geometry, Master Thesis, Department of Mathematics and Computer Science, Central Missouri State University, 1992.

2. W. D. Clayton, Euclidean Geometry and Transformations, Addison-Wesley, Reading, MA, 1972.

3. R. Kaya, “Area Formula For Taxicab Triangles,” Pi Mu Epsilon, 12 (2006), 213–220.

4. E. F. Krause, Taxicab Geometry, Addison-Wesley Publishing Company, Menlo Park, CA, 1975.

5. M. Özcan and R. Kaya, “Area of a Triangle in Terms of the Taxicab Distance,” Missouri J. of Math. Sci., 15 (2003), 178–185.

6. S. M. Richard and D. P. George, Geometry, A Metric Approach with Models, Springer-Verlag, New York, 1981.

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

8. M. Willard, Jr., Symmetry Groups and Their Applications, Academic Press, New York, 1972.

Mathematics Subject Classification (2000): 51F15, 51N25, 51Kxx

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

Ö. Gelişgen Department of Mathematics Faculty of Science and Arts University of Eskişehir Osmangazi Eskişehir, TURKEY email: gelisgen@ogu.edu.tr

S. Ekmekçi Department of Mathematics Faculty of Science and Arts University of Eskişehir Osmangazi Eskişehir, TURKEY email: sekmekci@ogu.edu.tr

A. Bayar Department of Mathematics Faculty of Science and Arts University of Eskişehir Osmangazi Eskişehir, TURKEY email: akorkmaz@ogu.edu.tr