By Dileepkumar R

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**Example text**

K + (q − 1)d} where q=ab. So, it remains to see that f is bijective in the respective cases C1 and C2. Suppose first that C1 holds and f (ui ) = f (vj ) for some i, j. Then by equation (14) we get k1 + (i − 1)d = k2 + (j − 1)ad ⇒ k2 − k1 = [(i − 1) − (j − 1)a]d. The last eqality contradicts C1. Thus f must be injective and hence is a required (k,d)arithmetic numbering of Ka,b in this case. On other hand, if under C2 we have f (ui ) = f (vj ) we get rd = k2 − k1 = [(i − 1) − (j − 1)a]d ⇒ (i − 1) − (j − 1)a = r ≥ a ⇒ (i − 1) ≥ ja which is a contradiction as 1 ≤ j ≤ b.

1, for any partition of k into two parts k1 and k2 with 0 ≤ k1 < k2 the map f of (13) is a required (k, d)-arithmetic numbering of C1,b . Hence, we assume that 2 ≤ a < b. 2. k + (a + b − 2)d}. Hence f (ui ) = f (vj ) yields k2 − k1 = (i − j)d (16) Therefore, if d | (k2 − k1 ), then (16) is a contradiction, and on the other hand, if k2 − k1 = rd, r ≥ a, then (16) yields rd = k2 − k1 = (i − j)d ⇒ i−j =r ≥a ⇒ i≥a+j a contradiction to the fact that 1 ≤ i ≤ a and 1 ≤ j ≤ b. Thus f must be a required (k, d)-arithmetic numbering of Ca,b .

2. 2. 1, for any partition of k into two parts k1 and k2 with 0 ≤ k1 < k2 the map f of (13) is a required (k, d)-arithmetic numbering of C1,b . Hence, we assume that 2 ≤ a < b. 2. k + (a + b − 2)d}. Hence f (ui ) = f (vj ) yields k2 − k1 = (i − j)d (16) Therefore, if d | (k2 − k1 ), then (16) is a contradiction, and on the other hand, if k2 − k1 = rd, r ≥ a, then (16) yields rd = k2 − k1 = (i − j)d ⇒ i−j =r ≥a ⇒ i≥a+j a contradiction to the fact that 1 ≤ i ≤ a and 1 ≤ j ≤ b. Thus f must be a required (k, d)-arithmetic numbering of Ca,b .