The floor function and the ceiling function are defined by respectively.
Proof. By the from the difinitions of the floor and ceiling functions, we have that This establishes claim (). Claim () is true, because where the second equality follows from the first claim ().
In the interval , there is one and only one integer. Because it is the largest among all integers below or equal to , it must be . Claim () therefore follows. We can verify claim () by applying claim () setting to and applying claim ().
The first inequlity in claim () is true by the second inequlity in claim (). Suppose that the second inequality in claim () is violated. Then the integer is no greater than but it is greater than ; a contradiction. To prove claim (), we replace in claim () with and apply claim () to obtain: Multiplying each part of this inequality by establishes the desired result.
Claims ()–() are direct consequences of the definitions of the floor and ceiling functions, except for the second implications in claims () and (). To verify the second implication in claim (), we establish its contraposition. Suppose that . Then is an integer no greater than , so that . The second implication in claim () can be verified analogously. ◻
Proof. Application of Theorem () and () yields that and It follows that we have that Becuase the middle term in this inequality is integer, it must be zero. This establishes (). We can obtain () by setting to in (). To verify () and (), apply () and (), respectively, with the fact that . ◻
Proof. Let , , and . Then it holds that , , and . It follows that , and that where the last equality holds because is integer, and . We also have that because . The equality () therefor follows.
For (), application of Theorem () yields that and By (), the right-hand side of these equalities are equal so that () holds. ◻
Proof. Setting to in Theorem yields the desired results. ◻
Proof. Write Then . Let and . Then we have that . Under the assumption of claim (), we also have that . It thus holds that and Claim () therefore follows.
To verify claim (), we apply Theorem () and claim () of the current thm along with the fact that under the assumption of claim (): ◻
The truncating division operator is defined by The remainder function is defined by
Note: This next theorem has been modified for clarification.
Proof. Let and . Then we have that and that If , we see that and . Otherwise, it holds that It follows that and . The equality () therefore follows.
For (), we have that where . If (i.e., ), it holds that and ; otherwise, it holds that so that and , as () claims. ◻
Proof. Suppose that . Then it follows from () that Thus, the equality in question holds. We can also analogously verify the equality for the case in which . ◻