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# Find some formulas for the integral of the step function [t]^2

1. For , prove

2. For , with , define

Draw the graph of on the interval .

3. Find all real such that

1. Proof. Let . Then is a partition of and is constant on the open subintervals of . Further, for . So,

The second to last line follows from this exercise (I.4.7, #6)

2. The graph is:

3. By inspection, we have, .

# Compute some integrals of step functions

1. Let , prove

2. Let , and define

Draw the graph of for .

1. Proof. Let be a partition of . Then, by the definition of the greatest integer function, is constant on the open subintervals of , so

The final equality follows from here (I.4.7, #5)

2. The graph is:

# Prove some statements about even and odd integers

We define even and odd as follows: an integer is even if for some integer , and is odd if is even. An immediate consequence of the definition, which we will feel free to use, is that an integer is odd if and only if for some integer .

1. Prove than an integer cannot be both even and odd.
2. Prove that every integer is either even or odd.
3. Prove that the set of even integers is closed under sums and products. What can be said about sums and products of odd integers?
4. Prove that if is even, must be even as well. Prove also that for integers and , if , then both and must be even.
5. Prove that every rational number can be written as the ratio of integers and at least one of which is odd.

1. Proof. Suppose otherwise, that there is some that is both even and odd. Then, there exist such that and (since is odd, is even, so ). Hence,

But , contradicting the fact that the integers are closed under addition (and subtraction)

2. Proof. Let be an arbitrary integer. Then, is a real number, so by I.3.12, Exercise #4 we know there exists such that

If , then is even and we are done.
Otherwise, if , then we have

But this implies since is the only integer in the interval. Thus, is odd.
Hence, every integer is either even or odd

3. Proof. Let be even integers. Then, there exist integers and such that , . Then,

where and are integers. Hence, and are both even

Next, we claim the sum of two odd integers is even, and the product of two odd integers is odd.
Proof. Let be odd integers. Then there exist integers and such that and . Then,

Hence, for some integer and for some integer . Therefore, is even and is odd, as claimed

4. Proof. Let with even. Since and we know from part (b) that must be either even or odd and from part (a) that it cannot be both. By part (c) the product of two even integers even, while the product of two odd integers is odd; hence, we must have even (otherwise would be the product of two odd integers and would have to be odd).

Next, if and are integers with , then is even by definition (since it equals for some integer ). So, by the above we know is even, say . Then . But then we have,

and by the same reasoning that lead to even, we also have must be even. Thus, both and are even

5. Proof. Let be a rational number with . If and are both even, then we have

with , and . Now, if and are both even, repeat the process. This will give a strictly decreasing sequence of positive integers, so the process must terminate by the well-ordering principle (see I.4.3, p. 34 in Apostol, Calculus, Volume I ). Thus, we must have some integers and , not both even with . This was the requested property