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# Solution of Term Exam 1

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Problem 1. Find formulas for , and in terms of . (You may use any formula proven in class; you need to quote such formulae, though you don't need to reprove them).

Solution. (Graded by Shay Fuchs) Using the formulas and and taking we get

Dividing the numerator and denominator by this becomes

Likewise using we get

Finally, dividing these two formulas by each other we get

Problem 2.

1. Let be a natural number. Prove that any natural number can be written in a unique way in the form , where and are integers and .
2. We say that a natural number is divisible by '' if is again a natural number. Prove that is divisible by if and only if is divisible by .
3. We say that a natural number is divisible by '' if is again a natural number. Is it true that is divisible by if and only if is divisible by ?

1. We prove this assertion (without uniqueness) by induction. If write (if ) or (if ). In either case the assertion is proven for . Now assume can be written in the form , where and are integers and . If then and so is a formula of the desired form for . Otherwise and so , and again that's a formula of the desired form for . This concludes the proof that every natural number can be written in the form , where and are integers and . Now assume it can be done in two ways; i.e., assume where , , and are integers and . But then and so and so . But is an integer and so is an integer. From it follows that and so and so the integer must be 0. Thus and so . But then the equality implies and so and we see that the pair is unique.

2. An integer is divisible by iff is an integer iff with an integer . Now if is divisible by then with an integer and then . So is also times an integer (the integer ), and so is also divisible by . Assume now that is not divisible by . By the previous part with integer and and with . Had been 0 we'd have had that is divisible by contrary to assumption. So or . In the former case , but then by the uniqueness of writing as it follows that , so cannot be written in the form , so is not divisible by . In the latter case and for the same reason again we find that is not divisible by . So if is divisible by so is , and if is not divisible by so is .

3. No it's not true. Example: is not divisible by but is divisible by .

Problem 3. A function is defined for and has the graph plotted above.

1. What are , and ?
2. Let be the function . What are , and ?
3. Are there any values of for which ? How many such 's are there? Roughly what are they?
4. Plot the graph of the function . (The general shape of your plot should be clear and correct, though numerical details need not be precise).
5. (5 points bonus, will be given only to excellent solutions and may raise your overall exam grade to 105!) Plot the graphs of the functions and .

1. By inspecting the graph, , and .
2. , and .
3. means . Denoting we must have , and inspecting the graph we find that . Thus . Inspecting the graph we find that there are two values of for which this happens and they are approximately and .
4. and 5.:

 g gf gg

Problem 4.

1. Define  '' and  ''.
2. Prove that if and then .
3. Prove that if then and .
4. Draw the graph of some function for which and .

1.  '' means that for every there is a so that whenever we have that , while  '' means that for every there is a so that whenever (i.e., whenever ) we have that

2. Let be given. Using choose so that whenever we have that . Using choose so that whenever we have that . Set and assume . If then and by the choice of it follows that . If then and by the choice of it follows that . So in any case, as required.

3. Let be given. Using choose so that whenever we have that . But then if then certainly so by the choice of we get . Thus . A similar argument shows that also .

Problem 5. Give examples to show that the following definitions of do not agree with the standard one:

1. For all there is an such that if , then .
2. For all there is a such that if , then .