Problem 1 asks to solve in the equation . 

This equation is equivalent to Since , there is a such that and multiplying by both sides of gives us . To finish the problem, it then suffices to find . For this, we use the Euclidean algorithm:

Now we work backwards from these equalities:

This means that, in , if we let then

(In effect, because )

And the value of we are looking for is

Problem 2 asks for a solution to the equation in

This can be done by direct computation. We have that 5 and 8 are solutions, since and

Remark: This is related to the fact, that you may have conjectured through the homework, that an odd prime is a sum of two squares iff We have The relation with the problem is that, in this equation becomes

or or where for the such that namely Note that  

Similarly, gives or But where i.e., Note that

Problem 3 asks to find the order of in i.e., the smallest such that

 There are two ways of solving this problem. One is to simply write down the powers of 3 until we reach 1:

This means that

The other way is perhaps less computational, but it requires a bit more theory. Recall form lecture Euler’s theorem stating that if then where is the number of numbers with such that and are relatively prime.

From the formula mentioned in lecture,

This means that is a divisor of 40, so it is one of These are the only powers we need to compute. This is faster than computing all the powers since, for example, We have:

and we have

Problem 1 asks to show that if three numbers are given, each a power of 2 or of 3, or a product of powers of 2 and 3, then either one of these three numbers is a square, or else the product of some of them is a square.

Each of these numbers, and their products, can be written in the form with A number of this form is a square if and only if both and are even. 

To a number we associate its “pattern of parities,” the pair Note that if has pattern and has pattern then their product has pattern

If one of the three numbers has pattern we are done. If two of the numbers have the same pattern, their product is a square, and we are done. So we may assume that the numbers have patterns and But then the product of the three of them is a square.

More generally, one can check that if positive integers are given (), and their prime factorizations together involve only primes, then there is a (non-empty) subset of these integers whose product is a square. Try to show this as an extra credit problem. 

Problem 2 asks to show that if 9 lattice points are given in there are two such that the midpoint of the segment they determine is also a lattice point. Here, a lattice point is a point all of whose coordinates are integers.

As before, associate to each lattice point its pattern of parity, Note that the midpoint of the segment joining and is It follows that this midpoint is a lattice point iff and have the same pattern of parity. 

But there are only 8 possible patterns: each of the three coordinates is either 0 or 1. Since 9 points are given, two must have the same pattern, and we are done.

Problem 3 asks for a list of 4 distinct integers without an increasing or decreasing subsequence of length 3. 

There are many possible ways of doing this. For example,

Problem 1 asks to show that if three numbers are given, each a power of 2 or of 3, or a product of powers of 2 and 3, then either one of these three numbers is a square, or else the product of some of them is a square.

Each of these numbers, and their products, can be written in the form with A number of this form is a square if and only if both and are even. 

To a number we associate its “pattern of parities,” the pair Note that if has pattern and has pattern then their product has pattern

If one of the three numbers has pattern we are done. If two of the numbers have the same pattern, their product is a square, and we are done. So we may assume that the numbers have patterns and But then the product of the three of them is a square.

More generally, one can check that if positive integers are given (), and their prime factorizations together involve only primes, then there is a (non-empty) subset of these integers whose product is a square. Try to show this as an extra credit problem. 

Problem 2 asks to show that if 9 lattice points are given in there are two such that the midpoint of the segment they determine is also a lattice point. Here, a lattice point is a point all of whose coordinates are integers.

As before, associate to each lattice point its pattern of parity, Note that the midpoint of the segment joining and is It follows that this midpoint is a lattice point iff and have the same pattern of parity. 

But there are only 8 possible patterns: each of the three coordinates is either 0 or 1. Since 9 points are given, two must have the same pattern, and we are done.

Problem 3 asks for a list of 4 distinct integers without an increasing or decreasing subsequence of length 3. 

There are many possible ways of doing this. For example,

The exercises in the handout are extra-credit. All extra-credit problems, other than the two due this Friday, are due the last day of lecture. (I may change this date, if need be, but let’s try it for now.)

The exercises in the handout are extra-credit. All extra-credit problems, other than the two due this Friday, are due the last day of lecture. (I may change this date, if need be, but let’s try it for now.)

Problem 1 asks to find integers such that  

One way of doing this is to follow the Euclidean algorithm to compute the g.c.d. of 13 and 8:

Now we work from these equations, starting with the last one and going up, expressing the remainders as linear combinations of the other two numbers:

Since we have

Since we have

Since we have

We have found that satisfy  

There are, of course, other ways we could have found these numbers, including trial and error: Simply list the multiples of 13 in order: 13, 26, 39, …, and note that 39 is 1 shy of a multiple of 8: giving the same solution as before.

Problem 2 asks for integers neither of which is 0, and such that

Perhaps the easiest solution is to make

Problem 3 asks for integers different from those in problem 1, such that

One way of finding these numbers is by adding the solutions to problems 1 and 2:

so

We could have also subtracted the solution to problem 2 from the solution to problem 1, to obtain

Or we could have squared the solution to problem 1: or

or

Problem 4 asks  for the number of injective functions from into

Let be 1-1. There are 5 possibilities for the value Whatever it is, can take any value in other than so there are 4 possibilities for The value can be anything in other than or so there are 3 possibilities for Finally, can take any of the remaining 2 values. This gives a total of injective functions.

Note that this is the same as the number of lists of length 4 without repetitions with values taken from the set

Problem 1 asks to find integers such that  

One way of doing this is to follow the Euclidean algorithm to compute the g.c.d. of 13 and 8:

Now we work from these equations, starting with the last one and going up, expressing the remainders as linear combinations of the other two numbers:

Since we have

Since we have

Since we have

We have found that satisfy  

There are, of course, other ways we could have found these numbers, including trial and error: Simply list the multiples of 13 in order: 13, 26, 39, …, and note that 39 is 1 shy of a multiple of 8: giving the same solution as before.

Problem 2 asks for integers neither of which is 0, and such that

Perhaps the easiest solution is to make

Problem 3 asks for integers different from those in problem 1, such that

One way of finding these numbers is by adding the solutions to problems 1 and 2:

so

We could have also subtracted the solution to problem 2 from the solution to problem 1, to obtain

Or we could have squared the solution to problem 1: or

or

Problem 4 asks  for the number of injective functions from into

Let be 1-1. There are 5 possibilities for the value Whatever it is, can take any value in other than so there are 4 possibilities for The value can be anything in other than or so there are 3 possibilities for Finally, can take any of the remaining 2 values. This gives a total of injective functions.

Note that this is the same as the number of lists of length 4 without repetitions with values taken from the set

Solutions follow.

Problem 1 introduces a relation on integers by saying that iff the difference between and is at most 3. In symbols, The problem asks to determine whether is transitive.

It is not. For a counterexample, consider We have and but

Problem 2 lets be the collection of finite subsets of a set and defines a relation on by setting iff is even. The problem is to show that is reflexive and symmetric.

To prove reflexivity, let be an arbitrary finite subset of Then so Since is even, we conclude that

To prove symmetry, let be finite subsets of such that is even. Since we have that is even, and

Problem 3 informs us that is a property about integers and asks us to choose among a few options for what is needed to do in order to prove that “For every integer there exists an integer such that is true.”

To do this, we need to start with an arbitrary integer and then find an integer that makes true. Of course, if we choose a different it could be that the that now works is different, i.e., it may well be that depends on Of the options listed in the problem, it is (e) that precisely captures this.

Note that if one manages to show (a) or (f), then one also succeeds in proving the required statement. However, both (a) and (f) are too strong in general, i.e., there are cases where the given statement holds but both (a) and (f) fail. None of the other options even guarantees the truth of the statement.

Problem 4 gives as an erroneous proof, and requires to identify the flaw. The statement being claimed is that all natural numbers are divisible by 3. The argument uses the well-ordering principle, and contradiction. Assuming that the statement is false means that there are natural numbers not divisible by 3. Letting be the set of all these numbers, this means that The well-ordering principle guarantees that there is a least element of and we call it  

Clearly, is not divisible by 3, so in particular

The argument then asks to consider Obviously, and therefore is not in If happens to be a natural number, then we are forced to conclude that and therefore since However, there is also the chance that in which case we cannot conclude that Thus, we cannot conclude that The flaw in the argument, of course, is in assuming that which is only assured if

Problem 5 asks to prove by induction that, for all positive integers we have

To argue by induction, we first prove the base case, when We have on the left hand side and on the right hand side as well. The inequality holds (in fact, it is an equality in this case).

Now we argue the inductive step. Assume that, indeed, 

We need to prove that also

To do this, note that  

The sum inside the first set of parentheses is at least by the inductive assumption. We are done if we manage to show that the sum within the second set of parentheses is at least  

But note that etc, so the second sum is at least 

where there are precisely many terms being added, all of them equal to This means that this sum adds up to Since this is what we needed, we have completed the inductive step, and therefore the proof.

Solutions follow.

Problem 1 introduces a relation on integers by saying that iff the difference between and is at most 3. In symbols, The problem asks to determine whether is transitive.

It is not. For a counterexample, consider We have and but

Problem 2 lets be the collection of finite subsets of a set and defines a relation on by setting iff is even. The problem is to show that is reflexive and symmetric.

To prove reflexivity, let be an arbitrary finite subset of Then so Since is even, we conclude that

To prove symmetry, let be finite subsets of such that is even. Since we have that is even, and

Problem 3 informs us that is a property about integers and asks us to choose among a few options for what is needed to do in order to prove that “For every integer there exists an integer such that is true.”

To do this, we need to start with an arbitrary integer and then find an integer that makes true. Of course, if we choose a different it could be that the that now works is different, i.e., it may well be that depends on Of the options listed in the problem, it is (e) that precisely captures this.

Note that if one manages to show (a) or (f), then one also succeeds in proving the required statement. However, both (a) and (f) are too strong in general, i.e., there are cases where the given statement holds but both (a) and (f) fail. None of the other options even guarantees the truth of the statement.

Problem 4 gives as an erroneous proof, and requires to identify the flaw. The statement being claimed is that all natural numbers are divisible by 3. The argument uses the well-ordering principle, and contradiction. Assuming that the statement is false means that there are natural numbers not divisible by 3. Letting be the set of all these numbers, this means that The well-ordering principle guarantees that there is a least element of and we call it  

Clearly, is not divisible by 3, so in particular

The argument then asks to consider Obviously, and therefore is not in If happens to be a natural number, then we are forced to conclude that and therefore since However, there is also the chance that in which case we cannot conclude that Thus, we cannot conclude that The flaw in the argument, of course, is in assuming that which is only assured if

Problem 5 asks to prove by induction that, for all positive integers we have

To argue by induction, we first prove the base case, when We have on the left hand side and on the right hand side as well. The inequality holds (in fact, it is an equality in this case).

Now we argue the inductive step. Assume that, indeed, 

We need to prove that also

To do this, note that  

The sum inside the first set of parentheses is at least by the inductive assumption. We are done if we manage to show that the sum within the second set of parentheses is at least  

But note that etc, so the second sum is at least 

where there are precisely many terms being added, all of them equal to This means that this sum adds up to Since this is what we needed, we have completed the inductive step, and therefore the proof.

1. Let be the function given by Show that is a bijection.
2. Define by setting and, if then, letting be the smallest prime such that then [Here, .] Compute for and show that is one-to-one. 

1. Let be the function given by Show that is a bijection.
2. Define by setting and, if then, letting be the smallest prime such that then [Here, .] Compute for and show that is one-to-one. 

Solutions follow.

Problem 1 asks us to prove that, for all natural numbers we have

There are several ways of showing this. I present two proofs, one using the well-ordering principle (least counterexample), and the other arguing by induction.

Proof 1: Using the well-ordering principle, one proceeds as follows: Suppose the result is false. This means that there is a natural for which the identity fails, and therefore (since the set of naturals for which the identity fails is non-empty) there is a least natural for which the identity fails. This means several things:

1. 
2. If is a natural number,

We begin by noticing that because meaning that the identity holds in this case.

It follows that is also a natural number. By item 2, we then have that

The plan is now to somehow transform this equality into the equality that item 1 claims must fail. To do this, the most direct route seems to be to add to both sides of the displayed equation. We have

We have indeed shown that item 1 is false. This is a contradiction, and we conclude that our assumption (that the identity fails for some natural ) is false, meaning of course that the identity holds for all natural numbers, as we wanted to prove.

Proof 2: Using induction, one argues as follows: First, we show that the identity is true when This is indeed the case, since

Now we assume that the identity holds for some number meaning that

We need to somehow conclude that also To do this, we can start with the identity for that we are assuming is true, and add to both sides. We obtain

Now we observe that and we conclude that the identity also holds for as we wanted to show.

By induction, we conclude that the identity holds for all natural numbers.

[Note that both proofs are very similar, as was to be expected.]

Problem 2 starts by defining a sequence by recursion: We set and, in general, for all naturals We are asked to prove that for all  

Once again, one can argue by either the well-ordering principle, or induction. I present here a proof using induction. First, we consider the case We have which is obviously larger than

Assuming now that we need to prove that as well. For this, we use that from which it follows, using the inductive assumption, that

This is what we needed to prove. By induction, we conclude that for all

Remark: One can also prove that induction that for all i.e., that the sequence is decreasing. To see this, note that Assuming that we then have that then and so But this means (using the recursive definition of the sequence) that By induction, the sequence is decreasing.

A fundamental result in calculus is that a decreasing sequence that is bounded below must converge. We can now easily compute In effect, let Then, of course, we also have Using the recursive definition of the sequence, we then have The equality quickly gives us that

As a matter of fact, we can show that if is any real number, and we define the sequence recursively, by setting for all then:

1. If then for all and the sequence is decreasing, meaning that for all
2. If then for all and the sequence is increasing, meaning that for all  
3. If then for all
4. In all three cases, 

Though the example presented here is simple, this technique of computing limits of sequences by studying their behavior using induction, is actually very useful in higher calculus (mathematical analysis).

Problem 3 asks us to identify the flaw in the following wrong argument:

Theorem. All natural numbers are equal.

Proof. We argue by induction on that if and then  

The case holds, because if and are natural numbers, then in fact

Suppose then that the result holds for and consider natural numbers with Then and, by the inductive assumption, it follows that But then  

The argument is presented as a proof by induction, with the base case and the inductive case indicated. If indeed both cases are argued correctly, the conclusion will follow. This means that the flaw is either in the proof of the base case or of the inductive case.

The base case, however, is proved correctly: If are natural numbers, then If we are also given that then we have and so indeed

It follows that the flaw must lie somewhere within the inductive case. Now, if then indeed Also, if then also And, finally, if it is indeed the case that we can apply the inductive assumption then, from we are forced to conclude that

It follows that the flaw is in assuming that the inductive assumption applies. Recall that the inductive assumption is:

If and then

And we are using it with and Since the only place we have left, and thus the place where the problem should reside, is in affirming that  

We see that this is, as expected, a problem, because if and then we cannot conclude that both as well. Of course, one of equals and but, for all we know, we could have and Then and the inductive assumption does not apply to since they are not both natural numbers.

Solutions follow.

Problem 1 asks us to prove that, for all natural numbers we have

There are several ways of showing this. I present two proofs, one using the well-ordering principle (least counterexample), and the other arguing by induction.

Proof 1: Using the well-ordering principle, one proceeds as follows: Suppose the result is false. This means that there is a natural for which the identity fails, and therefore (since the set of naturals for which the identity fails is non-empty) there is a least natural for which the identity fails. This means several things:

1. 
2. If is a natural number,

We begin by noticing that because meaning that the identity holds in this case.

It follows that is also a natural number. By item 2, we then have that

The plan is now to somehow transform this equality into the equality that item 1 claims must fail. To do this, the most direct route seems to be to add to both sides of the displayed equation. We have

We have indeed shown that item 1 is false. This is a contradiction, and we conclude that our assumption (that the identity fails for some natural ) is false, meaning of course that the identity holds for all natural numbers, as we wanted to prove.

Proof 2: Using induction, one argues as follows: First, we show that the identity is true when This is indeed the case, since

Now we assume that the identity holds for some number meaning that

We need to somehow conclude that also To do this, we can start with the identity for that we are assuming is true, and add to both sides. We obtain

Now we observe that and we conclude that the identity also holds for as we wanted to show.

By induction, we conclude that the identity holds for all natural numbers.

[Note that both proofs are very similar, as was to be expected.]

Problem 2 starts by defining a sequence by recursion: We set and, in general, for all naturals We are asked to prove that for all  

Once again, one can argue by either the well-ordering principle, or induction. I present here a proof using induction. First, we consider the case We have which is obviously larger than

Assuming now that we need to prove that as well. For this, we use that from which it follows, using the inductive assumption, that

This is what we needed to prove. By induction, we conclude that for all

Remark: One can also prove that induction that for all i.e., that the sequence is decreasing. To see this, note that Assuming that we then have that then and so But this means (using the recursive definition of the sequence) that By induction, the sequence is decreasing.

A fundamental result in calculus is that a decreasing sequence that is bounded below must converge. We can now easily compute In effect, let Then, of course, we also have Using the recursive definition of the sequence, we then have The equality quickly gives us that

As a matter of fact, we can show that if is any real number, and we define the sequence recursively, by setting for all then:

1. If then for all and the sequence is decreasing, meaning that for all
2. If then for all and the sequence is increasing, meaning that for all  
3. If then for all
4. In all three cases, 

Though the example presented here is simple, this technique of computing limits of sequences by studying their behavior using induction, is actually very useful in higher calculus (mathematical analysis).

Problem 3 asks us to identify the flaw in the following wrong argument:

Theorem. All natural numbers are equal.

Proof. We argue by induction on that if and then  

The case holds, because if and are natural numbers, then in fact

Suppose then that the result holds for and consider natural numbers with Then and, by the inductive assumption, it follows that But then  

The argument is presented as a proof by induction, with the base case and the inductive case indicated. If indeed both cases are argued correctly, the conclusion will follow. This means that the flaw is either in the proof of the base case or of the inductive case.

The base case, however, is proved correctly: If are natural numbers, then If we are also given that then we have and so indeed

It follows that the flaw must lie somewhere within the inductive case. Now, if then indeed Also, if then also And, finally, if it is indeed the case that we can apply the inductive assumption then, from we are forced to conclude that

It follows that the flaw is in assuming that the inductive assumption applies. Recall that the inductive assumption is:

If and then

And we are using it with and Since the only place we have left, and thus the place where the problem should reside, is in affirming that  

We see that this is, as expected, a problem, because if and then we cannot conclude that both as well. Of course, one of equals and but, for all we know, we could have and Then and the inductive assumption does not apply to since they are not both natural numbers.

Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.

To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.

By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.

Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties. 

To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.

To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,

To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.

(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )

Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).

That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all  naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with

Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.

To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.

By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.

Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties. 

To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.

To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,

To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.

(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )

Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).

That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all  naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with

Problem 1 asks to determine (with brief justifications) the truth value of the following statements about integers:

1. is False. To show this we provide a counterexample: Specific integers such that For example,

2. is False. To show this we need to exhibit for each integer an integer such that For example, Note that, although is a fixed integer once we know we are not giving a fixed value of that serves as a simultaneous counterexample for all values of

3. is True. To show this we exhibit for each integer a specific integer such that For example: Note that, although is a fixed integer once we know we are not giving a fixed value of that works simultaneously for all

4. is True. To show this, we exhibit specific values of such that For example:

Problem 2 asks to show by contradiction that no integer can be both odd and even. Here is the proof: Suppose otherwise, i.e., there is an integer, let’s call it such that is both odd and even. This means that there are integers such that (since is odd) and (since is even). 

Then we have that or But this is impossible, since 1 is not divisible by 2. We have reached a contradiction, and therefore our assumption that there is such an integer ought to be false. This means that no integer can be both odd and even, which is what we wanted to show.

Note that we have not shown that every integer is either odd or even. We will use mathematical induction to do this.

Problem 3 asks for symbolic formulas stating Goldbach’s conjecture and the twin primes conjecture (both are famous open problems in number theory).

Goldbach’s conjecture asserts that every even integer larger than 2 is sum of two primes:

Here, is the formula asserting that is even, namely, and is the formula (given in the quiz) asserting that is prime. Note we had to add existential quantifiers in order to be able to refer to the two prime numbers that add up to

The twin primes conjecture asserts that there are infinitely many primes such that is also prime. 

The difficulty here is in saying “there are infinitely many,” since the quantifier only allows us to mention one integer at a time, and writing something of infinite length such as is not allowed.

We follow the suggestion given in the quiz, and represent “there are infinitely many with [some property]” by saying “for all there is a larger with [some property].”

Problem 1 asks to determine (with brief justifications) the truth value of the following statements about integers:

1. is False. To show this we provide a counterexample: Specific integers such that For example,

2. is False. To show this we need to exhibit for each integer an integer such that For example, Note that, although is a fixed integer once we know we are not giving a fixed value of that serves as a simultaneous counterexample for all values of

3. is True. To show this we exhibit for each integer a specific integer such that For example: Note that, although is a fixed integer once we know we are not giving a fixed value of that works simultaneously for all

4. is True. To show this, we exhibit specific values of such that For example:

Problem 2 asks to show by contradiction that no integer can be both odd and even. Here is the proof: Suppose otherwise, i.e., there is an integer, let’s call it such that is both odd and even. This means that there are integers such that (since is odd) and (since is even). 

Then we have that or But this is impossible, since 1 is not divisible by 2. We have reached a contradiction, and therefore our assumption that there is such an integer ought to be false. This means that no integer can be both odd and even, which is what we wanted to show.

Note that we have not shown that every integer is either odd or even. We will use mathematical induction to do this.

Problem 3 asks for symbolic formulas stating Goldbach’s conjecture and the twin primes conjecture (both are famous open problems in number theory).

Goldbach’s conjecture asserts that every even integer larger than 2 is sum of two primes:

Here, is the formula asserting that is even, namely, and is the formula (given in the quiz) asserting that is prime. Note we had to add existential quantifiers in order to be able to refer to the two prime numbers that add up to

The twin primes conjecture asserts that there are infinitely many primes such that is also prime. 

The difficulty here is in saying “there are infinitely many,” since the quantifier only allows us to mention one integer at a time, and writing something of infinite length such as is not allowed.

We follow the suggestion given in the quiz, and represent “there are infinitely many with [some property]” by saying “for all there is a larger with [some property].”

Solutions follow.

Problem 1 concerns lists. We are given a collection of 5 objects, and we want to find:

1. The number of lists of length 2 where the entries come from these objects, and repetitions are allowed. We have that for each entry we can use 5 possible objects, for a total of lists.
2. The number of lists of length 2 where the entries come from these objects, and repetitions are required. Now, once we select the first entry (there are 5 possible ways of doing this), the second entry is completely determined. So we only have lists.
3. The number of lists of length 2 where the entries come from these objects, and repetitions are not allowed. There are two ways of counting: Either we simply notice that this is the total number of lists of length 2 (25), except for those that have repetitions (5), for a total of or else we note that whatever first entry we choose (5 possibilities), this eliminates one of the 5 possible objects from being the second entry (so now we only have 4 possibilities for this entry), for a total of

The extra credit problem asked for the number of lists of length 3 whose entries come from and repetitions are required. Again, there are two ways of counting. The easiest way is simply to count of lists of length 3 () and subtract from them those lists that do not have repetitions () for a total of lists. The second way counts the lists directly, but this is a bit more complicated. There are 5 possibilities for the first entry. If the second entry is the same, the third entry is arbitrary (so this gives us lists so far). If the second entry is different (4 possibilities) then the last entry is either the same as the first or the second (2 possibilities), for a total of lists. Hence, the total number of lists we are looking for is

Problem 2 asks us to explain in what sense the statement is equivalent to the statement and to prove that this is the case. 

Two statements are equivalent precisely when they have the same truth values, i.e., they are both true in precisely the same cases and both are false in precisely the same cases. To see that this is the case, the easiest method is to compare the truth tables for both statements and see that they are the same. This is indeed the case: Both statements are true precisely when latex \varphi$ and have the same truth value (be it true or false).

There is another way in which the equivalence can be analyzed. The statement is known to be equivalent to the conjunction Let’s analyze when this conjunction holds, which is the same as saying that both implications must be true. Note that holds if is true. However, if is false, then the implication only holds if is also false. This is to say, This statement is called the contrapositive of and our analysis just showed that the two are equivalent.

In the case that interests us, we see that is equivalent to which is equivalent, considering the contrapositives of both implications, to which is equivalent to which in turn is equivalent to as we wanted to see.

There is yet another way in which one can make sense of two statements and being equivalent, namely, that we can prove from and also prove from A careful study of this sense would lead us to study provability in the context of mathematical logic. 

Problem 3 asks us to show that if are sets and is an element such that and then also

To prove this, recall that If belongs to this symmetric difference, then there are two possibilities: Either or else

1. Assume that and recall that this means that but Then (since we are also given that ) we have that (by definition of ) and (again, by definition of ). But then and therefore (by definition of ), i.e., by definition of
2. Assuming that leads us to the same conclusion by essentially the same argument.

Since in both cases we have reached the desired conclusion, we are done.

Problem 4 asks to show that if and is divisible by then is a factor of

The problem essentially asked us to remember that “ is divisible by ” means the same as and that “ is a factor of .” All three statements mean that are integers and there is an integer such that

We are given that and There are then integers and such that and Then, multiplying the first equation by we get But and it follows that there is an integer (namely, ) such that This is the definition of which is what we needed to prove.

Problem 5 asks us to count the number of positive divisors of  

To find this number, note that and therefore any divisor will have the form where are integers with and

This means that the number of divisors is the same as the number of  lists with as explained. There are 4 possibilities for 3 for 2 for and 2 for for a total of lists, and therefore 48 is the number of positive divisors of

Solutions follow.

Problem 1 concerns lists. We are given a collection of 5 objects, and we want to find:

1. The number of lists of length 2 where the entries come from these objects, and repetitions are allowed. We have that for each entry we can use 5 possible objects, for a total of lists.
2. The number of lists of length 2 where the entries come from these objects, and repetitions are required. Now, once we select the first entry (there are 5 possible ways of doing this), the second entry is completely determined. So we only have lists.
3. The number of lists of length 2 where the entries come from these objects, and repetitions are not allowed. There are two ways of counting: Either we simply notice that this is the total number of lists of length 2 (25), except for those that have repetitions (5), for a total of or else we note that whatever first entry we choose (5 possibilities), this eliminates one of the 5 possible objects from being the second entry (so now we only have 4 possibilities for this entry), for a total of

The extra credit problem asked for the number of lists of length 3 whose entries come from and repetitions are required. Again, there are two ways of counting. The easiest way is simply to count of lists of length 3 () and subtract from them those lists that do not have repetitions () for a total of lists. The second way counts the lists directly, but this is a bit more complicated. There are 5 possibilities for the first entry. If the second entry is the same, the third entry is arbitrary (so this gives us lists so far). If the second entry is different (4 possibilities) then the last entry is either the same as the first or the second (2 possibilities), for a total of lists. Hence, the total number of lists we are looking for is

Problem 2 asks us to explain in what sense the statement is equivalent to the statement and to prove that this is the case. 

Two statements are equivalent precisely when they have the same truth values, i.e., they are both true in precisely the same cases and both are false in precisely the same cases. To see that this is the case, the easiest method is to compare the truth tables for both statements and see that they are the same. This is indeed the case: Both statements are true precisely when latex \varphi$ and have the same truth value (be it true or false).

There is another way in which the equivalence can be analyzed. The statement is known to be equivalent to the conjunction Let’s analyze when this conjunction holds, which is the same as saying that both implications must be true. Note that holds if is true. However, if is false, then the implication only holds if is also false. This is to say, This statement is called the contrapositive of and our analysis just showed that the two are equivalent.

In the case that interests us, we see that is equivalent to which is equivalent, considering the contrapositives of both implications, to which is equivalent to which in turn is equivalent to as we wanted to see.

There is yet another way in which one can make sense of two statements and being equivalent, namely, that we can prove from and also prove from A careful study of this sense would lead us to study provability in the context of mathematical logic. 

Problem 3 asks us to show that if are sets and is an element such that and then also

To prove this, recall that If belongs to this symmetric difference, then there are two possibilities: Either or else

1. Assume that and recall that this means that but Then (since we are also given that ) we have that (by definition of ) and (again, by definition of ). But then and therefore (by definition of ), i.e., by definition of
2. Assuming that leads us to the same conclusion by essentially the same argument.

Since in both cases we have reached the desired conclusion, we are done.

Problem 4 asks to show that if and is divisible by then is a factor of

The problem essentially asked us to remember that “ is divisible by ” means the same as and that “ is a factor of .” All three statements mean that are integers and there is an integer such that

We are given that and There are then integers and such that and Then, multiplying the first equation by we get But and it follows that there is an integer (namely, ) such that This is the definition of which is what we needed to prove.

Problem 5 asks us to count the number of positive divisors of  

To find this number, note that and therefore any divisor will have the form where are integers with and

This means that the number of divisors is the same as the number of  lists with as explained. There are 4 possibilities for 3 for 2 for and 2 for for a total of lists, and therefore 48 is the number of positive divisors of

Problem 1 requests a counterexample to the statement: An integer is positive if and only if is positive.

To find a counterexample to a statement of the form “an integer has property if and only if it has property we need an integer such that one of the properties is true of while the other is false. Note that if is positive, i.e., then i.e., is also positive. This means that the only way the statement is going to fail is if we have that is positive yet is not. 

We can now easily see that is a counterexample. (And, in fact, it is the only counterexample.)

Problem 2 asks to show that where and  

By definition, iff any element of is also an element of Let us then consider an element of By definition of this means that is an integer, and We need to show that i.e., that and that The first requirement is automatic, so we only need to verify that  

To see this, we use that By definition, this means that there is an integer, let us call it such that To show that we need (again, by definition) that there is an integer such that Now, since then or We immediately see that we can take This is an integer, and as needed.

Problem 3 asks for the cardinality of the set  

Recall that for any set then as shown in lecture. In particular, if , then Now let Then

Problem 1 requests a counterexample to the statement: An integer is positive if and only if is positive.

To find a counterexample to a statement of the form “an integer has property if and only if it has property we need an integer such that one of the properties is true of while the other is false. Note that if is positive, i.e., then i.e., is also positive. This means that the only way the statement is going to fail is if we have that is positive yet is not. 

We can now easily see that is a counterexample. (And, in fact, it is the only counterexample.)

Problem 2 asks to show that where and  

By definition, iff any element of is also an element of Let us then consider an element of By definition of this means that is an integer, and We need to show that i.e., that and that The first requirement is automatic, so we only need to verify that  

To see this, we use that By definition, this means that there is an integer, let us call it such that To show that we need (again, by definition) that there is an integer such that Now, since then or We immediately see that we can take This is an integer, and as needed.

Problem 3 asks for the cardinality of the set  

Recall that for any set then as shown in lecture. In particular, if , then Now let Then

Problem 1 asks to show that an integer is odd if and only if it is the sum of two consecutive integers.

Here is an argument: Let be an integer. We need to prove two statements, corresponding to the two directions of the “if and only if:”

1. If is odd, then there are two consecutive integers and such that
2. If is the sum of two consecutive integers and then is odd.

1. Assume that is odd. Then, by definition, there is an integer such that Note that so we can take and we see that and are consecutive, and

2. Now assume that where and are consecutive integers. Say, Then By definition, this means that is odd.

Problem 2 asks to show that if are integers and we have that and then also

To see this, assume that are integers such that and By definition, this means that there are integers and such that and Then But Let Then is an integer, and By definition, this means that

Problem 3 asks to show that if is a positive integer, then is composite.

As stated, this is false. For a counterexample, note that is a positive integer. However, is not composite according to the definition given in the book: 

An integer is composite if and only if there is an integer such that and

In any case, it is true that if is an integer and then is composite. To see this, note first that Now, if then and so Also, since then and therefore We have shown that and that By definition, this means that is composite.

Problem 1 asks to show that an integer is odd if and only if it is the sum of two consecutive integers.

Here is an argument: Let be an integer. We need to prove two statements, corresponding to the two directions of the “if and only if:”

1. If is odd, then there are two consecutive integers and such that
2. If is the sum of two consecutive integers and then is odd.

1. Assume that is odd. Then, by definition, there is an integer such that Note that so we can take and we see that and are consecutive, and

2. Now assume that where and are consecutive integers. Say, Then By definition, this means that is odd.

Problem 2 asks to show that if are integers and we have that and then also

To see this, assume that are integers such that and By definition, this means that there are integers and such that and Then But Let Then is an integer, and By definition, this means that

Problem 3 asks to show that if is a positive integer, then is composite.

As stated, this is false. For a counterexample, note that is a positive integer. However, is not composite according to the definition given in the book: 

An integer is composite if and only if there is an integer such that and

In any case, it is true that if is an integer and then is composite. To see this, note first that Now, if then and so Also, since then and therefore We have shown that and that By definition, this means that is composite.

Problem 1 is False. This is because the number 1 is neither prime nor composite. 

Problem 2 is False. This is because we can have in which case but For example, consider

For problem 3, start by writing the number as a product of primes: Plainly, any positive divisor of must have the form where or similarly, or ; or and or There are 8 possibilities for 5 for 2 for and 2 for This gives us a total of possible positive divisors.

Problem 1 is False. This is because the number 1 is neither prime nor composite. 

Problem 2 is False. This is because we can have in which case but For example, consider

For problem 3, start by writing the number as a product of primes: Plainly, any positive divisor of must have the form where or similarly, or ; or and or There are 8 possibilities for 5 for 2 for and 2 for This gives us a total of possible positive divisors.