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Introduction to Abstract MathematicsMA103

Solutions to exercises 2

1 (a) (i) We can write this set as A = { x N | x = 5 y for some y N }. But there aremany other ways; for instance :A = {m N | there is a natural number k such that m = 5 k },or A = { a | a N and b N : a = 5 b }.It is also correct to write: A = { 5 s | s N }.

(ii) This is the set B = { x Z | x4 < 1000 }.(b) The members of the set B are 5,4,3,2,1, 0, 1, 2, 3, 4, 5. (Note that 54 = 625 1000, so 5 and 5 are elements of B, but 6 and 6 are not.)Hence B = {5,4,3,2,1, 0, 1, 2, 3, 4, 5 }.

2 (a) This is true, since all of 1,3,9 are indeed odd integers.(b) This is false. This is because 0 cant be written as 2n 1 for some natural number n. We

can write 0 = 20 1, but n = 0 is not a natural number. So we have that 0 A, but0 / B. So it is not the case that for every x A we also have x B. Hence the statementA B is not true. There are other ways to express this argument, but they all amountto demonstrating that 0 is an element of A and not of B.

(c) This is true. Indeed, as stated in lectures, the empty set is a subset of every other set.( If you find this difficult to accept, in order to show that it is not true you would needto find an element of that is not an element of the second set. But since has noelements, no such counterexample can exist. )

(d) This is true. We need to show that every element of A is also an element of B. In otherwords, we need to show that every natural number n such that n is even is in the setB. Thus we need to show that, if n is an even natural number, then n2 is even. Thisis quite straightforward. For, suppose that n is an even natural number. Then there issome natural number k such that n = 2k and hence n2 = (2k)2 = 4k2. Since 4k2 = 2(2k2)and since 2k2 is a natural number, n2 is indeed even.

A general point: we have introduced the technical terms element, member and sub-set. So x S means that x is an element or member of the set S, whereas x S meansthat x is a subset of the set S ( so x is itself a set, all of whose members are members of S ),and these are very different things. In your work, you too must use these technical terms. Ifinstead you invent other terms, such as x is part of S, or x is below S, or x is within S,then dont expect anyone else to be able to guess whether you mean subset, or element, orsomething else entirely.

c London School of Economics, 2013

MA103 Introduction to Abstract Mathematics Solutions to exercises 2 Page 2

3 (a) Your answers here must engage with the definition of the set X. Im not just looking fora proper description of what it means for A to be an element of a general set X; I wantto know what it means for A to be a subset of the specific set X defined in the question.All you have to do is read the definition of X:

To say that A is an element of the set X = { A | A { 0, 1 } }means that A { 0, 1 }.It really is that simple, but its worth pausing to digest this. The elements of the set Xare themselves sets: specifically, they are subsets of the two-element set {0, 1}.To say that B is a subset of X means that B is a set, all of whose elements are elements ofX, i.e., B is a subset of X if all the elements of B are subsets of {0, 1}.

(b) If youve understood the answer to (a), then you can now list the elements of the set X.These are: , {0}, {1} and {0, 1} itself. In particular, {0} X.Alternatively, to test whether {0} is a member of X, we need to decide whether thestatement {0} { 0, 1 } is true or not. It is, and so {0} is an element of X.

(c) The set B = {0} is a subset of X if every element of B is an element of X. As B has onlyone element, there is only one thing to check: we have to ask whether 0 is an element ofX. It isnt: the elements of X are sets. So B is not a subset of X.

(d) We listed the four elements of X: , {0}, {1} and {0, 1}. Weve seen that {0} is nota subset of X, and similarly {1} and {0, 1} are not subsets of X. The only remainingpossibility is, and this is a subset of X is a subset of any set as well as an elementof X.

4 (a) The first statement can be written, for instance, as x Z, z Z such that x z 2x.

The second statement can be written as

n N, m N, n m.(b) A counterexample for the first statement is x = 1. For this x, we have x Z, but

2x = 2, and there is no z Z so that 1 z 2.(c) The second statement is true. The natural number n = 1 does have the property that,

for all natural numbers m, 1 m. ( The statement says that there is a natural numbern with a certain property. To prove it, you have to name a natural number n, namelyn = 1 in this case, and argue that it has the required property. )


(d) The negation of a statement of the form for all elements in some set, property A holds,is the statement there is an element in the set for which property A does not hold. Sothe negation of the first original statement is

There is an integer x such thatit is not the case that there is an integer z with x z 2x. ()

The second line is the negation of there is an integer z with x z 2x. The negationof a statement of the form there is an element in some set such that property A holdsis the statement for all elements in the set, property A does not hold. So the negationof there is an integer z with x z 2x is the statement for all integers z, it is not thecase that x z 2x. So we can rewrite the negation in () and obtain

There is an integer x such that, for all integers z, it is not the case that x z 2x.Now the final part of the previous statement is the negation of x z 2x. Notethat this is in fact a combination of two statements : x z and z 2x. And such anand-statement is false if at least one of the two parts is false. So the negation of x zand z 2x is the statement x > z or z > 2x.So as the final version of the negation of the original first statement we might write :

There is an integer x such that, for all integers z, either x > z or z > 2x.

In mathematical notation we can write this as :

x Z, z Z : x > z or z > 2x.

The original second statement was :

There is a natural number n such that, for every natural number m, we have n m.The negation of this existential statement is:

For all natural numbers n, it is not the case that,for every natural number m, we have n m,

which in turn is better expressed as:

For all natural numbers n, there is a natural number msuch that it is not the case that n m,

or, finally:

For all natural numbers n, there is a natural number m with n > m.

In symbols:

n N, m N such that n > m.The task of negating a statement with quantifiers can be seen as an automatic procedure:replace all the occurrences of with , all the occurrences of with , and negate thefinal statement (which may need some care).


5 The statement is False. Take the sets :A = { a, b, d, e }, B = { b, c, d, f } and C = { d, e, f , g },

where all of a, b, c, d, e, f , g are different. Then we have

A (B C) = A { d, f } = { a, b, d, e, f }(A B) C = { a, b, c, d, e, f } C = { d, e, f }.

So indeed, this is a counterexample to A (B C) = (A B) C.Many people doing abstract mathematics for the first time look at answers like the one aboveand wonder how on earth anyone could dream up this example. In fact, its impossible towrite an answer like that without doing some background rough work that doesnt appear inthe final answer. One of the main challenges in learning abstract mathematics is to developan attitude that allows you to obtain such answers eventually.

So lets have a look how you could deal with questions like this. Since the question is abouta small number of sets, a good way to get a feeling for what is happening is to draw Venndiagrams, as is done below. On the left the shaded area represents A (B C); while onright the shaded area represents (A B) C.

The shaded areas are not the same, so you strongly suspect that the sets they represent arenot the same. But these pictures do not constitute a proof ! For that you need to construct anexplicit counterexample. Of course, the pictures can be very helpful for that.

The counterexample at the beginning was obtained by making sure that each of the areas inthe Venn diagram contains one element.

However, many other answers are possible. For instance, looking back at the Venn diagramsfor A (B C) and (A B) C, we see that the subset of the set A that consists of elementsin A but not in B or C ( the top-left part of the diagram ) is a subset of the set A (BC), butnot of (A B) C. So, provided our example includes an element that is a member of A butnot of B or C, we should expect to have a counterexample.

For instance, set A = {1}, B = and C = . Then we have

A (B C) = A = A = {1};(A B) C = A = .


6 Very important: In this question, you are asked to prove certain things for all sets A, B andC. If you start your answer with: Let A = {1, 4, 5}, B = {2, 3, 4}, C = {1, 2, 6} (or any othersets that you happen to have chosen), and then work out A (B C) and (A B) (A C)for this example, finding them to be equal, then all you will have proved is that these threesets do not form a counterexample. You can never prove a statement about all sets by lookingat an example. Of course, you might be able to disprove such a statement by giving onecounterexample, as in the previous question.

(a) In order to prove that, for two sets X,Y, we have X Y, we need to prove that everyelement in X is also a member of Y. So we proceed as follows :

To prove : A (B C) (A B) (A C).Hence to prove : for all x A (B C), we also have x (A B) (A C).Proof : Take any x A (B C). That means that x A and x B C.The fact that x B C means that x B or x C.In the case that we have x B, then together with x A we have x A B, hencex (A B) (A C).And in the case that we have x C, then together with x A we have x AC, hencex (A B) (A C).So, in both cases we have x (A B) (A C), as we wanted to prove. 2

(b) In order to prove that two sets X,Y are equal, in general we prove that

X Y and Y X.And to show that X Y we proceed as in (a). Hence proving X = Y usuallyrequires proving

( for all x X, we also have x Y ) and ( for all x Y, we also have x X ).Note that the first part has already been done in (a), so we only need to prove the secondpart. We can write the following proof.

To prove : A (B C) = (A B) (A C)Since in (a) we already proved A (BC) (A B) (AC), we only need to prove :(A B) (A C) A (B C).Hence to prove : for all x (A B) (A C), we also have x A (B C).Proof : Take any x (A B) (A C). That means that x A B or x A C.In the case that we have x A C, then we have x A and x C. So in particular wehave x A and x B C, hence x A (B C).If we are in the case that x A C, then x A and x C. Again we find x A andx B C, hence x A (B C).So, in both cases, we have x A (B C), as was required. 2


7 For this kind of question, for which at first you even might have trouble understanding whatis going on, the first rule is dont panic. Take a pen and an empty piece of paper. And readcarefully the information in the question to see what is going on.

Often the best way to get a feeling for what is happening, and what kind of things you aresupposed to be looking at, is by making up an example. And in fact, parts (a) and (b) askyou to do just that. Quite often such a requirement to write down some examples is notpart of the question, but it is a good way ( if not the only way ) to start thinking about theproblem.

So lets start by doing (a) first.

(a) The question in (a) asks you to write down a specific example of the type of object weare looking at ( a 5 5 array of the numbers from 1 to 25 ). Following the question, weare supposed to write down

1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 25

Then you have to find the greatest number in each row. For the array above that givesthe sequence 5, 10, 15, 20, 25. And then the least number of those is s; so we get s = 5 forthe array above.

Next you are required to find the least number in each column, which leads to the se-quence 1, 2, 3, 4, 5. And from these you pick the greatest one, leading to t = 5.

So, as was predicted, the example array above is one in which s = t.

At this point, if you obtained s and t that were not the same, you should not just continueand ignore that. Clearly, in that case, something is going wrong : either there is an error in thequestion ( not impossible, but not that likely ), or you have made a mistake ( unfortunatelyfor you, that is much more likely ). So carefully read the question again, did you take theright example, did you obtain s and t in the right way, etc.? It doesnt make much sense tocontinue with parts (b) and (c) until youve found out ( and corrected ) your error.

OK, suppose you did part (a) as above. If there was no part (b), we would be tempted tothink that for all arrays of the type we are looking at we would get s = t. So if at this pointyou were asked to prove a relation between s and t for any array, you would be tempted totry to build a proof for s = t. And you would fail.

One reason you should be suspicious about the possible truth of the fact that s = t forall arrays, is that it is based on one very particular example ( namely the very regular oneabove ). To get somewhat more evidence, you should consider examples without such a nicestructure and see if you still find s = t. If that would always be the case, then that wouldstrongly suggest that maybe it is true that s = t for all arrays.

But, alas, it will appear that something different is the case.


(b) In this part you are asked to find an example of an array with a specific property ( s 6= t ).The one example we saw so far ( in (a) ) didnt have this property, so we need to dosomething different. And the question gives no further hints what things to look for.

Now there are a couple of things you can do. One is to see how to make some smallchanges to the array you already found. Or just start with a completely new example.You can try the first thing for yourself ( its not too hard ); I decided to just make a com-pletely new example. This time I tried to avoid making it too organised, and randomlyfilled in the numbers to get:

1 13 21 5 2420 6 2 15 1417 3 12 11 167 18 4 8 1925 9 10 23 22

For this example the greatest numbers in the rows give the sequence 24, 20, 17, 19, 25.This gives for the least one s = 17. And the smallest numbers in the columns are1, 3, 2, 5, 14. The greatest of these is t = 14. So indeed this second example satisfiess 6= t.( In fact, the first example was one of the very few unlucky ones in which s = t. Almostevery way you make this array you will get s 6= t. )

Note that in the first example above we have s = t and in the second example we have s > t.So it seems that indeed s t might be true for any array found this way.But examples are not proof.

For a proof ( as required in part (c) ) we need to make sure that s t is true for any examplewe could make up. In principle, we could investigate all the possible arrays, but there arefar too many of them to contemplate this approach.

OK, so how do we get a proof ? We are asked to prove something about two numbers s and t.So its a good idea to find out as much as possible about these two numbers.

First lets see how s is obtained from a given array. We first find the greatest number in eachrow, and then s is the least among those. But that still means that s is the greatest value in acertain row. In other words, there is a row in the array so that for each number a in that rowwe have s a.Is that all we can deduce about s ? No, s is the smallest among the greatest. So it follows thateach row that is not the special row from above contains a number b so that s b.That seems to be about all we can say about s.


We can reason similarly for the number t. This leads to the observation that there is a columnin the array so that for each number x in that column we have t x.And each column which is not this special column contains a number y so that t y.So now we should see if this gets us any closer to a proof of s t.Looking back at what we know about s and t, and in view of having to prove s t, thetwo observations involving s a ( for certain a ) and t x ( for certain x ) seem to be themost useful. If only we could find a and x so that a = x, or even a x, we would be done.But in fact we can do just that. The statement s a holds for all a in a certain row. Andthe statement t x holds for all x in a certain column. But whatever row and column wetake, there is always one number which is both in that row and that column. So let k be thatnumber for the special row and column related to s and t. Then we know that s k andt k. In other words, s k t, which immediately gives s t.And now the real answer

The long story above is not the answer to this question. Its a long description of how towork on this kind of problem, including things that you only think but not write, and thingsyou only would write on your rough work but not in the final answer. Below is more or lesswhat we would expect to see as a real ( and correct ) answer to part (c).

(c) To prove : For all arrays we have s t.Proof : The number s is obtained by taking first the greatest number in each row, andthen taking the least value among those. So there is a row for which s is the greatestvalue. Call this row A. So for all a in row A we have s a.A similar argument gives that there is a column B so that for all numbers x in column Bwe have t x.Now let k be the number written at the intersection of row A and column B. Then k isin row A, hence s k; and k is in column B, hence t k. From these two inequalitieswe can conclude s k t, and so s t. 2

This was a very challenging problem, and if you solved it then you deserve congratulations!

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