Math Tutorial: Navigating a|b vs. a/b Concepts

Introduction to Mathematical Thinking. Tutorial for Assignment 9 Look at task number 1 below. Distinguishing between these two notions: a|b and a/b, is important for avoiding errors. When there is a notation of a|b, it represents arelationship that is either true or false. a is divisible by a or b divides a. You may also see an expression of a number as the result of dividing a by b (a/b). When considering the integers, a|b makes sense. For the integers, a/b is not defined. The quantity a/b is undefined for all integersexcept those for which the result makes sense. But a/b is a division. The operation of division, however, is not one that works on theintegers. It's an operation on the set of rational numbers, or the set of real numbers. In the integers, addition, subtraction, and multiplication are the only operations.Division is not possible. What you can do is say whether a number is divisible by another number. Divisibilityis defined in terms of multiplication. The difference between a|b and a/b is that the first assumes the truth of a and b,while the second considers both possibilities. You have a notation for a number at a/b. This is an actual number. The correct answer is the one at the bottom, the fourth answer option in a row.Describe the relationship as accurately and concisely as possible. The last one is therelationship. The quotient of a divided by b is an integer if and only if a divided by b is itself aninteger. Let us summarize: the first variant denotes a rational number, which has a differentset of numbers than an irrational number. Variant two denotes the relation: b divides a, or equivalently, there exists an integer qsuch that a = qb. Whenever the notion of divisibility (a|b) is being considered and theintegers are being worked with, one must reduce that (a|b) abbreviation to this; thereis an integer switch at that point in the end. That point in the end means b|a. At the variant three, in the case where there is a divisibility (b|a), then, of course, theq that's at the end of variant two is the quotient. The q is used to stand for the wordquotient, but notice that this part at the end of the second variant says nothing of

division. Division is not an issue in this case. The focus is on the result of multiplyingtwo numbers. This concept makes sense when you consider the integers. So, we are not doing arithmetic in the sense of calculating variant four. Of course, it's obvious that all types of answers involve dividing one whole numberby another. The focus in this task is on what you should be doing within the systems of numbers. The rational numbers and the integers are two separate systems of numbers. Thereare two systems of numbers in use. With integers, you can perform addition, subtraction and multiplication. Withrationals, addition, subtraction, multiplication and division are possible. The systemsare different, but they're both systems of numbers. This task focuses on how you can create and manipulate integers. That's what number theory's about. Let's look at the second and third tasks . Second first. The problem with all such problems (0|7, 0|9, 0|0, 1|1, 7|44) is that they aredivisibility problems in which the divisibility property must be expressed in terms ofmultiplication. Remember that the property of divisibility is a characteristic of pairs ofintegers. Division of integers is not possible. Integers can be added, subtracted, ormultiplied. The definition of an integer can be stated as follows: a divides b (a|b) ifand only if there is a q, an integer, such that b equals q times a. The general strategyfor dealing with divisibility in all variants, therefore, seems to be to replace issues likea|b with a statement about multiplication. The point is that there is no operation in a|b, at a|b nothing, there is no arithmeticaloperation on the integers.

And remember that all of these examples are based on integers. So you have to express a|b in terms of integers. The language of the integers islimited to the operations of addition, subtraction and multiplication; it does notsupport division. Look at the variant a. The proof is immediate, and the reason for this is that the definition of divisibilityexplicitly excludes a number from being divisible by 0. A ≠ 0. It excludes thepossibility of A being 0. It is false, and that is why it's not true. Variant b. 9 does indeed divide 0. And to prove this point, we simply express it in terms of thedefinition. To show that q exists, you would have to first look at the definition in theupper right corner, which requires showing that 0 equals q times 9. Щf course thereis. Variant c. This variant is false. This statement is false for the same reason that statement awas false. The notion does not allow for a to equal 0, including the requirement thata be greater than or equal to 1. Variant d. This is true, and its proof can be demonstrated by writing it in terms of this. This iswhat you will be required to write (look below). In each case, if you look at theseexamples, we will have to reformulate the statement in terms of the definition. That'sall there is to it, really. Just reformulate statements in terms of the definition. Thedefinition is true. The equation q = 1 is true by virtue of the fact that q equals 1. Variant e. We know that there is no such q. You could argue this fact by noting that anypossible q would have to be less than seven, but it is a more efficient argument justto note that any such q would have to be less than 7. If you wanted to prove this inmore detail, you could assume that q represents all possibilities that have a chanceof being true. q equals 1, 2, 3, 4, 5, and 6. So you need to go to the sixth step inorder to reach 49. You could list all of the possible q`s if you wanted to. You couldleave it as it is. Variant f. This statement is true. The quantity q is equal to minus six, so that q equals negativesix. Variant g.

Here, too, you exhibit the q and it is negative seven. Variant h. At this variant you exhibit the q. And the q is 8. Variant i. For every positive integer n, you need to show that n 1 divides n. This is true. Thereason is that any number n in Z can be expressed as n times 1, and 1 divides everynumber. Variant j. The number 0 does not divide the number N, because for any N in Z, 0 times Nequals 0. Variant k. This is one we must be careful with, because if we are considering all of the integers,Zero is included. And you are not allowed to divide by zero, because that's excludedfrom the definition of a function. This is the one to keep in mind, because it includesthe Zero. Therefore, it is false. Notice that the pattern was simply a restatement of the definition of divisibility; oncethis fact is understood, each case follows easily. Number theory is elementary andsimple arithmetic, so it's not that there's anything very complicated going on here. Look at task number 3.

Variant a. The previous example shows that you can reduce each of a|a, a|0 to the definition ofdivisibility by using your knowledge of number theory. Divisibility is a property of pairs of integers, but it is not the same as division. Division is obviously related to the integers, but division does not exist amongintegers. The integers can be added, multiplied, and subtracted. Subtraction is theinverse of addition. However, division with integers is not possible. However, in order to discuss divisibility within the integers one must reduce it to afunction of essentially multiplication. How would you show that a divide 0, that's you got divisibility of 0 by a? You observethat the property of 0 allows you to multiply it by anything, so in particular, 0 satisfiesthe requirement for divisibility. There is a q in z, so it follows that 0 equals q times a,or q equals 0. Since a divides 0, then by definition a must equal 0. Becausemultiplying by 1 does not change the value of a, in the case of a dividing a, 1 times aequals a. The definition of divisibility is satisfied. Since q is equal to 1 in this case,the equation is satisfied. Thus a is divided by a. And the remaining variants are essentially the same idea. Variant b. a is divided by 1 if and only if a equals ±1. In this section, we will discuss twoimplications of our argument. First, let us assume that a is equal to ±1. Then all we need to do is show that there issomething that a equals q times a. If a divides one, then for some q, one equals q times a. However, if one equals qtimes a, then the absolute value of one is the absolute value of q times a, and hencethe absolute value of q times the absolute value of a. If , then the only possibility isthat the absolute value of q is equal to the absolute value of a, since these arepositive integers. This is because if q were not equal to a, then their product wouldnot be 1. Thus, if the absolute value of a equals 1, then a must be either 1 or -1. Variant c. If a divides b and c divides d then ac demands bd.We know that the divisors of b areq and r, which means that b is divisible by q times the definition of divisibility. The definition of divisibility is d equals r times c. Therefore, multiplying the twotogether gives bd is qr times rc. Rearranging them gives qr times ac. So by definitionac divides into bd. And the other variants are essentially the same idea. In each case, the question of

multiplication is reduced to the definition of divisibility. Because division is the inverseof multiplication, you can express division in terms of multiplication. They are nothing more than a translation of what you need to prove into divisibility.The first line of any of these arguments simply restates the assertion or claim youare trying to prove.