Tutorial for Assignment 10. Introduction to Mathematical Thinking

Introduction to Mathematical Thinking. Tutorial for Assignment Let us look at the first assignment. Prove that the intersection of two intervals is one of those intervals. If you think about the question in terms of diagrams, it becomes easier tounderstand. One interval could be entirely inside another, or the two intervals might overlap insome way. They could also be completely separate from each other. In addition, oneinterval can be entirely contained by another, or they can overlap. But let's see if we can do this in a more symbolic way. Let's say that a and b areopen intervals: a is ab and c is cd. By definition, any point in the intersection of a andb—so, x in the reals—must satisfy both inequalities: a is less than or equal to x, andc is less than or equal to x. Because of the way conjunction works, a set of x values exists such that themaximum of a and c is less than x and the minimum of b and d is also less than x. Itis the open interval between a and c, which is the maximum value of a and c,together with the minimum value of b and d. An interval is the distance between two pitches. It may be empty. If a set is disjoint, ithas no elements in common with another set. The empty set is the interval betweentwo numbers. It's still a set of numbers. So, we have concluded that the other thing isto see what happens at the endpoints. And let's just say this, in a similar manner, forclosed intervals and half open intervals. That is why unions are not intervals. For example, if we take the open interval 0, 1and form a union with the open interval 3, 4, then that set is not an interval.

Let us look at the next question. Here, we are asked to verify the alternativedefinition of least upper-bound. Variant a claims that b is an upper bound. Let us write that variant a asserts that b isan upper bound. The issue is whether condition b satisfies the requirement that it isat least one. Let us examine a result of this sort. Suppose that b is an upper bound for the set S ifand only if, no element c in S is less than b. The word least means there is nothing smaller, so it's a synonym for the originalconcept. So, there isn't a smaller one—it's a lower bound and upper bound. Let's clarify this point. The above statement is true if and only if for any c less than b,c is not an upper bound. That is, we can state the same thing in a different way. If c is less than b, then there exists some a in A such that a < c. If c is not an upperbound for A, that means there exists an a in A such that a ≥ c. It does not follow fromthe previous statements that A ≤ C. If c > A, then c is not an upper bound for A. Ifand only if, for every C less than B, there is an A such that a ∈ A and a < c. Let us look at the next task. This task demonstrates that the limit of [n] over [(n+1)squared] approaches 1 as napproaches infinity. There is a pattern to these questions. For example, if epsilon is greater than or equalto 0, then N is the number such that for any natural number n bigger than or equal toN. The absolute value of n over n plus 1, all squared, minus 1 is less than the epsilonvalue. From some point on in the sequence, the difference between n/n+1 and 1 is lessthan epsilon.

Now let's break down the equation we have written and see what it is really askingus to find. We need to find the integer n such that n>N implies that you writeeverything Over n+1 all squared and I've got n squared minus n squared minus 2nminus 1. That is less than epsilon, or a difference of less than epsilon. We need tofind an n such that n - 1 is greater than or equal to N. In other words, we need ngreater than or equal to N implies the n squared disappears, and we have 2n + 1over n + 1 squared. Thus, n squared disappears and we are left with -2n - 1 over n + 1 squared inabsolute value. When we take the absolute value, the -2n - 1 term becomes 2n + 1.This means that n plus one squared is positive, so we can ignore the n plus 1squared term and just look at the n squared term. What to do next? We are looking for a big N. So let us take N to be so large that N plus 1 squared over2N plus 1 is bigger than 1 over epsilon. Epsilon's small, so we'll assume we can doit. We use the symbol Epsilon to emphasize that in reality the quantity is very small. There is not necessarily a logical restriction on this, it just seems to be our intuitionsabout what the proof is doing. Thus, this number will be very large. For this term, N plus 1 squared over 2N plus 1, to be greater than 1 over epsilon, thenumerator must be made larger than its denominator. We can do this by making thenumerator larger than 1 over epsilon and setting it equal to 2n plus 1 over n plus 1squared. When we do this, we find that for n greater than or equal to N, 2n plus 1over n plus 1 squared is less than or equal to 2N plus 1 divided by 2N plus1squared—which is less than epsilon. That's it.

We just worked backwards. We looked at the goal and worked backwards to determine the necessary conditions.We found that we could always find an N with this property. Because the numerator grows faster than the denominator without bound, and sincethis growth is manifested in a squared versus linear relationship, the term asymptoteachieves its name. This expression arose as a result of this. This is a typical patternfor limit verifications, although not all of them follow this pattern. If you workbackwards, then you can ask yourself where to look and how far out to go in order tomake something happen. You often end up saying that something is bigger than 1over epsilon. This happens when there are quotients involved in the limit.