Introduction to Mathematical Thinking. Tutorial for Assignment
In problem 7, we must negate statements and put them in positive form.Part A asks us to prove that for all integers “x” in N, there exists an integer “y” in N such that X plus Y equals 1. The negation of this statement is: If an integer “X” isgreater than 0 and an integer “Y” is less than 0, then X plus Y does not equal 1. You can also write this last part as: If an integer X (greater than 0) and an integer Y (lessthan 0) are chosen such that X plus Y equals zero, then it cannot be true that both integers are equal to zero at the same time. For all values of “Y” less than zero, these do not change around. They stay the same way, because these simply tell us what Xs and Ys we’re looking at. And so we get X plus Y not equal to 0 here. Part C is the same as part B, except that you have an additional constraint on the function: it must be greater than 0. So epsilon greater than 0, negative epsilon less than X, less than epsilon. And when you negate that you get for all x epsilon greater than 0. But in this case you need to split up into two cases: For all values of x either x is less than or equal to negative epsilon or x is greater or equal to epsilon. Take precedence over logical connectors because they are connectors.Statements about mathematics connect statements about mathematics. This is a statement about mathematics, and that is a statement about mathematics. You could have put parentheses around the first part, and you could have put parentheses around this part as well. Now we move on to the last step. For all x and n, there exist y and n such that x plusy is equal to z squared. We aren’t proving these statements to be true; we are simply writing them down and then negating them so that you can see what they become.So here’s the last step: x+y is not equal to z squared. Negate the following statements and put each answer into positive form: (a) (forall x in N) (exists y in N) (x+y=1) (exists y in N) (forall x in N) (x+y=1) neg (x+y=1) (b) (forall x>0) (exists y <0) (x+y=0) (x,y are real number variables) (exists > 0) (forall y < 0) (x+y=0) (c) exists x (forall є >0) (- є<x< є) (x, є are real number variables) forall x (exists є>0) (x≤ -є) lor (x≥є) (d) (forall x in N) (forall y in N) (exists z in N) (x+y=z^2) (exists x in N) (exists y in N) (forall z in N) (x+y ne z^2) Let’s turn now to question eight. Abraham Lincoln is famous for the following statement: You can fool all of the people some of the time, and you can fool some people all of the time, but you can’t fool all of the people all of the time. At times, it is possible to fool some people all of the time. Some people can be fooled at all times,but not everyone can be fooled all of the time. Negating this statement is a simple matter of applying basic logic. If everyone cannot be fooled, then it must be true that someone will not be fooled. Thus, you have the formula FPT ⇒ ¬FPT—that is, ifsomeone cannot be fooled at all times, then that person will not be fooled for evenone instant. Again, we’re not putting these in parentheses, because this is a whole that quantifies binding the very tight binding of M-theory and then we have this junction conjunction here and this junction’s here and theirs less tightly binding. So looking at this one,existence becomes for all, for all becomes exist we get negation. Conjunction becomes disjunction and the negation here just disappears. The first part would be:At any time, there is someone you cannot fool. The second part would be: There is no one who can’t be fooled at some time. Let’s look at this example. There is no better way to say it than: you cannot always fool all of the people all of the time. Well, no. If we try and swap the sentence around, we run into an American melanoma type problem. So let’s not do that. And then finally: but that is easy; we can just say “you can fool all of the people all of thetime.” In any case, we’ve negated it and this one was nice and clean as well. Let F(p,t) mean “You can fool person p at time t” Lincoln’sstatements is: exists t forall p F(p,t) wedge exists p forall T F(p,t) wedge neg forall p forall t F (p,t) Negation: forall t exists p neg F (p,t) lor forall p exists t neg F (p.t) lor forall p forall t F (p,t) at any time, there is someone you can’t fool or for every person, you can’t always fool them. you can fool all the people all the time. The standard definition of a real function f being continuous at x=a is (forall є>0) (exists б>0) (forall x) [ |x-a|<б Rightarrow | f(x) – f(a)| < є] Write down a formal definition (positive form) for f being descontinuous at a. (exists є>0) (forall б>0) (exists x) [ |x-a|<б wedge | f(x) – f(a)| ≥є] The ninth term, or ninth derivative, is one of the most important and famous formulas in advanced mathematics. Figuring it out has taken hundreds of years of effort by mathematicians like Newton and Leibniz beginning in the 17th century. It took mathematicians hundreds of years to discover the notion of continuity and come up with this definition, which occurred late in the 19th century. This was a tricky thing todo; negating it is relatively straight forward, however, because we have rules fordoing this. When you negate it, what you get is that for all becomes exists and that greater than becomes a for all. There’s still a bunch of stuff in here that’s aconditional, an implication — so when you negate it, you get the antecedent conjoined with the negation of the consequent. If this is true, then the negation of it will be less than F1. If this is originally true, then the negation of it will be less than F1 as well. The interesting question is what on earth does this mean? One can read through it as we just did for all epsilon greater than 0, this is delta greater than 0, such that for all x. What does it mean? Well, this is captured in a symbolic language. In an algebraic formalism, it’s capturing something geometric. So, let’s see what it’s capturing. Let’s look at the original definition of continuity. It’s about functions. So let’s look at a function this way —draw a wavy line vertically, this is a real line, and then draw the real line vertically here. So this is the real line, and instead of writing horizontally as we usually do,write it horizontally. And the function f is going to take numbers here to numbers here. Now we’re trying to capture the notion of continuity at a. The epsilons are going to work on here, so let’s take an epsilon here and let’s look at F of a plus epsilon andf of a minus epsilon. So take an epsilon interval around f of a. And what this definition says, is that, given an epsilon in 1 of these intervals, find a delta. So here’sa plus delta, and a minus delta. Starting with an epsilon, which gives me an interva lhere, through any one of those, find a delta, which gives me an interval here. In order to make sure that all the values of the function are in this interval, we canfind an interval around A such that they’re all sent into here. So if we want to hit the target, imagine this is hitting the target, like throwing darts at a dart board. If we want to hit the target within a specified accuracy of f(A), do it by starting out within aninterval around a. So everything from the point where the function is discontinuous can be sent into a continuous function. And if you think about it long enough, you’ll realize that what that means is that the function is continuous at the discontinuity.There are no jumps. And to try and understand that, let’s look at what this guy meansin terms of a diagram. Now, the negation says, given some epsilon, there is some delta. In the previous case this was happening for all epsilons. We can find a delta that works in this case.And we look at the interval around there: F of A minus Epsilon. What it says is that there is an epsilon such that no matter what you take here, no matter which one you take here, there’s still an interval here for some epsilon such that no matter how small you make this delta, no matter how close you get here, you can always find apoint in here (in this case we found one of these points) that gets sent outside of there. The sum here is epsilon, and no matter what you do in this region, no matter howclose you are to a, something gets sent out here. In other words, there are pointsreally, really close to a that get sent outside this region. So there’s no way that youcan get all the points here. A gets sent to that, but arbitrarily close to a There are points that get sent away. So there’s discontinuity, because only a gets sent close to a. When you get really close to a, then some other points might get sent further away. So that’s a discontinuity. Thing’s jump; there’s a sort of jump from there to things here or here.
