Assignment
University
Stanford UniversityCourse
Introduction to Mathematical ThinkingPages
4
Academic year
2022
Staines
Views
10
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 Nsuch 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. Youcan 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 bothintegers are equal to zero at the same time. For all values of "Y" less than zero, these do not change around. They stay the sameway, because these simply tell us what Xs and Ys we're looking at. And so we get Xplus Y not equal to 0 here. Part C is the same as part B, except that you have anadditional constraint on the function: it must be greater than 0. So epsilon greaterthan 0, negative epsilon less than X, less than epsilon. And when you negate thatyou get for all x epsilon greater than 0. But in this case you need to split up into twocases: For all values of x either x is less than or equal to negative epsilon or x isgreater or equal to epsilon. Take precedence over logical connectors because they are connectors.Statementsabout mathematics connect statements about mathematics. This is a statementabout mathematics, and that is a statement about mathematics. You could have putparentheses around the first part, and you could have put parentheses around thispart 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 simplywriting 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.
Let's turn now to question eight. Abraham Lincoln is famous for the followingstatement: You can fool all of the people some of the time, and you can fool somepeople all of the time, but you can't fool all of the people all of the time. At times, it ispossible 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 simplematter of applying basic logic. If everyone cannot be fooled, then it must be true thatsomeone will not be fooled. Thus, you have the formula FPT ⇒ ¬FPT—that is, if someone 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 quantifiesbinding the very tight binding of M-theory and then we have this junction conjunctionhere 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. Conjunctionbecomes 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 isno 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 alwaysfool all of the people all of the time. Well, no. If we try and swap the sentencearound, we run into an American melanoma type problem. So let's not do that. Andthen 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.
The ninth term, or ninth derivative, is one of the most important and famous formulasin advanced mathematics. Figuring it out has taken hundreds of years of effort bymathematicians like Newton and Leibniz beginning in the 17th century. It tookmathematicians hundreds of years to discover the notion of continuity and come upwith this definition, which occurred late in the 19th century. This was a tricky thing todo; negating it is relatively straightforward, however, because we have rules fordoing this. When you negate it, what you get is that for all becomes exists and thatgreater 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 antecedentconjoined 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, thenthe negation of it will be less than F1 as well. The interesting question is what onearth does this mean? One can read through it as we just did for all epsilon greaterthan 0, this is delta greater than 0, such that for all x. What does it mean? Well, thisis captured in a symbolic language. In an algebraic formalism, it's capturingsomething geometric. So, let's see what it's capturing. Let's look at the originaldefinition 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 verticallyhere. 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 numbershere. Now we're trying to capture the notion of continuity at a. The epsilons are goingto 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 thisdefinition 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 intervalhere, 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 thetarget, imagine this is hitting the target, like throwing darts at a dart board. If we wantto 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 discontinuouscan be sent into a continuous function. And if you think about it long enough, you'llrealize 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 previouscase 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 thatthere is an epsilon such that no matter what you take here, no matter which one youtake here, there's still an interval here for some epsilon such that no matter howsmall 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 ofthere. 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 toa. When you get really close to a, then some other points might get sent furtheraway. So that's a discontinuity. Thing's jump; there's a sort of jump from there tothings here or here.
Negating Math Statements & Logical Connectors
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