can some one help me out with problem.. Explain using only the defenition and formal logic, wat would be needed to be done to show that , for a particular function f(x) and real number L, L is not the limit of f(x) as x approaches a. [hint : at least get as far as carefully negating the definition of the limit] that was the problem...i really dont know how to start this problem.i KNow the definition of limit which is lim-->a F(x) = L if for every number E > 0 there is a corresponding number G(gamma) >0 such that |F(x) - L| <E whenever 0< |x-a|<G plz ppl help!!!
im afraid i cant do it im sorry but i dont know how to negate that defenition...i will assume that the negation will be for every e<0 there is a number gama <0 such that |f(x) -L| >e whenever 0> |x -a| > gama ....probably that doesnt make sense..but does wat i can think of....i would really apriciate some help!
What do you know about negating logical sentences? The reason I ask is because the entire procedure for negating a sentence is entirely mechanical -- you just apply the steps. (They should be in your book -- if you tend to forget things like this, you should learn where in your book they are!) By the way, this looks like homework, so you should have posted in the homework help forums. :grumpy: (I'm moving it over there now)
well the book doesnt have any information on that.. the only defenition about negation is this the negation of P denoted ~P, is the propocition "not p" ~P is true exactly when P is false.. .. i used that defenition....wat do u recomend me to do ? any help will be apriciated it. thank you so much
i found this in another threat however i do not know wat he means by convergent sequences. Is something like when u trying to take the limit at an ASYMPTOTE of a fuction? i know that the limit doesnt not exist( or goes to infinitive i cannot recall) is that wat he means by convergent sequence? Let f:I->R and let c in I. I want to negate the statements: "f has limit L at c" and "f is continuous at c". Are these correct? f does not have limit L at c if there exists e>0 such that for some sequence {x_n} converging to c, |f(x_n)-L|>e for every n. f is not continuous at c if there exists e>0 such that for some sequence {x_n} converging to c, |f(x_n)-f(c)|>e for every n. edit: also, what is the negation of "f has a limit at c"?
If you have a separate question, a separate thread would probably be better. (Otherwise, everyone will get confused) Are you absolutely sure your book doesn't have any useful formulae like: [tex]\neg(P \vee Q) \equiv \neg P \wedge \neg Q[/tex] ? And similarly for the other logical operations (including quantifiers?)
would the negation be something like this? lim-->a F(x) = L IF not for every E>0 there isnt a corresponding number g>0 such that |F(x) - L| <E whenever 0< |x-a|<G....... i am not a math major im an ingeneer major ..so this class is really difficult to me..plz help! thank you very much Hurkyl ..u are really making me think
A function f with domain D has limit L at a point c in D iff For all e > 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e Is it clear to you that the above is the correct definition? Can you express this sentence symbolically? What kind of textbook are you using, and what kind of course is it? You can try to either formalize the above sentence, find its negation by mechanically applying logical rules, and then reinterpreting this new formal sentence into the semi-English that we generally use when doing math (the definition I gave above, for example, is in this semi-English), or you can look at that definition and just "do the logic in your head" to see what the negation would say. As a starter, the thing above says, "for all e > 0, something is true" so the negation would be, "for some e > 0, that thing is false." Does that make sense to you? Continue with that approach if it makes sense to you.
I guess what we need to do is to derive that the rules are for negating sentences! I had assumed you were in a course for which you would either be learning these, or expected to already know them, but I guess I was wrong about that. (I suppose I could just tell you the rules, but then you wouldn't understand as well) The thing I want to emphasize is that the technique is entirely straightforward -- at any point, there is exactly one rule that applies, so you apply it. (It is similar to differentiating a complicated expression in this regard -- at each step, there is an evident "right" rule to use, so you use it, and eventually you're done!) I'll work through two of them, and see if you can get the rest in exercises. [tex]\neg \forall x \in P: Q = \exists x \in P: \neg Q[/tex] (Q, of course, can be a complicated expression, but we don't care exactly what that expression is for the purposes of this rule! Just like the rule (f+g)' = f' + g' doesn't care about what f and g actually are) Suppose you want to negate the claim: For that all x in the class P, Q. Well, when someone is claiming that something is always true, all you have to do to prove him false is to find a single counterexample. That is: There exists an x in the class P for which not Q. So this latter sentence must be the negation of the former. [tex]\neg(P \implies Q) = P \wedge ~Q[/tex] Suppose you want to negate the claim: "If P then Q" Well, to disprove an implication like this, all you have to do is to demonstrate that the hypothesis is correct, but the conclusion is not. In other words, you want to prove: P and not Q Here's what you'd need to know to negate any logical sentence. I've left some blank for you to work out as an exercise. (If you want, you can post your results here, and I can check them!) Rule 1: [itex]\neg \forall x \in P: Q \equiv \exists x \in P: \neg Q[/itex] example: "It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black" Rule 2: [itex]\neg \exists x \in P: Q \equiv ?[/itex] example: "There does not exist a person that is 10 feet tall" is the same as "?" Rule 3: [itex]\neg (P \implies Q) \equiv P \wedge \neg Q[/itex] example: You would reject "If it's blue, then it tastes good" exactly when you would accept "It's blue and it doesn't taste good". Rule 4: [itex]\neg (P \wedge Q) \equiv ?[/itex] example: You would reject "He's big and he's tall" exactly when you would accept "?". Rule 5: [itex]\neg (P \vee Q) \equiv ?[/itex] example: You would reject "He's big or he's tall (or both)" exactly when you would accept "?". Rule 6: [itex]\neg \neg P \equiv ?[/itex] example: "It's not, not blue" is the same as "?"
Rule 1: example: "It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black" Rule 2: = Vx E P: ~Q example: "There does not exist a person that is 10 feet tall" is the same as "There exist all people that are not 10 feet tall" Rule 3: example: You would reject "If it's blue, then it tastes good" exactly when you would accept "It's blue and it doesn't taste good". Rule 4: ~PV~Q example: You would reject "He's big and he's tall" exactly when you would accept "He is not big or he is not tall". Rule 5: ~P^ ~Q example: You would reject "He's big or he's tall (or both)" exactly when you would accept "?"He is not big and he is not tall” Rule 6: P example: "It's not, not blue" is the same as "Blue" __________________ Now that i know a little bit more here is another try of the negation the negation of "f has a limit L at c" is "f does not have a limit L at c if for some number epsilon < 0 there isnt a corresponding number gama <0zero such that |f(x) -L| > epsilon whenever 0>|x-c|>gama i hope this make more sense now and its correct..let me know if im right Hurkyl thank you so much!!!
That's not quite phrased right -- you shouldn't have that "There exist" in front. That's not quite right -- I have two comments. First off, remember that "It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black", but not "There exists a non-umbrella that is not black" Secondly, it might help to apply the steps one at a time, rather than trying to do it all at once. In other words, start with not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G And apply the steps one at a time. E.G. you would take "there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G" to be your Q, and apply the "not for all" rule. Finally, it might help you to rewrite your "whenever" phrase as an "if, then" phrase.
here is another try not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G not for every number E>0 there isnt corresponding numbers G>0 such that if |F(x) - L| <E then 0< |x-a|<G not for every number E>0 there isnt a corresponding number G>0 such that if |F(x) - L| <E and is not the case 0< |x-a|<G i hope this one is right now..if it is, how would this look on a graph? would it be a graph that is not continous at an specific point? such as f(x) = 1/n thank u so much for ur help Hurky!
So you're starting with: not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G which is just: not (definition of continuity) The next line is supposed to be one step in the translation, right? So you have: not for every number E>0 there isnt corresponding numbers G>0 such that if |F(x) - L| <E then 0< |x-a|<G This is wrong. Next you have: not for every number E>0 there isnt a corresponding number G>0 such that if |F(x) - L| <E and is not the case 0< |x-a|<G This one just doesn't make sense. You have an "if... and..." going on, which doesn't make sense. Either you have an "if... then..." or you have just "... and ....". However, both of these fixes are wrong, i.e. both: not for every number E>0 there isnt a corresponding number G>0 such that if |F(x) - L| <E then is not the case 0< |x-a|<G and not for every number E>0 there isnt a corresponding number G>0 such that |F(x) - L| <E and is not the case 0< |x-a|<G are wrong.
A function f with domain D has limit L at a point c in D iff For all e > 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e The negation of this statement would be something like this? A function f with domain D does not have a limit at point C in D iff not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G not for every number E > 0 there is not a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| <E then is not the case 0< |x-a|<G Hopefully I am right this time..if not can u suggest me something how to fix this prove? Thank u so much
What is the difference between the following two: not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G not for every number E > 0 there is not a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G The first one is clearly just not (definition of continuity) and this is of course what you want to start with. The next line should be one step in the translation towards a sentence that is equivalent to not (def'n of continuity) but in some way more meaningful, i.e. it makes some positive claim. One problem with your second line is that the only difference from the first line is that you've thrown an extra "not" in there. Obviously then, the two sentences can't be equivalent. A bigger problem, however, is that all your sentences start with "not". You want to translate the initial not-sentence into a positive claim. For example, to turn the sentence, "it is not true that for all x, P is true of x" into a positive claim, you can say, "there exists an x such that P is not true of x." I really don't know how to help you with this problem. Maybe you can put, in your own words, what you think the goal of the problem is, and also, since you must have thought at some point that the answer in post 16 was right, tell us what makes you think its right. What makes you think that line two is equivalent to line 1? What makes you think that line three is equivalent to line two? I'll give you an example, see if you can understand it: Let Y be a metric space, with metric d. Let X be any topological space, with topology T. C(X,Y) denotes the set of continuous functions from X to Y. Now let F be a subset of C(X,Y). Then F is equicontinuous if and only if: for all x in X, for all e > 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. Symbolically: [tex](\forall x \in X)(\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] The negation of this sentence in words is: it is not that for all x in X, for all e > 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that it is not that for all e > 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that there does not exist a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, it is not that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that it is not true that for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that there exists f in F such that it is not the case that d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that there exists f in F such that d(f(x), f(x')) > e. Symbolically: [tex]\neg (\forall x \in X)(\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)\neg (\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)\neg (\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)\neg (x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(\neg(x \in U)\ \vee \ \neg (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)\neg (\forall f \in F)(d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)\neg (d(f(x),\, f(x')) < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \not < e))[/tex] [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \geq e))[/tex] and if you like, you can go one step further and say: [tex](\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \in (X-U)\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \geq e))[/tex]
question well the question is Explain using only the defenition and formal logic, wat would be needed to be done to show that , for a particular function f(x) and real number L, L is not the limit of f(x) as x approaches a. [hint : at least get as far as carefully negating the definition of the limit] the definition of limit which is lim-->a F(x) = L if for every number E > 0 there is a corresponding number G(gamma) >0 such that |F(x) - L| <E whenever 0< |x-a|<G so wat i was trying to do was "trying" to negate the definition of limit.. since you must have thought at some point that the answer in post 16 was right, tell us what makes you think its right. What makes you think that line two is equivalent to line 1? What makes you think that line three is equivalent to line two? Well i wanst trying to make them equivalent. i was just trying to negate the sentence steb by step. So at the end my negation of For all e > 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e was not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| <E then is not the case 0< |x-a|<G by the way in post 16 second line should have been not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G sorry
so is that a good negation of the defenition of the limit? A function f with domain D doesnt not have limit L at a point c in D iff not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| <E then is not the case 0< |x-a|<G am i right? wat am i doing wrong?
Yes, but what exactly does this mean. I.e. do you know why "it is not the case that for every number E > 0 there is a corresponding number G(gamma) >0 such that |F(x) - L| <E whenever 0< |x-a|<G" is not an acceptable answer, even though it is technically the negation of the definition. In other words, they want you to produce a sentence that is not only techincally correct, but also in some way meaningful or informative. Well you're way off then. You DO want them to be equivalent. You want to start out with a technically correct answer, "not (definition of limit)" and change it, step by step, into a sentence that is equivalent, and thus also technically correct, but in addition informative. So what exactly do you mean "trying to negate the sentence step by step?" What is it that you think you were doing in your post, because honestly it just appeared that you were randomly writing sentences, where each one differs from the previous by some random change in the logical connectives (like "not," "implies," "whenever," "and," etc.). You don't want to negate it step-by-step. Your first line, "it is not the case that (def'n of continuity)" is already the negation. Then you want to TRANSLATE it step-by-step into something more meaningful. What makes you think this is right? What is this last line supposed to be? Well it is supposed to be the negation of the definition of continuity. What makes you think it is? How did you even get to this line? Again, I simply don't know how to help you. I only suggest looking at the example I gave (I gave two versions, one in semi-English and one symbolic) and trying to figure out if you understand it. Then ask questions about the parts you don't understand. Make sure you know what the goal is in the first place, before you try to understand how my example reaches that goal. Remember, the goal is to take the definition, (def'n of continuity) and write its negation in a meaningful way. "not (def'n of continuity)" is technically correct, but not meaningful. However, starting with "not (def'n of cont'ty)" and translating it step-by-step, each step being a sentence equivalent to the one before it, you can produce a meaningful sentence that is also technically the negation of the definition.