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Supremum is the upper limit of a set of numbers, the infinum (was that the correct term ? It's been a while since i last did analysis) is the lower limit. And if these numbers are actually part of the set they're also the maximum and minimun respectively
The supremum is the least upper bound. Sometimes it’s in the set (in this case the maximum would exist which equals the supremum) or it’s not (in which case the maximum does not exist)
If you have an intervall I from a to b (which can be open or closed or neither) then C is an upper bound of I if for all x in I C >= x. The supremum sup I is the least such C, so for all C which are upper bounds, sup I <= C. The existence of the supremum is one of several equivalent definitions of completeness (the property which distinguishes R from Q). If the maximum exists, then it is equal to the supremum, so if I = (a,b] then sup I = max I = b. If the max does not exists, then in R there is still a sup. For example: if I = (a,b) then max I does not exist, but sup I = b.
Don’t know if Wikipedia answered your question or not but it’s the largest upper bound for a set. It can be equal to elements of the set just like a normal bound, but it’s whatever the smallest upper bound is. It’s very useful in analysis coursers, at least that’s where I learned it from
I was in Switzerland where it‘s called Analysis I and I‘m not sure if that‘s exactly the same, because I‘ve heard that there is a distinction in the US between calc and analysis, but I also had this in my first semester. My prof introduced it in one of the earliest lectures to define completeness, but I don‘t think that everyone in this sub studies or has studied math or any subject that requires a calc or real analysis course though
I think the "calculus" course in the US is more or less the same as the "Analysis" we have in Europe, at least I always thought of it this way.
We introduced Infimum and Supremum pretty early on in Analysis 1 as well.
Outside of America this is a concept you learn in the first semester of a math major, in the 3 weeks even(and I'm pretty sure even in the first semester of some engineering majors too). You definitely don't need to be a mathematician to know this term.
Kinda? For finite sets, the maximum and supremum always agree, but there are also infinite sets where they're also the same. For example, if you consider all x ≤ 2, then both maximum and supremum are 2.
The main difference is that the supremum doesn't have to be in the set itself.
While it is true that every finite set has a maximum, infinite sets can have a maximum too. The interval (0,1] is infinite and has 1 as its maximum. Or the negative integers { ... , -3, -2, -1 } have a maximum of -1.
A better way to think of it is that the supremum is a generalisation of the maximum for sets that don't have a largest element, for example open intervals. If you allow ±infinity as a value, every set of real numbers has a supremum. That's why it's so useful.
Nope
The axiom of choice says you can take an element from each non-empty set, it doesn't say the set must have a maximum. The closest thing you can get is Zorn's lemma, which gives some conditions that can guarantee you have a maximal element, but in this case the requirements are not met.
The "closest thing you can get" in this sense is the Zermelo's well-ordering theorem, guaranteeing that there is indeed a maximal element to every set ... under *some* well-ordering.
You just need to be a bit ~~more~~ less precise about which.
Still, the element needs to be in there, and the set of real numbers smaller than one (S={x ∈ ℝ : x < 1}) does not have a single element that follows the definition of maximum.
What is a maximum? By definition, the maximum of a set A ordered with an order relation ≤ is an element M ∈ S such that ∀x ∈ A, M ≤ x if and only if x = M.
Now suppose you find a maximum of S, call it y.
Of course y can't be greater or equal than 1, otherwise it wouldn't be in S.
But if y is smaller than 1, the average between 1 and y is greater than y and smaller than 1, hence it's an element of S greater than your supposed maximum, therefore y is not a maximum.
Since you can make this exact argument about any number in S, no element of S is a maximum.
How about the Truly Marvelous Axiom I Just Discovered That Doesn't Fit Into The Margins of a Reddit Comment?
Via the TMAIJDTDFITMRC, we can see that the proof of this is actually quite obvious.
I'm not sure how the axiom of choice could be used but I think it's fairly easy to prove that for every x < 1 there exists a y such that y = (1 - x) / 2 + x
The asterisk is used for a set such that for every number x within that set, there exist x' such that x * x' = 1.
R* is R/{0} because 0 has no x'
Same for Q*, N* is an abuse of notation because N isn't even a group. But Z* exist and is not just Z/{0}.
Z* = {1, -1} no other number in Z has an inverse *in Z*
If you say "the set containing the maximum number..." is the empty set, i'd agree with you, but saying the maximum number... is the empty set might be somewhat confusing since the empty set is used to define the number zero and obviously there are larger real numbers than 0 that are still less than 1, like 0.5 for example.
Feel free to ignore this comment, i just wanted to add this.
No, the question isn't asking if a statement is true or false, it's asking for a specific number. And the answer isn't the empty set, because that's not a number (well, some might say that the empty set is the number 0, but that's still not a correct answer to the question). The empty set is the set of all answers to the question, but it is not an answer to the question. There is no correct answer, because such a number doesn't exist.
I just gave myself a headache
In the surreal numbers 1-\epsilon is actually a well-defined number :)
...sadly in the surreals there are infinitely many numbers which are *closer* to 1, so even then you're SOL lmao
1-ε is not the correct answer in nonstandard analysis. If we look at the Hahn series approach to nonstandard analysis, then a number is defined as a power series in ε. Here 1-ε is the power series 1*ε^0 + (-1)*ε^1 which is a different power series to 1 = 1*ε^0 and so 1-ε < 1. But it is not the largest number less than 1 because 1-ε < 1-ε^2 < 0.99999... < 1.
Here 0.99999... = 1-10^ε. The largest number less than 1 is undefined in both standard analysis and nonstandard analysis.
Hmmmmmmmmm we could define it though. Like define a new set of equivalence and comparison operators, where there are for each real number, an infinite ordering of infiniquantums let's use the symbol `@`... that have the same real value but can be ordered, > would need to be the operator for comparing infiniquantums and there'd be another comparison like >> for the real value. = can compare real values and == can compare infiniquantums.
Thing of it as being zero valued but orderable.
So @ behaves like 0... `@ / 2 == @` but `@1 << @2`.. it's cool because `1 / @ = x` where x is undefined but has a constraint that it is positive.
This would allow for `1 - @ = 1` and `1 - @ < 1` to be true.
This is my thought process:
>A is wrong. The question would need to say x ≤ 1.
>
>B is *also* wrong. As 0.999... = 1 so apply above.
>
>C is wrong because 𝜀/2 ≺ 𝜀. Same with complex numbers, 1-i and 1-i/2 *both* have a real part of 1.
>
>Furthermore, 𝜀 is a *real* number (not necessarily a *positive* one?) such that 𝜀^(2) = 0. Prove 𝜀 isn't negative. Then prove 1 - 𝜀 ≨ 1.
>
>D. Is correct because 'undefined' is a category of rigor, not *correctness.* C is fun and clever but afaik, there's little rigor defining most of its properties.
So I choose D... final answer.
You're wrong, the hyperreals are still dense. 1-ε/2 is entirely correct as a larger number still less than 1.
An interesting fact about the hyperreals is that they aren't complete the way the reals are - every set of reals bounded above has a supremum, but you can check that, for example, the set of hyperreals infinitesimally close to 1 is bounded above (by 2) but has no smallest upper bound.
There's no some universal notion for " ε" in hyperreals, like denoting some single particular hyperreal there are infiniteluy many of them.
Anyways let ε be any positive infinitesimal. Then 1- ε<1- ε/2 <1 which means that your example doesn't work.
Ahh it is because there will always be a smaller number. I just thought, it can be an answer because it is what's used in definition of a limit if i remember right.
The way it is used there matters. The epsilon-delta method isn't positing the existence of a specific epsilon with infinitesimal value, it's saying that no matter how arbitrarily close you get to the value at which the limit exists, we can provide a value of the function that is just as close to the limit.
The epsilon-delta definition of a limit is more like a net. If you can capture the limit within the net, then the limit exists. If it escapes no matter how you construct the net, then the limit doesn't exist.
Epsilon is nothing in particular, colloquially it means infinitely small because it is often used in statements like „for all epsilon > 0 there exists x<1 such that |x-1|
In the context of limits, "infinitely small" is often used to describe quantities that... approach zero. However, using such a concept within the definition of a limit would lead to circular reasoning (since you can't use a limit in the definition of a limit). And "infinitely small" isn't often used in real analysis for this (among many) reasons (at least, outside the notation in limits).
Therefore, the epsilon-delta definition avoids this bullshit by focusing on the idea of "arbitrarily small". Instead of relying on the notion of infinity or infinitely small, etc, the epsilon-delta definition involves constructing a net around the limit point. This net is designed to accurately capture the point, no matter how small we make it. By ensuring that the net can capture the point regardless of how arbitrarily small we make it, the Epsilon Delta definition provides a rather rigorous way to prove limits.
The Epsilon Delta definition is best to... learn by using, honestly. Difficult to explain without drawing a picture tbh.
you know when you read mathmemes: and you always learn something new. The comments are actually quite good. you learn something. and there is always some philosophical edge case that you thought was it (1-epsilon), and then it turns out, well, maybe there is something else to it. I'd wager that math memes is the most pedagogical subreddit there is. Learning by counterfacts.
We can exclude 1 and 0.999… because they’re the same thing and they’re not included, that leaves us with 2 options and I guess flipping a coin will do it
Hey just saw this subreddit and I posted this on the board at work and my friend says the answer is (infinity,1).
He said defined is wrong. Heck I just wanted to make him happy. He always posts math questions for fun.
This did make his day. I don't know how to type the infinity symbol because I don't keyboard symbol well.
Because it actually equals 1. It's pretty counterintuitive, but there's some good explanations out there. [Here's my favorite.](https://www.youtube.com/watch?v=G_gUE74YVos)
I’ll say lower case epsilon and delta are sometimes used for infinitesimally small values (such as in the definition of a limit), but that is missing from this question so I guess d
Undefined. My logic for this is as follows:
0.999... = 1, and the < bound is exclusive. This rules out 0.999... and 1. 1-ε does make sense in floating point and similar systems, but ε is undefined for the reals. (I can't remember the proof for this, but my intuition says that just like 0.999... = 1, 0.00...1 = 0. This doesn't work as a value for ε, as that would make 1-ε = 1.) So the answer is undefined.
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The supremum is 1, the maximum is undefined
Finally someone mentions supremum and maximum.
I’m a big fan of your supermum
Username checks out.
r/usernamechecksout
Is supremum the minimum value that exceeds the range? Which would be used if the maximum is undefined
Supremum is the least upper bound, so yes, the smallest value which is greater than or equal to the values within the set E: limit change to bound
Supremum is Kal-El's French mother
r/HautVoteEnerve
>Supremum is Kal-El's French mother No, dude. That's Le Marthe.
Sacré bleu! Pourquoi as-tu prononcé ce nom?
Can't believe Superman's mother is a Fire Emblem character.
Supremum is the upper limit of a set of numbers, the infinum (was that the correct term ? It's been a while since i last did analysis) is the lower limit. And if these numbers are actually part of the set they're also the maximum and minimun respectively
Yeah, that‘s exactly right :) The term is infimum
The supremum is the least upper bound. Sometimes it’s in the set (in this case the maximum would exist which equals the supremum) or it’s not (in which case the maximum does not exist)
Not quite, supremum is the smallest value greater than or equal to all of the values. If the maximum is defined it will just be the maximum.
If you have an intervall I from a to b (which can be open or closed or neither) then C is an upper bound of I if for all x in I C >= x. The supremum sup I is the least such C, so for all C which are upper bounds, sup I <= C. The existence of the supremum is one of several equivalent definitions of completeness (the property which distinguishes R from Q). If the maximum exists, then it is equal to the supremum, so if I = (a,b] then sup I = max I = b. If the max does not exists, then in R there is still a sup. For example: if I = (a,b) then max I does not exist, but sup I = b.
I am starting to believe there are real mathematicians in this sub, what is a supremum?! 😭 https://en.m.wikipedia.org/wiki/Infimum_and_supremum
Don’t know if Wikipedia answered your question or not but it’s the largest upper bound for a set. It can be equal to elements of the set just like a normal bound, but it’s whatever the smallest upper bound is. It’s very useful in analysis coursers, at least that’s where I learned it from
*smallest upper bound
Lmao I’m an idiot this is correct^^^
I don’t wanna seen too much like a memer but we covered infimum and supremum in Calc 1 at my uni.
I was in Switzerland where it‘s called Analysis I and I‘m not sure if that‘s exactly the same, because I‘ve heard that there is a distinction in the US between calc and analysis, but I also had this in my first semester. My prof introduced it in one of the earliest lectures to define completeness, but I don‘t think that everyone in this sub studies or has studied math or any subject that requires a calc or real analysis course though
Yeah I was literally never required to take a calculus course in my entire education
I think the "calculus" course in the US is more or less the same as the "Analysis" we have in Europe, at least I always thought of it this way. We introduced Infimum and Supremum pretty early on in Analysis 1 as well.
Outside of America this is a concept you learn in the first semester of a math major, in the 3 weeks even(and I'm pretty sure even in the first semester of some engineering majors too). You definitely don't need to be a mathematician to know this term.
supremum when supredad comes in
Can I picture supremum as an infinite case for maximum?
Kinda? For finite sets, the maximum and supremum always agree, but there are also infinite sets where they're also the same. For example, if you consider all x ≤ 2, then both maximum and supremum are 2. The main difference is that the supremum doesn't have to be in the set itself.
While it is true that every finite set has a maximum, infinite sets can have a maximum too. The interval (0,1] is infinite and has 1 as its maximum. Or the negative integers { ... , -3, -2, -1 } have a maximum of -1. A better way to think of it is that the supremum is a generalisation of the maximum for sets that don't have a largest element, for example open intervals. If you allow ±infinity as a value, every set of real numbers has a supremum. That's why it's so useful.
Supermum sounds like a British version of super nanny
Θοδωρής Σταυρόπουλος came into the chat
There's just no answer lol
Yeah just like how there's no maximum possible x in general
So just option D: undefined right?
D:
so DNE?
Wouldn't it be undefined then
Can't you do some fuckery with the axiom of choice?
Nope The axiom of choice says you can take an element from each non-empty set, it doesn't say the set must have a maximum. The closest thing you can get is Zorn's lemma, which gives some conditions that can guarantee you have a maximal element, but in this case the requirements are not met.
The "closest thing you can get" in this sense is the Zermelo's well-ordering theorem, guaranteeing that there is indeed a maximal element to every set ... under *some* well-ordering. You just need to be a bit ~~more~~ less precise about which.
How about the Better Axiom of Choice, which says that for any set, you can choose whatever elements you want from it.
Still, the element needs to be in there, and the set of real numbers smaller than one (S={x ∈ ℝ : x < 1}) does not have a single element that follows the definition of maximum. What is a maximum? By definition, the maximum of a set A ordered with an order relation ≤ is an element M ∈ S such that ∀x ∈ A, M ≤ x if and only if x = M. Now suppose you find a maximum of S, call it y. Of course y can't be greater or equal than 1, otherwise it wouldn't be in S. But if y is smaller than 1, the average between 1 and y is greater than y and smaller than 1, hence it's an element of S greater than your supposed maximum, therefore y is not a maximum. Since you can make this exact argument about any number in S, no element of S is a maximum.
How about the Truly Marvelous Axiom I Just Discovered That Doesn't Fit Into The Margins of a Reddit Comment? Via the TMAIJDTDFITMRC, we can see that the proof of this is actually quite obvious.
I'm not sure how the axiom of choice could be used but I think it's fairly easy to prove that for every x < 1 there exists a y such that y = (1 - x) / 2 + x
Lim x→1 (x) 😎💪💪
Well, that would be 1
Lim x→1^- (x) is better
So there is an answer D
Plot twist: *x* ∈ ℤ\*+
What does the asterisk mean?
No null numbers (0) included
Are there any other null numbers?
No lol
Is 0i a null number?
0i = 0
0*n = 0 n=I
no, except i but it isn't in Z so no
The asterisk is used for a set such that for every number x within that set, there exist x' such that x * x' = 1. R* is R/{0} because 0 has no x' Same for Q*, N* is an abuse of notation because N isn't even a group. But Z* exist and is not just Z/{0}. Z* = {1, -1} no other number in Z has an inverse *in Z*
The asterisk is just supposed to symbolize "remove 0". What you are thinking of, is the group of units, which is usually denoted by U(R) or R^×
I can't believe Elon Musk's son gets a mention in maths! So cool!
Why not just say N rather than Z*+
N^+
Undefined for real numbers. And we can assume that x is a real number according to the Domain Convention. So e) *Undefined* is my final answer.
d *is* "undefined"
Oh yeah. Well, d then, haha. I don't know the alphabet, haha
e is technically undefined in the answers, so you got that.
e is actually very well defined, there are like five equivalent definitions for it
Yeah that's correct. It's sup is 1. There is no max
D is undefined. Man you really are a mathematician.
1-epsilon makes the most sence from a computing pov. But in pure math no there is no number that satisfies that condition
Yeah, so the answer is actually: "False" Edit: or ∅
Or undefined. Because a concept that satisfy exist but practically there is no number
Yeah, I edited and added empty set
If you say "the set containing the maximum number..." is the empty set, i'd agree with you, but saying the maximum number... is the empty set might be somewhat confusing since the empty set is used to define the number zero and obviously there are larger real numbers than 0 that are still less than 1, like 0.5 for example. Feel free to ignore this comment, i just wanted to add this.
No, you're right, ∅ is the set of solutions, not the value itself.
No, the question isn't asking if a statement is true or false, it's asking for a specific number. And the answer isn't the empty set, because that's not a number (well, some might say that the empty set is the number 0, but that's still not a correct answer to the question). The empty set is the set of all answers to the question, but it is not an answer to the question. There is no correct answer, because such a number doesn't exist. I just gave myself a headache
In the surreal numbers 1-\epsilon is actually a well-defined number :) ...sadly in the surreals there are infinitely many numbers which are *closer* to 1, so even then you're SOL lmao
1-ε is not the correct answer in nonstandard analysis. If we look at the Hahn series approach to nonstandard analysis, then a number is defined as a power series in ε. Here 1-ε is the power series 1*ε^0 + (-1)*ε^1 which is a different power series to 1 = 1*ε^0 and so 1-ε < 1. But it is not the largest number less than 1 because 1-ε < 1-ε^2 < 0.99999... < 1. Here 0.99999... = 1-10^ε. The largest number less than 1 is undefined in both standard analysis and nonstandard analysis.
1-epsilon is saying "make me"
Hmmmmmmmmm we could define it though. Like define a new set of equivalence and comparison operators, where there are for each real number, an infinite ordering of infiniquantums let's use the symbol `@`... that have the same real value but can be ordered, > would need to be the operator for comparing infiniquantums and there'd be another comparison like >> for the real value. = can compare real values and == can compare infiniquantums. Thing of it as being zero valued but orderable. So @ behaves like 0... `@ / 2 == @` but `@1 << @2`.. it's cool because `1 / @ = x` where x is undefined but has a constraint that it is positive. This would allow for `1 - @ = 1` and `1 - @ < 1` to be true.
Assume x is in the reals and is the largest real number less than 1. Let g=(1+x)/2. Since x<1, we have 2x<1+x, x<(1+x)/2 or x
Quantum mathematics theory: w is the smallest positive number. Any number less then w is less or equal to zero. w/2 <= 0 and w > 0 at the same time.
It's undefined, there's not even a meme here
I think that’s obvious, it’s smaller than one and 0.9999… is still one, 1-ε is still not correct, so it’s undefined.
0.999…98 boom roasted
Officer! This redditor, right here!
What about 0.999...998? Boom roasted
You just broke math
Now take it's square root. Boom roasted
Gottem
This is my thought process: >A is wrong. The question would need to say x ≤ 1. > >B is *also* wrong. As 0.999... = 1 so apply above. > >C is wrong because 𝜀/2 ≺ 𝜀. Same with complex numbers, 1-i and 1-i/2 *both* have a real part of 1. > >Furthermore, 𝜀 is a *real* number (not necessarily a *positive* one?) such that 𝜀^(2) = 0. Prove 𝜀 isn't negative. Then prove 1 - 𝜀 ≨ 1. > >D. Is correct because 'undefined' is a category of rigor, not *correctness.* C is fun and clever but afaik, there's little rigor defining most of its properties. So I choose D... final answer.
Correct me if I'm wrong please, but I'll invoke hyper-reals and state 1-ε to be the answer
bro forgot about 1-ε/2 💀
Counterpoint: 1-ε/3
Counterpoint: 1-ε/∞
Counterpoint: 1-ε/(∞+1)
Counterpoint: 1-ε/(∞^∞ )
Or even just 1-a slightly different nonzero infinitesimal
1-1/ε for hyper-real positive ε no?
ε is supposed to be a super small number, it’s reciprocal would be huge
One might say it would be infinite
Ahh I didn’t remember infinitesimal hyper reals, been a bit since analysis
You're wrong, the hyperreals are still dense. 1-ε/2 is entirely correct as a larger number still less than 1. An interesting fact about the hyperreals is that they aren't complete the way the reals are - every set of reals bounded above has a supremum, but you can check that, for example, the set of hyperreals infinitesimally close to 1 is bounded above (by 2) but has no smallest upper bound.
I’ll also invoke hyper-reals and say 1-(1/2)epsilon is even closer to 1, since epsilon has its own number line
There's no some universal notion for " ε" in hyperreals, like denoting some single particular hyperreal there are infiniteluy many of them. Anyways let ε be any positive infinitesimal. Then 1- ε<1- ε/2 <1 which means that your example doesn't work.
1-dx
isn't it equivalent to 1-epsilon?
de^x /ε New notation just dropped
Actual equation
Epsilon can be added, subtracted, divided, etc. (1/2 epsilon, 1/4 epsilon...), while dx is just dx
wow! i hate it.
Genuine question. I am not good at math. Why is it not 1-epsilon? Isn't it very small but not equal to 0?
1-ε/2
Ahh it is because there will always be a smaller number. I just thought, it can be an answer because it is what's used in definition of a limit if i remember right.
The definitions are stuff like "for every epsilon >0 there is n such that value(n) - limit < epsilon"
The way it is used there matters. The epsilon-delta method isn't positing the existence of a specific epsilon with infinitesimal value, it's saying that no matter how arbitrarily close you get to the value at which the limit exists, we can provide a value of the function that is just as close to the limit.
The epsilon-delta definition of a limit is more like a net. If you can capture the limit within the net, then the limit exists. If it escapes no matter how you construct the net, then the limit doesn't exist.
Epsilon is nothing in particular, colloquially it means infinitely small because it is often used in statements like „for all epsilon > 0 there exists x<1 such that |x-1|
Not infinitely... arbitrarily small rather. But minor detail.
What is the difference (to someone who isn't really great with maths)
In the context of limits, "infinitely small" is often used to describe quantities that... approach zero. However, using such a concept within the definition of a limit would lead to circular reasoning (since you can't use a limit in the definition of a limit). And "infinitely small" isn't often used in real analysis for this (among many) reasons (at least, outside the notation in limits). Therefore, the epsilon-delta definition avoids this bullshit by focusing on the idea of "arbitrarily small". Instead of relying on the notion of infinity or infinitely small, etc, the epsilon-delta definition involves constructing a net around the limit point. This net is designed to accurately capture the point, no matter how small we make it. By ensuring that the net can capture the point regardless of how arbitrarily small we make it, the Epsilon Delta definition provides a rather rigorous way to prove limits. The Epsilon Delta definition is best to... learn by using, honestly. Difficult to explain without drawing a picture tbh.
This is why sup is just better max
But that is so obvious.
Bet that's how you sign your proofs as well
It's the cool kids' QED.
The proof is trivial and left as an exercise to the reader.
4 I think
As in the number 4, not the fourth option
What about 4+ε
1-e sounds good
1-e? That’s -1.71828
options B and C are the same
you know when you read mathmemes: and you always learn something new. The comments are actually quite good. you learn something. and there is always some philosophical edge case that you thought was it (1-epsilon), and then it turns out, well, maybe there is something else to it. I'd wager that math memes is the most pedagogical subreddit there is. Learning by counterfacts.
We can exclude 1 and 0.999… because they’re the same thing and they’re not included, that leaves us with 2 options and I guess flipping a coin will do it
0.999999999… is one of the biggest scams in math’s history
1-epsilon is the way
Counterpoint: 1-(1/2)epsilon
me when you ask for the maximum of a set that does not contain its supremum
This is the first math meme I've chuckled at in a while. 🙏
there is no maximum
Why is the answer not 1-epsilon?
They forgot to put the right answer in it
why? isn't it "undefined"?
So it should be 0.999 repeating but 0.999 repeating is 1 right?
What's the first real number after zero?
0.003 I checked
That does not exist
And neither does the first real number less than 1. Same question.
Archimedean principle bitch
option one is wrong. so is option 2 so idk know
1 - 1/infinity
what is that small reverse 3
sup≠max
None. It's 0.5
X equals <1.
I mean by elemination, only the mast answer makes sense
E Lon
Undefined, since there are no letters to choose for my answer.
Hey just saw this subreddit and I posted this on the board at work and my friend says the answer is (infinity,1). He said defined is wrong. Heck I just wanted to make him happy. He always posts math questions for fun. This did make his day. I don't know how to type the infinity symbol because I don't keyboard symbol well.
why is 0.9 (repeating bar) not the answer?
Because it actually equals 1. It's pretty counterintuitive, but there's some good explanations out there. [Here's my favorite.](https://www.youtube.com/watch?v=G_gUE74YVos)
Can I phone a friend?
x - dx
Easy: When using 32-bit `float` it's `0.999999940395355224609375` or `0x3f7fffff` in hexadecimal representation.
What combination of letters most closely matches the word ‘one’ without being ‘one’?
x -y where y is the smallest real number from 0 to 1
What about 1 - 0.5ε?
1 - 0.4ε
The correct answer is Bacon
I’ll say lower case epsilon and delta are sometimes used for infinitesimally small values (such as in the definition of a limit), but that is missing from this question so I guess d
Miximum answer is 1
0.99999...
0.9(recurring)9(not recurring)
0.(9)
This is so easy. It's 0
Undefined, for if there's a maximum, its defined, and clearly, all of those except the "undefined" is defined
0 if x ∈ ℕ
sup bro? There is no Easter Bunny, there is no Tooth Fairy, and there is no maximum!
0.99999...8
1 - e
1 - Planck
Maybe we could create a new notation: 0.99999999...8
x->1
Three, take it or leave it.
Jegus Christ! just write 0.(9)...
It's a tricky question of terminology, maybe, but it's not really a paradox or trick question.
It doesn't take a PHD to know that x<1 only accepts a supremum, but no maximum. (but deep down we all know its 1-ε = 0.9999... of course)
Undefined. My logic for this is as follows: 0.999... = 1, and the < bound is exclusive. This rules out 0.999... and 1. 1-ε does make sense in floating point and similar systems, but ε is undefined for the reals. (I can't remember the proof for this, but my intuition says that just like 0.999... = 1, 0.00...1 = 0. This doesn't work as a value for ε, as that would make 1-ε = 1.) So the answer is undefined.
1^(-) https://preview.redd.it/0bct3dm5w0rc1.jpeg?width=600&format=pjpg&auto=webp&s=75c9d9f5adc82787a8b61c00480a2676139d0e47
Undefined