For every rational number there's an irrational one, and visaversa.
reddit.com/r/math/comments/fd73hl/are_there_t...Posted Mar 5, 2020, 9:02 PM UTC
15788
reddit.com
At some level I still just don't understand people who argue about this sort of thing on Reddit. It's not like it's controversial - every single actual mathematician will tell you the same thing. If you truly believe you have a contradictory proof then that would overturn most of modern set theory, analysis, and topology. So why do you think r/math is the place to litigate it? Shouldn't you be writing up the single most significant paper ever instead?
74
I think it's more that they just don't have any experience with formal mathematics, so they don't realize their purported proof is actually wrong. Instead they think they're commenting something obvious and that everyone else just isn't as smart as they are.
62
People who do mathematics for a living aren’t as smart as random internet guys? Sounds right to me I guess.
13
I think the issue suggested is that they're so unfamiliar with the field that they don't even realize that their impression of things runs counter to what mathematicians actually say. They're sure they're right and haven't realized they don't have any actual knowledge on the topic. Dunning-Kruger in action.
52
Lol yeah I was being just a bit sarcastic.
4
When I first started studying math on my own, I thought I had come up with a proof of the continuum hypothesis after getting halfway through an introductory set theory book.
The obvious truth was that I was so out of my element that I didn't even realize how out-of-my-element I was. Luckily I didn't take it further than my own thoughts, but when I learned how wrong I was it was a very humbling experience.
6
Given that his claim contradicts literally the foundation proof for the study of infinite things I would say he probably doesn't know much about the subject at all. I suppose he's confident because his misunderstanding points to a very simple proof of why he's right. If there were one irrational between every pair of rationals (and vice versa) then their cardinality would clearly be the same. He's forgetting (or failing to understand) that there are infinity many of each in any interval so no simple bijection is possible like that.
I think this is probably because he's heard "there is an irrational between any two rationals" which is true but means "at least one" despite sounding like it means "exactly one".
28
This is likely the biggest contributor. It even echoes a technique a physics professor of mine used to use for counting cells in weird lattices by offsetting the vertices from their original positions. It’s exactly the technique you would think of for the finite case.
9
[deleted]
2
For every two irrationals, there are one or more rationals inbetween, and visa versa. Therefore: 1) the number of rationals =or> number of irrationals, and visa versa: 2) the number of irrationals =or> number of rationals. Both these being true means the only possibility is the number or rationals = number of irrationals.
So unfortunately you are wrong, but I will still review your attempted proof when time permits.
How generous of him to find the energy to explain these complicated matters to worldly minds. We should all be thankful.
56
His “visual proof” is even better. “Look. I’ve ordered some arbitrary variables and they alternate because that’s how I wrote them down.”
46
In his defense, that does visually capture exactly what he's thinking. I know because many years ago I had the same thoughts, as I'm sure a number of us have.
Of course, a good educator would not present the proofs of the density of the rationals and irrationals without soon after explaining that there are more irrationals than rationals anyway, to nip these sorts of misconceptions in the bud.
12
I didn't get the impression that anyone in that thread was unclear what he was saying. They were just pointing out that his premise was wrong.
And prior to his visual proof, we have his original proof:
Look, pick any two irrational numbers and there is a rational between them, no? There's proof for you it's >0.5 to randomly pick a rational. I added the greater than as 0 and 1 are both rational, so there's at least one more rational than irrational number in the set.
And then these comments:
My proof is in a standard and logical format
So unfortunately you are wrong, but I will still review your attempted proof when time permits.
I'm just not willing to cut him all that much slack specifically because he was so combative and arrogant.
Edit: Added some more and cleared up some verbiage.
1
Oh I know... I'm the primary person he was "debating"!
And I was expecting him to come up with something like that, I just didn't think he was ever going to try, given how convinced he was of his Truth.
2
All of these infinites, it's all the same bro!
2
All infinite sets are the same cardinality. I construct a bijection from X to Y as follows:
To start, choose any x from X and any y from Y. Let f(x)=y. Then choose the next x and the next y. Since the sets are infinite, there's always a next. QED.
8
S O E L E G E N T!
2
I have always wondered what happens to stubbornly wrong teenage math enthusiasts when they get to college and have to learn, and pass a test on, the topic they had an extended argument about on reddit. Is it humbling or do they stubbornly fail that test?
I was curious so I looked at the wrong guys profile. Unfortunately I discovered he isn't the stubborn high school know-it-all that I assumed from his r/math comments. He's an adult climate change denier who is vocally bigoted against transgender people. At least that fits well with his mathematical stupidity.
Edit: He says I was mistaken about this. Bad at math or not, I have no reason to believe he's lying about his own beliefs, so I retract my earlier statement about him. Edit again: Never mind, I took a quick skim through his profile again and the guy definitely harbors some denialism and bigotry.
50
I can anecdotally answer your question. The people I’ve met that fit that description seem to most often be a bit arrogant and know-it-all until they quickly realize they are no longer among people who can’t tell they’re bullshitting. They then either become reasonable fastidious students or drop the subject entirely due to having zero actual interest other than seeming smart and powerful. There are a select few, at any university it seems, that persist in their stubborn asshole ways far into adulthood. I like to imagine it has something to do with an inferiority complex and a lot of fear of being wrong.
21
Can somebody really get through a math degree while believing that there is a bijection between the rationals and irrationals? I mean I'm an engineer and even I know that's wrong.
5
I mean, “get through” and “math degree” might be kinda strong words there.
7
Lol
1
On top of that, a lot of the time know-it-alls are only that way because it's the only option that they have to feel relevant. Once life stops being entirely high school drama, they (formerly we) get some room to breathe and find sources of identity that aren't so toxic and fragile.
4
I think that outlier is my professor
1
Haha unfortunately that’s not as uncommon as it should be.
2
Quick nitpick; the adjective is "transgender". The suffix "-ed" carries the connotation that it's possible to "transgender someone", or that someone is cis until they've been "transgendered".
15
Oh I went back and forth on that. Thanks. It's fixed.
10
I'm not sure about that last sentence. We say people are brown-eyed, blonde-haired, or long-legged. That doesn't imply anything about those traits being changeable.
2
But nobody assumes that brown-eyed people have had their eyes painted brown or something, whereas lots of people have strange views about what it means to be transgender. Anti-trans activists have even taken to using the word "transed" to emphasize that they really do think people are coerced into being trans.
The use of "transgendered" mostly just seems to be an innocent mistake, but I suspect there is some connection with misconceptions about people "becoming" trans. I can't think of any other adjectives that people commonly append -ed to by mistake.
5
Isn't it a matter of consistency? "Gendered" as an adjective requires -ed.
1
He's an adult climate change denier who is vocally bigoted against transgender people.
Look, I don't care about you lot brigading the post or making jokes, but don't outright lie. I have never been bigoted and am not a climate denier. I am not sure what comments you were referring to, I assume you missed the context or misread them. Thanks.
-8
Muh cow farts. Meat damaging the environment is such bullshit (pardon the pun). Stop reading The Guardian.
12
I then gave an explanation and linked a peer reviewed article. Regardless, discussing the impact of meat production is not climate change denial.
-13
Your article also agrees that meat is bad for the environment. It points out that cutting back somewhat on meat and dairy may be better than cutting out meat altogether and increasing dairy. This is because dairy also comes from cows.
10
Okay, that's fair.
Wait, no. I just looked again. Maybe you don't like hearing it stated as such but you have a history of bigoted remarks about trans people, climate change (faux-)scepticism, etc. If you don't like hearing people say that, you should work on making it stop being true instead of trying to make people stop pointing it out.
12
You're a biggot. I can make stuff up too.
-1
We know, this whole thread is about you making up stuff
9
[deleted]
1
vocally bigoted against transgender people.
what does this have to do with anything?
-17
Kinda like Crank magnetism. Usually if someone believes one dumb thing (rational numbers are as dense as irrational numbers), they usually happen to also believe other dumb things (climate change denial, transphobia).
33
Bigots are usually dumber than dirt. I really don't think that's controversial.
14
R-{0}. There’s a dense set. Between any two numbers in it I can find a rational. Thus they are the same set.
Quod Erat Demonstrandum
30
A simple proof:
For any rational number, I give you pi.
For any irrational number, I give you 3.
Therefore there exists a bijection between the rationals and the irrationals, please give me my Fields Medal.
28
A simpler proof:
I have a piece of paper. On it I drew a blackboard bold RQ and a blackboard bold R\Q. Between them I wrote "a bijection."
Therefore there exists a bijection between the rationals and the irrationals.
9
Here, my favorite unicode characters: ℝ, ℚ
5
That's a bijection between the reals and the irrationals (ehich actually exists).
The first should be blackboard bold Q.
2
Ah right. Whoops.
1
For every action, there's an equal and opposite table.
22
I think this is my favorite part:
For every two irrationals, there are one or more rationals inbetween, and visa versa. Therefore: 1) the number of rationals =or> number of irrationals, and visa versa: 2) the number of irrationals =or> number of rationals. Both these being true means the only possibility is the number or rationals = number of irrationals.
It's an exciting new variation of wrongness in the style of the classic "every probability is 50%! Either it happens, or it doesn't!"
17
The downvotes of fools are like delicious candy to me.
Here's an archived version of the linked page.
^^Quote ^^| ^^Source ^^| ^^Send ^^a ^^message
5
Visaversa Visavers Visave Visav Visa You've had your visa revoked for bad grammar
5
without looking, it's possible the OP speaks another language where this approximates the pronunciation
3
This is the same thread as a the non-standard analysis word salad guy
5
I think his confusion comes from interpreting "for every pair of irrational numbers there is a rational number between them", as somehow "for every pair of consecutive irrational numbers there is a rational number between them". Obviously the idea of "consecutive" irrational numbers is ridiculous on its own, but that's the only way I can understand how he only needs that one statement to "prove" that the number of irrationals and rationals are equal to each other.
Obviously the counter-argument without even resorting to anything technical is to say, sure, you have this pair of irrationals and a rational between them, but there were uncountably many irrationals between that pair as well.
4
In the finite case, if you want a red ball between every two blue balls, and vice versa, then you really must have a line that goes RBRBRBRBRBRB, and that gives you an obvious bijection (pair each red with the following blue). I suspect that something like this picture is what people have in mind with this misconception.
4
Furthermore, even in the infinite case, the following might "feel" true to beginners:
--There is at least one irrational between any pair of rationals, so there are "many" irrationals if we can find one for *any* pair of rationals.
--There is at least one rational between any pair of irrationals, so there are "many" rationals if we can find one for *any* pair of irrationals.
--Therefore the amount of rationals must be "about the same" as the amount of irrationals, since the above works in both directions.
1
The moral: If someone is making a claim that A and B have the same cardinality but they have to embed them in the same set and look at their ordering, there's probably a fallacy somewhere.
1
Rationals & irrationals. Say that I use the half circle to project numbers outside (-1,1) into that interval. Now, let's project rationals r = n/m, n and m integer, with |r|>1, in (-1,1). If a projected rational number accidentally replaces an irrational number then there is one less irrational number in the result set A. The rational numbers will never eliminate all the irrational numbers in (-1,1). The resulting set A will always contain irrational numbers.
The other way around: let's project all irrational numbers outside (-1,1). Then the projection of all irrational numbers outside (-1,1) will eliminate all the rational numbers from (-1,1). There are with this procedure no rational numbers in A left.
Can this be performed / checked with Bishop's constructive mathematics? I think not. There's most likely no constructive proof of it. Constructive mathematics (Bishop's variant) is based on rational numbers.
If the proof is not constructive mathematics, i.e. a procedural approach, then how can one be sure?
I don't know.
1
Oh yes btw the Dunning Kruger concept is the escape module for bad teachers.
1
[deleted]
1
I think this comment could use a bit of context: https://www.reddit.com/r/math/comments/fd73hl/are_there_two_irrational_numbers_that_contain_no/fjhyqta/?context=7
0
With the understanding that a) every square root of a number that is not a a perfect square is an irrational number, and b) the difference between a rational and an irrational number is irrational, it becomes fairly straightforward to map every rational to an infinite number of irrational numbers like so:
Rational number N mapped to the set of irrationals N - sqrt(x), where x is the set of all primes (for example). As every member of one infinite set is mapped to a infinite subset of a different set, it's clear that the cardinalities cannot be the same.
0
it's clear that the cardinalities cannot be the same.
Why? It's very easy to come up with a bijection between N and a partition of N^2 into infinite sets, like so:
f(n) = {(n, m) | m ∈ N}
But N and N^2 are both countable.
4
You're right. Would pointing out that the set of irrationals is not fully covered by my treatment help? No it would not, as f(n) = {(n, 2m) | m ∈ N} wouldn't account for the odds, and yet wouldn't change the countable status of either set.
In fact, there's another disproof of my failed attempt above, which is to map a set to infinite unique partitions of itself.
Summary of the lesson learned: a bijection into infinite partitions doesn't disprove the existence of a bijection, which was what I wanted to do.
1
Another way to think about it - if you have some understanding of cardinal numbers - is that you showed that the cardinal of the irrational numbers is at least the cardinal of the rational numbers (which is ℵ0 the cardinal of the natural numbers) times the cardinal of prime numbers (which is also ℵ0), but since ℵ0*ℵ0=ℵ0, you just showed irrationals have equal or bigger cardinal to rationals.
2
This subreddit serves no purpose other than to ridicule people. It's pathetic.
-26
Ah. First day here, eh?
24
No, I know the common response is "but these people are cranks and you can't talk them out of it". Just tell these people that they don't understand it and walk away. There's no reason to devote all this effort to publicly shaming them on a subreddit.
-15
Bold advice from someone with a 900 point post on r/iamverysmart.
27
And a post to this subreddit from just four days ago, which was removed because of R4. Smells like he's real salty to me and doesn't give a shit about his hypocrisy.
17
There's a difference between making an attempt at math and being shamed for it, vs just saying bluntly that you are smart.
-17
Ridiculing is trying to make someone bad. We just laugh at them, that's different.
Beside the type of person you see here isn't the honest to god mistakes, it's very cocky people that keep their ground in the most stupid ways and for the longest.
I said some pretty stupid stuff on reddit but never feared to be featured in that sub
12
I'll tell you this: I've been "featured" on this subreddit at least twice, maybe 3 times, and I've also passed graduate level mathematics courses. One of those times I had to report the post about me, but the posts about me I believe all got upvoted significantly. It's very discouraging to be "featured" on this subreddit. Maybe you should think of the effect on the person, effectively classifying them as a crank. My way of dealing with being classified as a crank is that I don't give this subreddit any credence. This is just a subreddit full of people trying to put other people down and it's horrible that such a subreddit exists.
1
You really think that making a post about someone on this subreddit isn't ridiculing them? If someone is making some mathematical effort at least they are trying. They might figure it out eventually. Putting them up as someone who does horrible math is just not necessary.
-1
R4: what starts as a genuine misunderstanding common to many young mathematicians turns into a stubborn insistence that the density of the rationals in the reals proves that they are equal in cardinality.
It does not.