Ok, here's another silly math problem. This one is from elementary school:
1/3 = 0.333333... (don't know how to write the upper bar)
2/3 = 0.666666...
3/3 = 1/3 + 2/3 = 0.333333... +0.666666... = 0.999999...
:wink:

Ok, here's another silly math problem. This one is from elementary school:
1/3 = 0.333333... (don't know how to write the upper bar)
2/3 = 0.666666...
3/3 = 1/3 + 2/3 = 0.333333... +0.666666... = 0.999999...
:wink:
oh please! not the 0.9999... = 1 thread again!
I am sure you must have had this before my time.
What was the most intelligent interpretation of this one?
x=0,999999....
10x=9,999999....
9x=9
x=1
Jesus Christ. We had a huge argument over this shit.
If you do not believe that 0.999... = 1, then do not believe it. The world is not going to end.
there should be a thread clearly stating that
0.99999999...=1 is agreed and argued by most of the people here.
9 times .9.... isn't 9Originally Posted by Zelos
also x can't be ONE and point nine nine nine... two different things.
watch my step by step and you see there is nothign wrong with it
this is mathematical proof that 0,99999..... is 1
lince!: there should be a thread clearly stating that
0.99999999...=1 is agreed and argued by most of the people here.
0.99999999...=1 is agreed
Thank you very much!
agreedOriginally Posted by M
that used to trip me out when I was younger, and to some extent it still does. For instance, when I graph the equation y=(x^21)/(x1) it makes a perfectly clear line, and y=2 when x=1, although if I actually evaluate the equation at x=1 then it comes out to 0/0. So I guess basically it has to be solved with limits...we say the limit is 2 as x approaches 1. Intuitively, it seems like .99999... is the same sort of thing some how: its different from 1 but functionally the same.
Well yeh, it is the same. The demonstration is quite simple and showed by Zelos before. .9999999... = 1
There's no doupt. And yes, we could see it like a limit, even if the demonstration doesn't need them.
this basically proves that an infinitely small value is equal to 0.
I don’t know if there is a formal mathematical definition for “infinitely small,” but most people would probably agree that “infinitely small” and “zero” have the same meaning.Originally Posted by ZebraFiesta
I don't know what percentage of the population use the term "infinitely small" or "infinitesimal" on a regular basis. Those who do would disagree with the statement "infinitesimal is the same as zero". The conceptual difference may be infinitely larger than the difference in magnitude, though.
I suppose I should have said "most people here".
Ha! In the spirit of this thread, I say this is silly, in fact it is wrong.Originally Posted by ZebraFiesta
Zero can be defined in several different ways, not all of them obviously connected (but,conceptually, they are). Here's one definition: zero is the additive identity, such that x + 0 = x, and x  x = 0 (these equalities are the same, of course). Does that look like an "infinitesimally small" number to you? It certainly doesn't to me.
Just like infinity, in all its guises, zero is a slippery customer, but for sure it's not "infinitely small".
And may we please, please have no more nonsense about whether 0.99... = 1, it's been done to death
@Zelos:
How do you get to that:
10x = 9.999999999999...
9x = 9
?
Because
9x = 8.99999999999...
and we have the same problem again!
But the answer with the "infinitely coming close to 1 but not being 1" is not
valid, since 1/3 + 2/3 = 3/3 = 1.
It is intriguing for me, when I consider that longer...
Oh, for eff's sake, read what Zelos said. if x = 0.99... and 9x = 10x  x, use your calculator again, not that it's relevant, calculators don't handle infinite decimal places.Originally Posted by mastermind
Look, as I'm feeling grumpy, and this is not particularly directed at you, Mastermind, let me get this off my chest. This is a math subforum. Ask a question, and it may be answered. Start a thread and it will be looked at, and it may be criticized. If you have no math skills at all, fire away with questions, but please, please do not attempt to demolish all mathematics without fully understanding the beast you are contending with.
Threads like this are infinitely tedious, there is nothing left to say.
My, this is almost like "divide by zero."
Just for fun, why don't I argue that 0/0 = 1. After all, x/x = 1 for an infinite set of numbers.Originally Posted by e_atom
... then why are you still talking?Threads like this are infinitely tedious, there is nothing left to say.
I think you are right.
However, the solution is relatively simple. Just ignore the thread and don't try to correct the incorrectable.
The End...
...
(or is it?) :wink:
You have a problem with this? But seriously  do you?Originally Posted by Absane
Before you get on a high horse and say "division by zero is undefined" tell us division of what by zero is undefined.
As you say, just for fun!
because I'm a wellknown selfinflated windbag, that's why. Next question?Originally Posted by M
Heh.. not really.Originally Posted by Guitarist
0.9999999 is not 1! it can be considered 1 but it is not! The problem lies with this:
1/3 is 0.33333
2/3 IS NOT 0.66666
2/3 is 0.66667
1/3 + 2/3 = 3/3 = 1
0.33333 + 0.66667 = 1
Solved
Agreed.Originally Posted by shawngoldw
No it's not1/3 is 0.33333
Agreed, again2/3 IS NOT 0.66666
No. Your calculator is misleading you by rounding up2/3 is 0.66667
As you say, "solved", but not in your favour, I'm afraid.1/3 + 2/3 = 3/3 = 1
0.33333 + 0.66667 = 1
:D Solved :!:
well 1/3 and 2/3 should never be taken out of fraction form. But if they are taken out then you need to round
rounding is a convienience, i've never HAD to round and in this case neither do you. in this case rounding has resulted in your decimal expansion no longer being equal to your fraction.
0.3 is not 1/3.
0.33333 is not 1/3.
0.333333333333333333333333333333333333333333 is not 1/3.
0.33333... recuring to an infinite number of decimal places is 1/3.
yes, an infinite number of decimal places... but in the original post how was it added? i saw 0.333333 + 0.666666. I don't know about you, but I sure can't add 2 infinite numbers.
it wasOriginally Posted by shawngoldw
3/3 = 1/3 + 2/3 = 0.333333... +0.666666... = 0.999999...
hmm... i just typed in 0.3333333... + 0.6666666, each to 30 places, into my calculator and hit enter, and got 1. My thoughts on this are basically that .99999... is not 1, simply because it is not the numeral "1," but it can be effectively equal to the numeral 1. Probably someone's already said that in some way, but I thought i'd go ahead and add my thoughts...
just realizing... probably where some of the discrepancy lies is in whether we're talking about .99999... being 1, or .99999... equaling 1...
neither can i, but these numbers aren't infinite.Originally Posted by shawngoldw
umm... yes they are... If you call it 1/3 it isn't but if you convert it it is 0.333 with infinite 3s
Then, my friend, you don't know the meaning of infinite. Wallaby is quite right; however you write 1/3 (or 1, for that matter) it is most decidedly finite.
You also tried to draw a distinction between "equals" and "is". This was a more profound statement than you perhaps realize, which we can talk about if you want. But, on the evidence, I wonder if your mathematical skills are up to it?
so if it isn't infinite, how many decimal places is it to?
an infinite number of decimals.... all adding up to a finite value....
1/3 = 0.3 +0.03 +0.003 +0.0003 +...
= sum(3/10^n); n = 1 ... inf
It's a converging series.
The only reason anyone has any problem accepting that 1/3 = 0.3333... or 0.9999...=1 (and they do...they are exactly equal) is because we use a BASE 10 NUMBER SYSTEM.
0.333... is represented the way it is because of that base 10 number system. 3 is not evenly divisible by 10.
But what if we used a base 9 number system? (08 instead of 09)
1/3 = 0.3 now.
1/3 x3 = 1.
0.3 x 3 = 1.
Neat and tidy. Just because we switch over to our comfortable base 10 REPRESENTATION of these exact same values doesn't change the fact that the math still works out exactly the same. 1/3 x 3 = 1 whichever way you slice it. 0.333... is another way of saying 1/3 and they represent the exact same value.
I don't get it. (forgive my ignorance...I'm limited to high school AP Calc AB here).Originally Posted by Guitarist
Ha! Ignorance alone is infinitely(!!) forgivable (and correctable). Ignorance + arrogance are neither. We've seen too much of that here, not from you, though.
Let me try and explain: we don't need a definition of infinity to know that it's a really, really big number. The number of 3's in 0.33... is a really, really big number, but 0.33... = 1/3 is not. Simple as that.
Here's the slightly technical version: an ordered set is said to be bounded above if there is an element greater than or equal to any element of the set. The real line R is clearly unbounded, and convention dictates the use of the symbol ∞ for the (nonexistent) upper bound of R. The set [0,1] is bounded above by 1 and below by 0; as 0.33... and 0.99... are elements of [0,1], therefore by definition cannot = ∞.
Now let f:[0,1] → Z+ (positive integers) be an enumeration ("counting") of the decimal places in all a in [0,1]. Obviously f(a) = ∞ for all a, including 0.33... and 0.99... = 1.00...
Do you see the difference between these last 2 paragraphs?
That's very approximate  don't use it at school!
Nice explanation Guitarist.
In slightly less mathematical semantics: Let's say you approximate 1/3 by just considering N decimal places. If you choose N=1, you have just one decimal and your approximation is 0.3. The more decimals you consider (increasing N) the higher the value of your approximation will be, and the closer you will get to the value of 1/3. In other words you approach this value from below. However, you will never exceed it! Your result can never be larger than 1/3, even as the number of decimal places goes to infinity. This is because each addition you make becomes smaller and smaller as N increases. Your approximation of 1/3 is "bounded", which means, for example, it will never be larger than 1 or smaller than 0. An infinite number (and this is really just semantics), however, is by definition "unbounded", which means there is no definite number that is larger. There sure are a heck of a lot numbers larger than 1/3 (like 1, 2, 3...). Agree?
Actually, I imagine you're pretty well equipped if you remember any of that stuff.I'm limited to high school AP Calc AB here
Unfourtunetly, AP calculus courses tend to concern themselves with passing the AP exam instead of teaching calculus. I remember that all of my homework from that course came from the AP calc exams. Proofs were very limited.Originally Posted by M
But this isn't just limited to AP calculus. Generally, all math courses in public schools (primary and secondary) are concerned with getting kids a passing grade instead of really learning mathematics.
Yay! That's why we older guys hang out here! To try and teach a little simple math by offering proofs (often of a very sketchy kind, in my case). Often one receives no thanks for this here. But then, did you thank your profs, Absane? I didn't then  I do now, mentally, as it were.
I just ended my intro course to Graph Theory. At the end of the semester, I actually shook my professor's hand and told him it was a great course. I am also looking forward to doing some undergraduate research in graph theory with him.Originally Posted by Guitarist
OK, it's Saturday night (coming up), here's a bit of fun.
Somebody show me the fallacy here. Earlier I defined f:[0,1] → Z+ as an enumeration of the decimal places in all a in [0,1]. I said that obviously f(a) = ∞ for all a, including 0.33... and 0.99... = 1.00...
So let's now refine f slightly: f<sub>exact</sub>(a) terminates the enumeration on a exactly, i.e. f<sub>exact</sub>(1/2) = 2, f<sub>exact</sub>(1/4) = 3, etc. To save typesetting, let f<sub>exact</sub>(a) = g(a).
Then for all g(a) in R+, there is some a in [0,1]: g is surjective.
g will also be injective iff, for some a, b ∈ [0,1], g(a) = g(b) in R+ ==> a = b. This is certainly not true; g(0.33...) = ∞ = g(0.66...) = ... Likewise, we must have a ≠ b ∈ [0,1] ==> g(a) ≠ g(b). However, g(0.99...) = ∞, but g(1.0) = 1.
Therefore g is not a bijection i.e. is not both injective and surjective, and therefore has no inverse. As the relation equality requires a bijection between two elements, I conclude that 0.99.. ≠ 1.
Needless to say, this is seriously flawed. Tell me why (for a sweetie)
Thanks, Guitarist, though M's explanation was easier to grasp.
g:I>N where I = [0,1]. For all x in I, there must exist y in N where y is unique (that is, a = b implies g(a) = g(b)). However, g(0.999...) is not defined by the definition of your function since infinity is not "in" N.Originally Posted by Guitarist
Also, consider g(pi/4). This is an irrational number in I. But, there exists no elements in N that pi/4 maps to in your function. Hence, g is not a function.
A function like h:Q<sub>t</sub>>N, where Q<sub>t</sub> is the set of terminating rationals, is defined by your rules. Clearly, 0.999... is not in Q<sub>t</sub>. The function h is not injective, however (if we follow the same rules as g) as h(1/2) = h(1/10) = 2.
That's the fella  a sweetie for Absane. I said halfway down that the function g is not injective, and then used injectivity to "prove" 0.99... ≠ 1!
Hey guitarist
My issue with your function is that it is not well defined in the first place  decimal expansions are equivalence classes.
I'm not quite sure what you're saying here. My jocular function g: R → Z+ I defined as an enumeration of the minimum number of decimal places required to find an exact value for some a in R. For sure, g induces a partition on R, elements of which are of the form [a], [c], ... these being the elements of R whose exact decimal expansions coincide.
This is nice; to be outrageous, let's call this partition = {[a<sub>i</sub>]} as the set part(R). Then evidently, part(R) = Z+, up to isomorphism, as per the definition of g.
g([a]) = x in Z+ will be well defined iff g is indifferent to the choice of which member of [a] it selects; clearly this is the case here. Let [a] = {1/3, 1/6, 1/7..}, then g([a]) = ∞ in Z+. This is in fact, a perfect example of a well defined surjection.
River_rat, this is not directed at you, I know that you know this stuff, but I'm just finishing a thought train.
Let me put it another way: any function f: X → Y is well defined on X iff for each x in X there is at most one f(x) = y in Y and for some f(x) = y in Y, there is at least one x in X. In other words, f isn't spraying itself all over Y, neither is it ignoring any element of X.
Hey guitarist
What i meant was that a decimal expansion is actually an equivalence class of sequences.
Let (a<sub>n</sub>) and (b<sub>n</sub>) be two sequences in {0, 1, ..., 9}. Define ~ by saying that (a<sub>n</sub>) ~ (b<sub>n</sub>) iff Sum(10<sup>n</sup>a<sub>n</sub>) = Sum(10<sup>n</sup>b<sub>n</sub>). Then each equivalence class defines a unique real number [(a<sub>n</sub>)] in [0, 1] which we can choose to write as [0.a<sub>1</sub>a<sub>2</sub>a<sub>3</sub>...]
The problem is that [0.5000000...] ~ [0.4999999999...] and your function thus gives f(0.5) = 1 = f(0.499999...) = infinity (I dont recall reading anything about a minimal expansion in your first post  which is where the problem arises).
Look, my post was intended as a joke, whereby I set out to fallaciously prove that 0.99.. ≠ 1. I thought that, unlike socalled normal jokes, this one would pass through Customs.
But, feeling pedantic, and defending my joke (how sad is that?) let me say I am very unhappy with this:
Unless you and I are using the square brackets differently (unwise in this context), [0.500...] ~ [0.499...] is meaningless. For sure, 0.499.. ~ 0.500... (where the ~ is the vanilla equality, that is if we are sane).Originally Posted by river_rat
But this, by the definition of equivalence, places 0.500... and 0.499... in the same equivalence class. For sure, we may conjecture that a ∈ [a] and a ∈ [c], then by the partition that induces the equivalence relation we must conclude that [a] = [c].
Unless I've gone totally bonkers (not impossible), an equivalence relation between equivalence classes has no meaning  they're defined as the disjoint classes of equivalent members, so by definition the classes can not be equivalent.
I deserve a flogging for that
[0.49999999....] = [0.50000000] using the convention i mentioned earlier in my first post and thus they represent the same rational number etc etc etc
0.9999...=4! dumbasses..d
I got into yet another debate about 0.999... = 1 on another forum. I don't know why I do this to myself... but every time I do it, I can I can feel my blood pressure rising.
I'll give an example of what I had to deal with:
And his signature:See all above arguments. I can find an infinite number of .999... that have more 9's than just .999... does. It's all using the properties of infinity.
Really... what do I say?Georgetown University
Biology and Math Double Major
Giardia lamblia researcher =P
Earlier he also said this:
Again, what do I make of all this?All i said was that .999 to infinity is not the same as another .9999 to infinity because one infinity can have more 9's on the end of it than another. This is something that is known by everyone. No two infinites are alike. infinity != infinity. I obviously didn't bother reading everything you wrote because it wasn't necessary (I too have been studying math and have a 4.0 math gpa in all my math classes at georgetown: calc III, number theory, ordinary linear differential equations, nonlinear differential equations, partial differential equations, linear algebra, and a credit from another university in abstract algebra), all I said was that since .999... does not exactly = 1, you can find another .999 that is inbetween the first one and 1. We are splitting hairs here. I said what you guys did was fine in applied math, but if you want to get really technical about it the only thing that equals 1 is 1. hence my asymptote argument. You can jump over one a few times, but you can't go through it and you can't land on it.
Well for a math major this guy is blowing a lot of smoke.
There are two options here absane  get technical and state that the infinity in question is omega  the first infinite ordinal  as you are dealing with sequences and thus need everything to be countable. So his statement about infinity not equalling infinity is balderdash etc
Secondly ignore the guy  i smell a troll in need of attention and only a troll has to use his credentials as leverage in an argument.
That's what I was thinking... I even said to him at one point "I question your ability to do mathematics."Originally Posted by river_rat
After talking to him for a while, I found out he was talking about something different than I was. Further, his choice of words was very poor. But he also mentions that he's more interested in biochemistry than mathematics...There are two options here absane  get technical and state that the infinity in question is omega  the first infinite ordinal  as you are dealing with sequences and thus need everything to be countable. So his statement about infinity not equalling infinity is balderdash etc
Secondly ignore the guy  i smell a troll in need of attention and only a troll has to use his credentials as leverage in an argument.
However, I still cannot figure out what he ever meant by "infinity + 1 > infinity." He was talking about how the lim(x>oo) x^4 is BIGGER than lim(x>oo) x^2. So, gave this example:
I was referring to his claim that 0.999... > 0.99.. because the former has "more 9s." I think he is part of the boat that thinks 0.000...001 exists where there is an infinite number of zeros.lim(n>oo) [Sum(9/10^i)  Sum(999/1000^i)] = 0 where the sum is carried from i = 1 to n. If your assessment that the latter is bigger than the former, how is it that the limit is zero?
I thought I was doing just great in the discussion until I this guy showed up. He was starting to convince everyone else in the discussion that there is actually debate among mathematicians about 0.999... = 1.
My head is going to explode.
But, I am reminded of a quote:
"Against stupidity the gods themselves contend in vain."  Friedrich von Schiller
Ok here we are about to enter murky waters so hold onto your hat. In a strict sense infinity + 1 > infinity if you interperate infinity to be any one of the host of infinite ordinals numbers that are around. But here the order relation has nothing to do with the actual size of the number but with its order type (so this almost follows from the definitions of the ordering on the ordinals). I can go into the actual theory if you want to use a rather large club in this arguementOriginally Posted by Absane
If you need a hand in the debate pm me the address  else ask away here.
Thank you for that bit. The guy I was talking about has left the debate, but I am still stuck with 2 members that continuously call mathematicians like Euler wrong (I brought him up because he proved 0.999... = 1). I got to say... even though my blood boils every time I see they have replied to something I have said, I am learning quite a bit.Originally Posted by river_rat
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