I was just wondering
Yes, if you mean .999...
My views are slightly different here.
If you consider 0.999999... you must specify what metric space you are working in. If you have a complete one (likeR ), 0.9999... is simply a number in R. Although it is arbitrarily close to one, it does not equal one.
I guess it really depends on your definition of equality. If you take it as x-y<E => x=y, but I think my analysis professor would shoot me if I were to write it up on the blackboard next week.
Remember that .999... represents an infinite number of 9s not an increasing number but an infinite number. Of course since nothing else was specified I am speaking of the real number system.
Can you specify the number between .999... and 1?
Did you make an effort to look at my proof?
I did, and it relied on the statement that if you can approach something by less then any epsilon, it equals it.
But any analysist will tell you that this is only indeed true for a limit, and not for any quantity.
For example, by that reasoning an real number equals a rational number, because every epsilon interval has infinitely many rational numbers in it.
Anyone have a third opinion?
not a rigor presentation, but quite interesting
1/9 = 0.111111......
2/9 = 0.222222......
8/9 = 0.888888......
9/9 = 0.999999......= 1
Let x = 0.99999...
[uni]ε > 0: 1 - ε < x < 1 + ε
The rest of the proof is by contradiction. Suppose x [x=] 1.
From the axioms of an ordered ring and the fact (1/2) exists in R, one can prove:
x < (1/2)(1 + x) < 1
Now, if we let ε = 1 - (1/2)(1 + x) > 0
Then we have x < 1 - ε. This contradicts Integral's proof that [uni]ε > 0: 1 - ε < x
Therefore, our assumption that x [x=] 1 was incorrect, and thus x = 1.
This method of proof predates limits, and is fully rigorous. It predates limits, and was essentially known to the ancient Greeks as the Method of Exhaustion.
The theorem explicitly stated is:
If x and y are real numbers, and:
[uni]ε > 0: y - ε < x < y + ε
Then we may conclude x = y.
and this is something that actually gets used in analysis; sometimes it is much easier to prove the hypotheses of the above theorem than to set up and prove the appropriate limit.
Incidentally, IMHO, everything is much more clear if we appeal to the actual formula for mapping decimals to reals:
0.9999.... = &Sigmai=1..∞ 9 * 10-i
And we know the sum of this infinite geometric series to be:
= (9 * 10-1) / (1 - 10-1) = 1
To me 10,000 equals 1 is I'm just considering the penny differences in price of buying 2 houses, I mean 9 and 10 are not the same number but if it's practical to simply use two 10's then same enough, incidentally this is probably why I'm not very good with math.
Yes, my opinion is that you had better go have a long talk with your analysis teacher.
If it is true that |x-y|< epsilon for ALL epsilon, then x= y.
Yes, it is true that, given any real x and positive epsilon, there exist an infinite number of rationals, y, such that |x-y|< epsilon.
However, if they exist a rational, y, such that |x-y|< epsilon for ALL positive epsilon, then x is rational and, in fact, x= y.
The DEFINITION of the "decimal expansion" of a real numbers, say
a.d1d2d3... is that it is the infinite sum a+ d1/10+ d2/100+ d3/1000+... which is, in turn, the limit of the sequence a, a+ d1/10,
a+ d1/10+ d2/100, ...
In particular, the DEFINITION of a decimal expansion such as
0.9999.... is 0+ 9/10+ 9/100+ 9/1000+ ... which is a geometric series and easily summable to 1.
0.999... = 1 in the standard definition of the real numbers.
I see the proofs and they make sense, but it just seems odd that an infinite number can be equal to a rational number like that. Oh well looks like its just something I'll have to get used to. Thanks guys.
I think that it depends on the way you look at the expressions.
0.999... = 1 - 1 / infinity
Integral, you are off course completely correct. I'm sorry, but I was confusing the matter with some other proof.
What I stated about a limit is true in general. But off course [rr] is a complete metric space, so off course the limit of 0.999... exist and the sum of the aforementioned series is 1.
I blame all that chemistry they make me take in the first year, it has a way of messing with your brain
Discussion closed I guess , 0,999.... DOES equal one.
as a physicsist, 0.999..... does equal 1 to all intents and purposes.
Some of my lecturers used to use some really bizarre approximations.
"look, that 4.3*10^-5 is about 10^-5"
"well, pi^2 is about 10..."
the list goes on
approximations make life worth living
To Dimitri Terryn: Yeah, chemistry will do that to you- especially if you inhale!
To jonnylane: As a mathematician, 0.999... IS 1 there is no "to all intents and purposes" about it, it's not an approximation.
I was providing a physical view on things.
Physicists make clumsy mathematicians and vice versa
What??? Mathematicians make clumsy physicists? How dare you!
Ooops, I didn't mean to knock that cyclotron over!
You should realize that infinite, repeating numbers such as 0.9999 are only representations of division in number set in a give base (such as base-10 that we are all so familiar with), so it should not be such a surprise that it ends up being equal to one.
For example, how do you get 0.333......? Divide 1 by 3. You keep continueing with the digits because the same remainder keeps coming up. However, in the base-3 number system, one third is just 0.1
oops, there goes that abacus.
Of course Newton was a mathematician, but he was also a physicist.
I just came back from my chemistry exam.
I think that mathematicians can be good physicists and vice-versa, if only there is proper motivation and interest.
My university still has the combined first year mathematics/physics, and I'm very happy they keep this up, even though there is pressure to make the physics program less "abstract"
I will say one thing though, physicists make clumsy chemists
.999 repeating is actually rounding 1/9, isn't it?
Notice I say rounding, exchanging numbers for a repeating fraction means you round that fraction, since you can't possibly right an infinite amount of 9's...so isn't the question does 1/9 = 1?
I just summed it up in stupid terms, but this, I think [hope ], this is the center of the question.
EDIT: I suppose you can all disregard this post, I have heard the argument for .999 repeating = 1 before but I used the wrong fraction. Bah, my example kinda works for the .999 repeating =1 argument as 9/9 would be .999 repeating.
1/9 = .111...
Where the 3 dots represent an infinite repetition of the preceding pattern.
so no .999 is not rounding 1/9,
.999 = 999/1000 these are equaly valid representations of the same point on the number line. the first is the decimal represention of the fraction, both are rational numbers.
when we say .999... = 1 this says that a decimal point followed by an infinite number of 9s is the same point on the number line as 1.
This is not an ever increasing number of nines but an INFINITE number.
I have the following (simple) proof, i don't you if you will like it (but i can tell that a school student will surely find this proof lot more comprehensive then yours).
Represent 0.999... as a serries.
0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...
Or (in a better way) :
0.999... = 0.9*10^0 + 0.9*10^-1 + 0.9*10^-2 + ...
It is obvious that this represents an infinite serries that has a known value, which can be calculated using the equation
S = A/(1-R) ----(1)
Where S is the sum, A is the first element (?) in the serries, and R is the ratio between any element and the one before it.
Applying on (1)
S = 0.9/(1-0.1)
A = 0.9/0.9 = 1
There is also this result (That some people consider as a proof).
x = 0.999...
10x = 9.999...
10x-x = 9.999... - 0.999...
9x = 9
x = 1
0.999... does not equal 1.
What is the first number after 1? Is it 1.1? How about 1.01? No? How about 1.0000000001? No? Is it 1+(1-0.999...) or 2-0.999...? This is the issue that all so-called 'proofs' neglect. Anytime infinities are used in math they will create an ambiguous relationship.
That doesn't make sense, because even if 0.999... =\= 1, then you STILL can't answer "What is the first number after 1?" so that is meaningless.
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