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06-02-2004, 09:56 PM
Ok, so Cecil did a pretty good job explaining why 0.999~=1, now I'm here to explain why it doesn't

First, I'm gonna give you a little proof :)

x=0.999~
If you multiply by 10
10x=9.999~
Subtract x from each...
9x=9 (because x=0.999~)
divide both by 9 to find x
x=1

However, there is one problem with that proof, and that is it treats infinity as if it is not an actual number. What we must remember is that infinity if a number, just a number we cannot express.

When we multiplied both sides by ten, it was 9.999continuing infinitely-1. In an equation, infinity must remain the same number, so multiplying by ten would remove a decimal at the end just like 0.99*10=9.9, NOT 9.99. Therefore, this argument is false.

Regarding fractions, 1/3 DOES NOT=0.333~. That is the best way we can put in it non-fractional form. The only true equal of 1/3 is 1/3. Saying 1/3=0.33~ is like saying 22/7=pi. It's pretty damn close, but not quite right (perhaps the most accurate way of expressing 1/3 would be to say 0.333~+0.000~1, meaning that it ends in a four.)

Anyway I hope this has been helpful, and the original column can be found here (http://www.straightdope.com/columns/030711.html) . Now I just have to finish my explanation of Stairway to Heaven, last time the forum deleted it when I tried to post...

Call me Frank
06-02-2004, 10:18 PM
I'm a believer. I have no factual or mathematical basis on which to base my belief, but I just can't accept that .999....=1. However, I do understand why the non-believers say .333...=1/3. IF .999... does equal one, then .999 divided by 3 would be equal to 1/3. But that is, of course, contingent upon .999... being equal to 1.000...

Moe
06-02-2004, 11:07 PM
oh boy, here we go again.

not that I'm knocking your OP. Just that this has been discussed ad infinitum (snicker) here and looks like it will continue to be. Personally, I always enjoy them so please, carry on.

Hyperelastic
06-02-2004, 11:09 PM
You are confusing equality with identity. Look at it this way. If x=0.999..., then no matter how small a quantity you name, I can show that the difference between x and 1 is smaller still. Thus there is no limit to the closeness of 0.999... and 1. That meets any reasonable definition of equality you care to propose.

Mbossa
06-02-2004, 11:26 PM
Infinity is not a number. (http://www.galactic-guide.com/articles/8R69.html)
Infinity is not a number. (http://members.cox.net/mathmistakes/infinity.htm)
Infinity is not a number. (http://mathforum.org/dr.math/faq/faq.large.numbers.html)

If you disagree, could you please expand on why you believe infinity to be a number?

perhaps the most accurate way of expressing 1/3 would be to say 0.333~+0.000~1, meaning that it ends in a four.

No. The decimal expansion of 1/3 does not end in a four. The decimal expansion of 1/3 does not end. Assuming that the "~" means an unterminated string of zeros, "0.000~1" is simply nonsensical. How can you tack a 1 on the end of something that doesn't even have an end?

You're correct in saying that 22/7 does not equal pi. In fact, I'd argue that 22/7 is nowhere near pi (althought this is purely subjective). We can see that by performing the decimal expansion to a few decimal places (where "~=" means "approximately equal" - not a precise concept, admittedly, but adequate for our purposes):

pi ~= 3. 1416
22/7 ~= 3.1429

However, you're incorrect in saying that 0.333... does not equal 1/3 (assuming that "..." means something along the lines of "an unterminated string of 3s"). Saying "1/3 = 0.333..." is the same as saying "the sum of 3/10n, where n takes the value of each positive integer, is 1/3", which can be proven to be true. I can't be bothered proving it, but I'm certain someone else can.

Derleth
06-03-2004, 02:21 AM
0.333~ is a limit. So is 1.000~ and 0.5000~ and every other number in the set of the reals. It is not 'infinity' or 'infinite' or any other incomprehensible babble you come up with. It is a limit, and therefore it works like every other real in the set of all reals.

0.999~ is a limit as well, and so it works the same way as everything else. It obeys all the same rules without exception. One of the rules of the set of the reals is that for two real numbers to be unequal, there must exist a real number strictly between them. (That is, it must be strictly less than one of the two and strictly greater than one of the two and unequal to both of the two.) Another rule is that you can only obtain the additive identity (0.000~) from x-y if x is equal to y. So, what real number lies between 0.999~ and 1.000~?

1.000~
-0.999~
----------
0.000~

Well, look at that: 0.000~ is the result of the subtraction, so 0.999~ must equal 1.000~.

This is not surprising. 0.999~ can be seen as a convergent series, and a convergent series is equal to its limit. The limit of summing 9e-n as n goes from 1 to infinity is 1.000~, so 0.999~ must equal 1.000~.

This equality is not approximate. There are no approximations in mathematics.

Mathochist
06-03-2004, 03:03 AM
0.333~ is a limit. So is 1.000~ and 0.5000~ and every other number in the set of the reals.

Maybe they need the definition of R.

The set of real numbers, R, is the unique complete extension field of the rational numbers. This means it's constructed from the set Q as a subquotient of all sequences of elements of Q.

Given a sequence {xn}n in N, if for every e>0 there exists a k(e) such that for all n,m>k, |xn-xm|<e we call the ssequence Cauchy. Roughly, a sequence is Cauchy if all the numbers in the sequence get close together as n increases. Take the collection of Cauchy sequences of elements of Q.

Now, given two Cauchy sequences, {xn}n in N and {yn}n in N, interleave them to define {zn}n in N. That is, z2k=xk and z2k+1=yk. If z is also Cauchy, we consider the sequences x and y to be equivalent. The set of Cauchy sequences of elements of Q modulo this equivalence relation is R. That is: every real number is instantiated as an equivalence class of sequences of rational numbers, such as the sequence of its decimal approximations.

That there is a unique field structure on R extending that of Q is left as an exercise for the reader. ;)

Xiphos
06-03-2004, 05:45 AM
I think I speak for everyone when I say... huh

SCSimmons
06-03-2004, 07:53 AM
Regarding fractions, 1/3 DOES NOT=0.333~. That is the best way we can put in it non-fractional form. The only true equal of 1/3 is 1/3. Saying 1/3=0.33~ is like saying 22/7=pi.

No, it's not. It's like saying that 4/1 - 4/3 + 4/5 - 4/7 + 4/9 ... (and so on endlessly) equals pi. Which it does, as does .33333 .... = 1/3.

zut
06-03-2004, 09:21 AM
As a public service:
if .999,,,,=1 does .999,,,,+.999,,,= 2 or 1.999,,,8? (http://boards.straightdope.com/sdmb/showthread.php?t=256497)
Why doesn't .9999~ = 1? (http://boards.straightdope.com/sdmb/showthread.php?t=187902)
And probably some other I've missed. Go thou, and read.

ultrafilter
06-03-2004, 11:32 AM
In any Archimedean field, .9... = 1. The reals are an Archimedean field, so in the reals, .9... = 1. If you don't understand what that means, don't try to debate it.

Giles
06-03-2004, 11:35 AM
The only true equal of 1/3 is 1/3
So, 2/6 is a different number, then? And 3/9 must be something else, and 4/12 a fourth number, and so on?

kabbes
06-03-2004, 11:45 AM
AAAAARGH!

So what is the decimal representation of 1/3 in base 12 then? And what do three of these sum to? So why do you think that the same number in different bases magically becomes different?

This has been done so MANY times before that it's like the "drive on the parkway" question. If you're serious about debating it, go find each and every argument in those threads and THEN come back to us with arguments against them.

SCSimmons
06-03-2004, 12:15 PM
I think I speak for everyone when I say... huh

Mathochist just repeated the first two sentences of Derleth's post immediately preceding, translated into 'mathese'. The reason he did this can be inferred from his user name. :D

C K Dexter Haven
06-03-2004, 12:16 PM
The problem here is a difference between the theoretical and the practical.

In the theoretical realm, the mathematician has no problem dealing with infinite decimals. In the practical realm, of course, there's no such thing. After a couple of dozen decimal points, you're way past a point where anyone can measure or verify or validate anything, and the non-mathematical mind can't grasp the idea of "infinite decimal."

So, the mathematical mind thinks on a theoretic plane and obviously .9999... = 1 as .333... = 1/3. The practical mind thinks that .9999999999[a zillion more 9's]999 stops and is obviously different from 1, as .33333333333333333333 is obviously different from 1/3.

sturmhauke
06-03-2004, 12:40 PM
I'm a believer.

Stop right there. This isn't religion, it's math. There is no room for belief here, only logic.

Now then, here's a proof that 0.9... = 1

Let us assume that 0.9... != (not equal to) 1. But there are an infinite number of 9s, so it must be infinitely close to 1; that is, there is no number x such that

0.9... < x < 1

Now let x = square root (0.9...)

Because 0 < 0.9... < 1, therefore 0.9... < x < 1

But this contradicts the statement that there is no x such that 0.9... < x < 1, which folllows directly from the assumption that 0.9... != 1

Therefore, 0.9... = 1

Giles
06-03-2004, 01:16 PM
There is no way to understand what 0.999... means unless you understand limits of sequences. It is defined as the limit of the sequence:
{0.9, 0.99, 0.999, 0.9999, 0.99999, ... }, adding an extra 9 at the end of each item in the sequence to get the next member of the sequence.

The limit of this sequence is 1, because you can get as close as you like to 1 by going far enough along the sequence, even though no member of the sequence is equal to 1.

(Similarly, the limit of the sequence:
{1/2, 1/4, 1/8, 1/16, 1/32, ...}
is 0, even though no member of the sequence is equal to 0).

Of course, many sequences do not have limits: examples of sequences without a limit are:
{1, 2, 3, 4, 5, ...}
{0, 1, 0, 1, 0, 1, ...}

It's fairly natural to assume that you can do the usual sort of arithmetic in sequences, e.g.:
0.333... + 0.333... + 0.333... = 0.999...
but because we are talking about infinite sequences here, you need to do some pretty rigorous mathematics to show that you can do operations like this. However, you can only do it with sequences that have limits -- but since these infinite decimal fractions always have a limit that's not a problem

ultrafilter
06-03-2004, 02:56 PM

Exapno Mapcase
06-03-2004, 07:50 PM

All the Charter Members have seen the 0.999... does not equal 1 threads ad nauseam. So only new Members and Guests are posting this nonsense.

There already is an Questions we refuse to answer (http://www.straightdope.com/faq/lamequestions.html) page in the FAQ. And it sorta works. I mean, we don't see many people ask these questions anymore.

So, as part of the registration agreement process, we force people signing up to go through a page called Questions Already Answered Too Damn Many Times. And they have to check each box, stating that they will not ask the third word that ends in "gry" or... that 0.999 is not equal to 1. And there will be a link to the relevant threads, which will be locked so that they cannot post to them.

Think of it! We can beat ignorance to a pulp before it appears!

I may cry.

06-03-2004, 09:58 PM
Ok Ok, I retract my statement of 1/3=whatever i posted. I did the math and it makes no sense. I DO NOT however retract my statement 0.999~/=1. Anyway keep up the discussion, even it is going way over my head....

06-03-2004, 10:01 PM
Ahh damn just read exapno's pos... and I think that's a great idea. I had no idea this had been posted b4 (although I should have guessed).

Oh and also, please allow guests to edit posts, I'm planning on subscribing anyway and it just ends up making ppl double post like this.

ultrafilter
06-03-2004, 10:20 PM
I DO NOT however retract my statement 0.999~/=1. Anyway keep up the discussion, even it is going way over my head....

So you've accepted that .3... = 1/3, right?

I think you'll agree that .3... + .3... + .3... = .9..., and that 1/3 + 1/3 + 1/3 = 1. Doesn't that imply that .9... = 1?

Derleth
06-04-2004, 01:39 AM
Oh and also, please allow guests to edit postsMan, not even Charter Members can edit posts. And the mods think this is a feture.

FWIW, I agree: If people were allowed to edit posts, you'd get jokers posting inflammatory nonsense, garnering huge amounts of follow-ups to what was originally written, and then altering their post to make everyone else look stupid. A five-minute editing window might allay this, but everyone has access to the Preview button.

So, do you now accept that 0.999~ = 1.000~?

Derleth
06-04-2004, 01:54 AM
The limit of this sequence [0.999~] is 1, because you can get as close as you like to 1 by going far enough along the sequence, even though no member of the sequence is equal to 1.Careful: Statements like this have been confusing in the past. It's better to say that if the sequence has a limit, the sequence as a whole is equal to the limit. That the sequence is infinite isn't relevant, as summing a sequence isn't a process in the usual sense.

(Similarly, the limit of the sequence:
{1/2, 1/4, 1/8, 1/16, 1/32, ...}
is 0, even though no member of the sequence is equal to 0).How does this work? I have an intuitive grasp of limits and other basic calculus, but I don't get how this sequence can sum to zero.

Mbossa
06-04-2004, 08:00 AM
(Similarly, the limit of the sequence:
{1/2, 1/4, 1/8, 1/16, 1/32, ...} is 0, even though no member of the sequence is equal to 0).How does this work? I have an intuitive grasp of limits and other basic calculus, but I don't get how this sequence can sum to zero.
The sequence doesn't sum to zero - it approaches zero. If I'm not mistaken, the sequence sums to one.

Giles
06-04-2004, 08:00 AM
Derleth, I think you are confusing "sequence" and "series". (It's easy to do, and I just checked in a mathematics dictionary to make sure I got it right.) When you are tralking about sums, you are talking about a series. A series is the sum of the terms given in a sequence. For example, the sequence that I gave:
{1/2, 1/4, 1/8, 1/16. 1/32, ... }
can be treated as a series, where the partial sums (the sums of the first n terms) are the sequence:
{1/2, 3/4, 7/8, 15/16, 31/32, ... }
And, in this case, it is a convergent series, with limit 1.

SCSimmons
06-04-2004, 08:15 AM
Right. And the value of .99999 ... is the limit of the series {.9, .99, .999, .9999 ...}, which is 1. And Skadi, if you want, I'll show you the formal epsilon-delta proof of that; but if the discussion is going over your head already, you may not want to see it. (Though it's just college freshman calculus stuff, so if you're planning on getting to that educational level, if may be worth your while.) (Actually, a college math prof here posted that they'd removed that from the coverage in that course at his school, which leaves me flabbergasted. Teaching calculus without teaching the basic theoretical underpinnings of limits???)

carterba
06-04-2004, 09:44 AM
If .999... != 1, then there is some real number y such that .999... < y < 1. What is y?

Mathochist
06-04-2004, 10:06 AM
Right. And the value of .99999 ... is the limit of the series {.9, .99, .999, .9999 ...}, which is 1. And Skadi, if you want, I'll show you the formal epsilon-delta proof of that; but if the discussion is going over your head already, you may not want to see it. (Though it's just college freshman calculus stuff, so if you're planning on getting to that educational level, if may be worth your while.) (Actually, a college math prof here posted that they'd removed that from the coverage in that course at his school, which leaves me flabbergasted. Teaching calculus without teaching the basic theoretical underpinnings of limits???)

Indeed, which floored me when I learned I wasn't supposed to teach it in my Multivariate courses, specially since they hadn't seen it in one variable.

Of course, all this means is that I circulate it as samizdat.

ultrafilter
06-04-2004, 10:51 AM
The sequence doesn't sum to zero - it approaches zero. If I'm not mistaken, the sequence sums to one.

The sum is 1, and the product is 0. How weird is that?

ultrafilter
06-04-2004, 11:05 AM
The sequence doesn't sum to zero - it approaches zero. If I'm not mistaken, the sequence sums to one.

The sum is 1, and the product is 0. How weird is that?

Lord Ashtar
06-04-2004, 11:18 AM
I was discussing this with my dad and brother last night (my brother and I saying that 0.999...=1 and my dad saying they aren't equal). My dad had an interesting observation that I couldn't answer, but I thought someone else here could.

If two number are equal, you should be able to divide one by the other and get 1. So he did some long division for 1 divided by 0.999... and ended up with a remainder. He reasons that even though the answer will be 1.000... the fact that you have a remainder shows they are not equal.

Any takers?

ultrafilter
06-04-2004, 11:21 AM
I was discussing this with my dad and brother last night (my brother and I saying that 0.999...=1 and my dad saying they aren't equal). My dad had an interesting observation that I couldn't answer, but I thought someone else here could.

If two number are equal, you should be able to divide one by the other and get 1. So he did some long division for 1 divided by 0.999... and ended up with a remainder. He reasons that even though the answer will be 1.000... the fact that you have a remainder shows they are not equal.

Any takers?

You can't divide by an infinite number of digits--you need to use a finite number.

Giles
06-04-2004, 11:38 AM
I was discussing this with my dad and brother last night (my brother and I saying that 0.999...=1 and my dad saying they aren't equal). My dad had an interesting observation that I couldn't answer, but I thought someone else here could.

If two number are equal, you should be able to divide one by the other and get 1. So he did some long division for 1 divided by 0.999... and ended up with a remainder. He reasons that even though the answer will be 1.000... the fact that you have a remainder shows they are not equal.

Any takers?

What algorithm did he use to divide 1 by 0.999... ?

Let me suggest one, given that 0.999... represents the limit of an infinite sequence {0.9, 0.99, 0.999, ... }:

Given an infinite sequence {S1, S2, S3, ...} with a limit other than 0, the result of dividing 1 by the limit of that sequence is the limit of the sequence {1/S1, 1/S2, 1/S3, ... }.

That gives you the sequence of recurring decimal fractions:
1/0.9 = 1.1111 ...
1/0.99 = 1.01010101 ...
1/0.999 = 1.001001001001 ...
1/0.9999 = 1.0001000100010001 ...

It's pretty clear that the limit of that sequence is 1, so 1 divided by 0.999 ... is equal to 1. QED

John W. Kennedy
06-04-2004, 11:51 AM
I was discussing this with my dad and brother last night (my brother and I saying that 0.999...=1 and my dad saying they aren't equal). My dad had an interesting observation that I couldn't answer, but I thought someone else here could.

If two number are equal, you should be able to divide one by the other and get 1. So he did some long division for 1 divided by 0.999... and ended up with a remainder. He reasons that even though the answer will be 1.000... the fact that you have a remainder shows they are not equal.

Any takers?It's the same old problem, not understanding the difference between an infinite number of nines and a very large number of nines. The "remainder" is, by definition, smaller than any positive number, and is therefore zero.

Mathochist
06-04-2004, 12:39 PM
The sum is 1, and the product is 0. How weird is that?

Not very. The sequence of partial products of any summable sequence of real numbers is 0. For an infinite products to converge to a nonzero limit the sequence of factors must converge to 1.

Lord Ashtar
06-04-2004, 01:06 PM
You can't divide by an infinite number of digits--you need to use a finite number.

Figured as much. My dad's main argument was that if you can't write it down, you can't prove it for certain. I asked him to show me a case anywhere where 3+6!=9 and he accused me of asking him to prove a negative.

Sigh...

Lord Ashtar
06-04-2004, 01:07 PM
Of course, that should read as 3 + 6 != 9 rather than 3 + 6! = 9. :smack:

Giles
06-04-2004, 01:31 PM
My dad's main argument was that if you can't write it down, you can't prove it for certain.
Some mathematicians would share your father's belief -- they don't accept anything which depends on infinity. But for them, the expression 0.999... would be meaningless, because it depends on infinity and limits and such stuff -- not different from 1, just meaningless. (I suspect that they have trouble with other stuff, like the square root of 2 and pi, as well, because working with these involves infinity as well -- not accepting infinity means not accepting a large part of mathematics).

Mathochist
06-04-2004, 01:43 PM
Some mathematicians would share your father's belief -- they don't accept anything which depends on infinity. But for them, the expression 0.999... would be meaningless, because it depends on infinity and limits and such stuff -- not different from 1, just meaningless. (I suspect that they have trouble with other stuff, like the square root of 2 and pi, as well, because working with these involves infinity as well -- not accepting infinity means not accepting a large part of mathematics).

Well, it's subtler than that. The so-called "constructivists" insist that everything be "constructible". They're cool with the sequence of partial sums to the extent that any given element of it can be constructed, but they have misgivings about the whole sequence. Constructivists have their own version of the real numbers, which works sufficiently like everyone else's that they don't disagree until you get to some really weird things like the Banach-Tarski paradox.

The flip side of this is that with the rise of structuralism in the philosophy of mathematics and the rise of topos theory in categorical mathematics, there is a very real sense in which constructivists aren't doing the same thing at all. "Standard" mathematics is performed in one topos (commonly called Set) while constructivist mathematics is performed in another. One of the major differences, for instance, is that the logical structure of the constructivist topos is not a Boolean algebra, but a Heyting algebra. This means that the law of the excluded middle is rejected. P^!P is not trivially true, so !!P is not equivalent to P, although !!!P is equivalent to !P. For further reading, you could try to read Saunders Mac Lane and Ieke Moerdijk's Sheaves in Geometry and Logic: A First Introduction to Topos Theory (http://www.amazon.com/exec/obidos/tg/detail/-/0387977104/qid=1086374537/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/103-7538132-9083003?v=glance&s=books&n=507846), but it might be simpler just to get a Ph.D. in math at a category-friendly school.

Derleth
06-04-2004, 05:05 PM
Derleth, I think you are confusing "sequence" and "series". (It's easy to do, and I just checked in a mathematics dictionary to make sure I got it right.) When you are tralking about sums, you are talking about a series. A series is the sum of the terms given in a sequence. For example, the sequence that I gave:
{1/2, 1/4, 1/8, 1/16. 1/32, ... }
can be treated as a series, where the partial sums (the sums of the first n terms) are the sequence:
{1/2, 3/4, 7/8, 15/16, 31/32, ... }
And, in this case, it is a convergent series, with limit 1.This makes sense. Thanks.

Interestingly, it was obvious that it must converge to 1: Representing the fractions as binary fractions gives you:
{ 0.1, 0.01, 0.001, 0.0001, 0.00001, ... }
or, as a sequence of partial sums:
{ 0.1, 0.11, 0.111, 0.1111, 0.11111, ... }
which is very similar to 0.999~ in base 10.

tim314
06-05-2004, 02:38 PM
The sum is 1, and the product is 0. How weird is that?

1 + 0 = 1
1 * 0 = 0

The sum is one and the product is zero. I'm not sure why you consider this any weirder when it happens for a sequence than when it happens for two numbers.

Everyone seems to have already thoroughly explained why 0.999... = 1. (But I can't resist adding my 2 cents). I think the key is in understanding that a repeating decimal is defined to be the limit of a sequence.

However, if I could try to put the flaw in the original poster's reasoning into plain English, I would say this:
He assumes that infinity minus one is not equal to infinity. (So if there were infinity nines to start with, there would be less than infinity in the end.) But this is false, since infinity minus one is still infinity. If you consider that impossible for "numbers", then it simply means infinity isn't a number.

This argument isn't really relevant, though, because really 0.999... = 1 because that's what it is defined to be. We define repeating decimals to be the limit of sequences, and in this case the corresponding sequence is one whose limit is one. The mistake some people make is in thinking that it is not necessary to have a separate definition for repeating decimals -- that they instead can be defined in the same way as terminating decimals. In other words, people think that because 0.237 has a meaning, then the same definition can be used to define 0.3333... But if you think about how you define 0.237, this isn't the case. To determine what fraction corresponds to 0.237, you count the number of digits after the decimal point, call that number n, and then define this decimal expression as being equal to the fraction whose numerator is the integer that appeared to the right of the decimal point (in this case, 237), and whose denominator is 10^n. Thus, in this case the fraction is 237/1000. My point is that this procedure fails for non-terminating decimals, because you can't count the number of digits after the decimal place. So a new definition is required to give these expressions meaning, and the definition that mathematicians have agreed to use automatically makes 0.999... = 1.

Mathochist
06-05-2004, 02:56 PM
He assumes that infinity minus one is not equal to infinity. (So if there were infinity nines to start with, there would be less than infinity in the end.) But this is false, since infinity minus one is still infinity. If you consider that impossible for "numbers", then it simply means infinity isn't a number.

Well, as covered in an earlier thread it depends what you mean by "infinity". In Conway's numbers, the appropriate notion of "infinity" is the first transfinite number "omega", and omega-1 is actually less than omega. On the other hand, 1-1/omega is "infinitesimally" less than 1 so the "infinitesimally less than" argument might seem to go through. Even so, Conways numbers are a proper ordered extension field of R and so have .999~ = 1.

I Love Me, Vol. I
06-05-2004, 03:31 PM
One question: What is

.999~ + .111~ ?

ultrafilter
06-05-2004, 04:24 PM
One question: What is

.999~ + .111~ ?

1.111~, or 10/9.

RM Mentock
06-06-2004, 10:37 PM
Ok Ok, I retract my statement of 1/3=whatever i posted. I did the math and it makes no sense.
I had to go back to the OP to check, but you said 1/3 = 1/3