2=1

[Rh0ads]

I'm not sure if you've seen this before but here is a really funky equation:

b+b=b
2b = b
2b=1b
2b/b = 1b/b
2=1

I know there's something wrong with that, but you figure it out!

[crutchlb]

:-)

How can b+b=b ?

ie: 1 + 1 = 1

just not possible to begin with.

'b+b' could equal 'a' though.

[gravyboy]

No way can b+b = b!!!!

b+b=2b (or not to be, but that's a different story . . .)

thus tken to the extreme

b=b . . . pointless and we should be shot for even answering such needless hyperbole . . .

[Lior]

Hi
well...everybody knows that b+b<>b unless b=0
so lets take the only case where the first equoation is correct: b=0 :

b+b=b
2b=b
2b=1b
2=1 (Here you devided both sides by b, and b=0, you cannot devide by 0) !!!!

That's it.

[Starman]

You could try this:

let: a=b

multiply by a: a^2=ab

add(a^2-2ab):a^2 + (a^2-2ab) = ab+ (a^2-2ab)

simplify: 2(a^2 - ab) = a^2 - ab

divide by (a^2 -ab): 2(a^2 - ab))/(a^2 -ab) = (a^2 - ab)/(a^2 -ab)

Therefore: 2 = 1

[gravyboy]

On the basis that a^2=2ab then a^2-2ab=0 . . . .

So why add this?

AND


a^2 + (a^2-2ab) = ab+ (a^2-2ab)

does not simplify to

2(a^2 - ab) = a^2 - ab

It simplifies to

a^2-ab=0!

[Starman]

ah! you spotted it :7)

1. I think I can add what I like - even 0

2. Perhaps 'simplify' was not the right word to use

3. The actual failure is at the division line. As you correctly pointed out (a^2 -ab)=0, so the division is not allowed.

[marnitzg]

Prove that 0.9999(recuring) = 1

Here goes:
let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

Well....?

[Starman]

Yep, 0.9999(recur) = 1, I tried typing loads of 9's into Excel and it came out as 1.

:7)

[parksie]

Originally posted by marnitzg
Well....?
Since 0.9999 recurring is infinitely close to 1, then for all reasonable situations it is 1 ;)

[Fried Egg]

0.9999(Recurring) is NOT equal to 1.


Reasons:

1) You could substitute x = 0.6666(recurring) into the proof and get the same thing (or any other recurring number between 0 and 1).

2) The proof is not reversable. i.e. you cannot go from x = 1 to x = 0.9999(recurring).

[Starman]

So 0.6666 = 0.9999 ?

[Fried Egg]

the proposed proof true.

It must be flawed.

[Starman]

I think I've got it!

0.9999(recur) / 3 = 0.3333(recur) = 1/3

-----------------------------------------

Seriously, I did think that the problem with 0.9999 x 10 was that at the point where the parrallel lines meet there would be a 9 missing, but can someone please explain what I have done now ?

[marnitzg]

1) You could substitute x = 0.6666(recurring) into the proof and get the same thing (or any other recurring number between 0 and 1).


Substitute 0.66666(rec)
let x = 0.6666(recuring)
then 10x = 6.6666(recuring)
then 10x-x = 6.6666(recuring) - x
then 9x = 6
then x = 6/9

Which doesn't equal 1. In fact it equals 0.666(rec). Fancy that. Therefore your theory doesn't hold.

[marnitzg]

Originally posted by Starman
I think I've got it!

0.9999(recur) / 3 = 0.3333(recur) = 1/3

-----------------------------------------

Seriously, I did think that the problem with 0.9999 x 10 was that at the point where the parrallel lines meet there would be a 9 missing, but can someone please explain what I have done now ?

You've just given another proof that 0.999(rec) = 1
Its also another answer.
If 1/3 = 0.333(rec) then *3
1 = 0.999(rec)

[Guv]

Fried Egg: Substituting recurring .66666 in the Marntzg proof results in proving that recurring .666666 is equal to 2/3 (You end up with 9*X = 6). Nothing wrong with that.

I am not sure that reversability is required for a proof to be true. It is not obvious to me that you cannot do the reverse proof.

At any rate, proofs involving limits do not really prove statements like recurring .999999 = 1. They prove that substituting 1 for the recurring decimal does not lead to paradoxical or erroneous results.

To be very picky, mathematicians use statements like recurring .999999 approaches 1 as the number of nines increases without bound, which is not quite equivalent to saying that recurring .9999 equals 1.

[Fried Egg]

I stand corrected on my first point but I still think that the irrevsersability of the proof is an example of its falacy.

You cannot get from

9x = 9

to

10x-x = 9.9999(recuring) - x

when you've already established that x = 1.

Surely that's a flaw?

Fried Egg,
The first time I saw that recurring proof, I was pissed off too.

They don't say, "in the limit, it's equal; they say it's equal". Hard to swallow when it looks like there is a missing 9 when you multiply by 10.

Or that something funny is going on with converging/diverging series.

Bottom line: Whether this is true or false, it doesn't profit me. But it sure can waste time. Maybe there is profit in that.

[Guv]

VirtuallyVB: Do now worry about a missing nine. It is not missing. Dealing with infinite sets, infinitely long numbers, et cetera has a lot of pit falls.

For example: Consider the set of all positive integers and the set of all positive even integers. Each set has the same number of members!!! The definition of an infinite set is a set that can be put into a one to one correspondence with one of its subsets. If throwing away all the odd integers does not decrease the number of members, why worry about one lousy nine?

BTW: Mathematicians hedge a bit when they talk about limits. They may claim that using a limiting value (in certain situations) is valid and leads to no contracitions, without actually saying that the function is ever equal to its limit. Saying that a function is equal to its limit is a kind of mathematical slang used as shorthand for all the business of "approaches .... as ... grows without bound"

Fried Egg: I have never seen reversability as a requirement for the validity of a proof. It just does not seem necessary.

Surely I can prove that a given number is the root of a 100th order polynomial, but the polynomial cannot be reconstructed from that single root without knowing the other 99 roots.

e is the limit of (1 + 1 / N)^N as N grows without bound. Given 2.71828..., can you reconstruct the expression from which is it derived?

It seems to me that there must be provable statements which do not contain enough information to reverse the proof.

[simonm]

The real issue is this:

Does 1 - 0.9999... = 0

Or to put it another way,

Is there anything left when you take away 0.9999... from 1.

You might say that 0.9999... is infintely close to 1 and therefore is 1, or you might say there is an infintiely small quantity left, that nethertheless, is greater than zero.

[HarryW]

I'd say 1 - 0.999999... = 0.00000000...

which is 0

[simonm]

Is there such a number as 0.000...01 ?

If there is, then 0.999... is not equal to 1.

If there isn't, then (perhaps) it is.

[Starman]

Maybe it's a problem with the decimal system. I can appreciate that 1/3 can never be properly represented in decimal, nor can 2/3 (6/9).

Is there a fraction which would evaluate to 0.9999... ?

I tried 3/2 x 2/3 and that's how I came upon my earlier revelation.

[noble]

bottom line
2=2
1=1

and move on with your life :P

[marnitzg]

Originally posted by Guv
Fried Egg:

To be very picky, mathematicians use statements like recurring .999999 approaches 1 as the number of nines increases without bound, which is not quite equivalent to saying that recurring .9999 equals 1.

Actually the answer isn't a limit. .999... actually does equal 1. I have been doing some research and I found the exact same proof in my calculas book (Calculas: 1 and several variables eighth edition).

[Guv]

Marnitzg: I am surprised that the proof would appear in a calculus text. It is not a calculus-like proof. While it looks valid to me, I am not sure that it would be considered valid by a serious mathematician on one of his picky days.

I remember an entirely different type of proof from the calculus courses that I took many years ago. Recurring .99999 is actually a geometric series.9/10 + 9/100 + 9/1000 . . .

Sum = A * (1 - R^N) / (1 - R), where A is first term, R is ratio, & N is number of terms.

Recurring .9999 is a special case: Sum = (1 - R^N), since A = (1 - R)
As N grows without bound, R^N approaches zero, and the sum approaches one. The memory of this proof is the reason I viewed one as the limit of the recurring nines.

The above is a valid proof. I suspect that your Calculus text was not as rigorous as it should have been.

[marnitzg]

Well, I found the proof in the second chapter - limits and continuity! Before that proof they do a few examples on limits so maybe they want to show that .999 is not a limit?

[nickster]

Originally posted by Starman
You could try this:

>let: a=b

>multiply by a: a^2=ab

>add(a^2-2ab):a^2 + (a^2-2ab) = ab+ (a^2-2ab)

>simplify: 2(a^2 - ab) = a^2 - ab
a = b
Therefore, a^2 = a * a = a * b = ab
Hence, a^2 - ab = 0
and 2 * 0 = 1 * 0 = 0

>divide by (a^2 -ab): 2(a^2 - ab))/(a^2 -ab) = (a^2 - ab)/(a^2 -ab)

Division by zero is undefined. Above line is nonsense as (a^2 - ab) = 0.

> Therefore: 2 = 1

No it is not.

Regards,

Nick

[Starman]

Thanks Nick,
Actually I have already admitted to this in a post just before marnitzg offered 1 = 0.99999.

Can you shed any light on my present confusion of:

0.9999(recur) / 3 = 0.3333(recur) = 1/3


I hadn't bothered to type the last part of this before as I thought it was an obvious step but 3 x 1/3 = 1

so 0.9999(recur) = 1 (another way)

I do not believe that the two are equal, but feel there is a fault in my 'maths' .

[nickster]

Starman,

This is a notation thing, it boils down to the fact that one third cannot be expressed exactly as a decimal.

Whichever number system you use, you are always going to have to express approximated fractions - for example, I don't think that 0.1 can be expressed exactly in binary (I'm not sure, but there are some decimals that can't be, which induce floating point errors in computer processors).

0.9999 does not equal 1,
0.9999 recurring forever does = 1.

Think of it this way - 1/3 cannot be expressed exactly so can be approximated to 0.333 - the more 3's you add, the closer you get to 1/3. You only ever express 1/3 exactly if you go on forever. The same applies to 0.9999 - if you go on forever, you are expressing 3/3 = 1. Question is one of how you represent the number (in my opinion). I'm pretty sure that there is a number theory proof where you can apply limits to show this.

Regards, Nick

[simonm]

1/1 + 1/2 + 1/4 + 1/8 + 1/16 .... = 2

[noble]

i have a question......

is it necessarily correct to use an equal sign when
writing a series?

and i have to disagree with the point that 0.99999
recurring forever equals one. 0.9999 recurring forever
equals 0.9999 recurring forever and 1 equals 1.
0.9999 recurring forever "approaches one and for
simplicity can be represented as 1". In a math problem,
sometimes you will have a number subtracted by
another number and they will be so close to one that
you can just disregard it when multiplying or dividing
by it. By saying that the difference between those two
numbers is 'equivalent' to one, it will probably destroy
any chances of argument :P

ps, if i could draw the wavy equal signs i could rewrite
0.99999 recurring forever (place wavy equal sign here) 1

[nickster]

I still say that 0.9999 recurring forever is equal to one. I am not progressing it in terms of decimal places as it repeats an infinite number of times, so the complete number is not approaching anything as it is not changing.

The changing number of places is a conceptual representation to help people understand the transition from a finite digit-length number to an infinite digit-length one. The two are different as I tried to explain earlier.

Regards,

Nick

[Starman]

Thanks Nick,

That sounds a reasonable explanation.
I'm afraid that I still don't believe that 0.9999 recurring forever will quite be 1 (a bit hare and tortoise I suppose), but you have confirmed my fear of the decimal system. I will continue as long as I can in feet and inches and pray for a return to miles to the gallon, lbs and ounces and 20 shillings to the pound. If awkward for calculations these measurements at least teach that there is more than one way to measure a quantity, and show that there is a reason to learn your times tables.

[simonm]

If 1 = 0.9999...

What is the first finintely describable real number below 1 (or 0.9999...)?

Or is it, as I suspect, that the next finitely describable number below 1 is not nameable?

[Fried Egg]

Guv,

Some complex operations cannot easily be reversed but surely it is not unreasonable to expect a simple operation like subtraction to be reversable. If you subtract number x from number y to get a result z, you would reasonably expect to be able to work backwards to add y to z to get x.

Therefore, starting with x = 1, you cannot get 9 + x to equal 9.9999(rec) without the prior assumption that 1 = 0.9999(rec).

Simonm,

You cannot name the smallest real number below 1 for the same reasons you cannot name the largest number.

[noble]

nick:
The only reason I tend to
assume it is not equal to one is that sometimes in my
field, i deal with extremely small or extremely large
numbers that approach a value but can be not
necessarily 'rounded' in order to simplify an operation.
Consider a bridge that flexes an almost infinitestimal
amount to the human eye, can we say it bends nothing
and 10 years from now have a collapsed bridge?

Your point is very good and it's a good view to not
look at 0.99999(recur) as a changing number as it
simply infinite i'm simply firm in my belief that it is
'equivalent to' 1 and not 'equal to' 1. :P

[nickster]

noble,

Subtle difference is that with your bridges you are talking about finite changes. 'Almost infinitesimal' still implies quantity. This goes completely against the concept of infinity and infinitely small. If an electron has a collision with your bridge and transfers some of its momentum to it, there is still a finite amount of momentum acting on the bridge (which is greater than infinitely small). Whether it is insignificantly small is another matter.

If you express 0.9999 as a geometric progression (gp) from x=1 to infinity (which you can do as there are an infinite number of recurring places) then taking limits gives you 1.
I am saying that 0.999 recurring infinitely is the same as taking limits on the gp, as all that you are doing by expressing it as a gp is to fragment the number, but you need to repeat it to an infinite number of terms to get the whole number. When you take limits, you sum every component of the number. I don't think that we are going to find common ground here (except that we both agree that they are equivalent!)

Regards,

Nick

[nickster]

Starman,

Every base system has infinite fractions, they would just be attached to different fractions, that's all. For example, in base 12 (using 0123456789AB) 0.1 = 0.12497249... (goes on and on I think)

I personally prefer Denary, although I can see a lot of advantages to going over to Hex....

For scientific measurements, everything works quite nicely with metirc measurements. Fact of the matter is that people understand best pretty much what they learn first. I hated radians until I started doing more advanced physics/mathematics and they make stuff much easier there as you always get a few pies [or should that be pi's :o) ] that cancel out. I still translate mph to km/h and have little comprehension of kW in terms of car engine power (has to be bhp), although just for every other form of power it has to be in kW!

Best to stick to whichever you find easiest, it can always be converted if required.

Regards,

Nick

[Guv]

Noble has the right idea.0.9999 recurring forever equals 0.9999 recurring forever and 1 equals 1. 0.9999 recurring forever "approaches one and for simplicity can be represented as 1".The definition of limit does not claim that a function is equal to it limit. It merely states that the difference between the function and its limit can be shown to be smaller than any finite value.

You must be careful about setting a function equal to its limit. For example. Limit[ (1 + 1 / n)^n ] approaches e (2.71828...) as n grows without bound. Limit[ 1 + 1 / n ] approaches one.You cannot say that 1^n = e for infinitely large n, which is what you get if you set functions to their limits here and try to specify what is happening for infinite values. Mathematicians avoid statements about what happens at infinity.

Fried Egg: I have never heard that a proof must be in some sense reversible in order to be valid. Is there a text somewhere that makes this claim? How do you define the reverse of a proof? Such a definition would have to be on a case by case basis. Certainly a single definition would not cover all the possible methods of proof. It seem particularly difficult to define the reverse of a Reducto Ad Absurdum proof. What is the reverse of Wiles proof of Fermat's Last Theorem? What about disproving a statement by showing a counter example? Can such a disproof be reversed? I do not remember any book or text that showed a proof, and then went on to validate it by showing it in reverse.

I can prove that a particular number is not a prime by showing that it is the product of two particular numbers. 7 * 13 = 91, therefore 91 is not prime. What does it mean to reverse this proof? Can you determine the factors from the conclusion that 91 is not prime? The forward proof is simple, how do you do it backwards without using information from the forward proof? 91 / 7 = 13 is the only approach I can think of and that uses data from the forward proof. Godel proved that certain axiomatic systems are either inconsistent or incomplete. What would be the reverse of such a proof?

Most ordinary geometric proofs seem reversible. Given the Pythagorean relationship, I suppose you could prove the triangle to be a right triangle. Many other proofs are probably reversible. After all, if a forward going proof seems valid, you would hardly expect a reverse proof to be invalid. It seems reasonable to assume that most valid proofs are in some sense reversible. It does not seem reasonable or necessary to require reversibility for validity.

Others: The proof that recurring .9999 equals one seems reasonable, but does not seem valid to me in spite of what some calculus text might say. I do not question its validity due to lack of reversibility.

I think the proof is invalid because of the multiplication of recurring .999 by 10 and subsequent subtraction of the recurring nines. These operations seem reasonable, but they also seem undefined. It does not seem valid to do arithmetic on infinitely long decimal fractions. There must be some subtle unstated assumptions about such operations. Reasonable assumptions about transfinite and infinitesimal numbers are often incorrect.

The proof seems valid, especially since there is a valid proof showing that the limit of recurring .99999 is one (this proof does not rely on decimal notation). Without the other proof, I would be nervous about accepting the one involving arithmetic on infinitely long decimal numbers. I would worry about the possibility of such methods leading to erroneous results in another context.

BTW: In base 16, you can similarly prove that recurring .FFFFF equals one. Does this prove that recurring .99999 in decimal equals recurring .FFFFF in hex? For all finite examples, the two recurring fractions are unequal. Note that you can certainly prove that they approach the same limit (namely one).

[simonm]

Fried Egg,

You may have a point. Perhaps your reversal of the proof indicates that an infinitely small quantity is lost thereby allowing you to 'prove' 1 = 0.9999...
Probably in the step that multiplies 0.9999... by 10 to get 9.999...

At the end of the day though, 1 and 0.9999... can still be considered, for all intensive purposes, to be equivelant (even if they aren't, in the strictest sense, equal).

[marnitzg]

Well, I still say that is isn't a limit. Plain and simple maths shows this to me. I believe that something is wrong but I can't find anything wrong with wither proof.

Originally posted by Guv

BTW: In base 16, you can similarly prove that recurring .FFFFF equals one. Does this prove that recurring .99999 in decimal equals recurring .FFFFF in hex? For all finite examples, the two recurring fractions are unequal. Note that you can certainly prove that they approach the same limit (namely one).

No because they are different bases.

Sorry, this one of those brainless days again. I take back the above statement.

[Starman]

marnitzg,

>No because they are different bases.


I may be wrong but I have always thought that 1 was 1, no matter what base you are counting it in.

[simonm]

Guv,

The definition of limit does not claim that a function is equal to it limit. It merely states that the difference between the function and its limit can be shown to be smaller than any finite value.

The statement above is saying the same thing that Fried Egg was saying (albiet in a different way).

Fried Egg said that there is an infintely small quantity lost in the proof that 1 = 0.9999(...) and not a small fininte quantity.

In this specific case, the reversability of the proof highlights the loss of this infinitely small quantity. He did give this impression that 'reversability' is a requirement of a proof (which is obviously not) but I think what he meant was that, in this case, reversing the proof demonstrated a flaw in the proof.

[parksie]

Originally posted by Starman
I may be wrong but I have always thought that 1 was 1, no matter what base you are counting it in. What if it's base 1/2? :p

[Guv]

Serious mathematicians try to avoid proofs based on the notation used for expressing numbers.

The above is one reason why I question the validity of the proof involving the multiplication of recurring .9999 by ten. Another reason to distrust this proof is the requirement for arithmetic on infinitely long decimal numbers, which surely requires some supporting assumptions and/or definitions. The use of intuitive notions (rather than formal methods) in the 17th, 18th, and 19th centuries led to problems in mathematical logic, which is the reason mathematicians tend to be very picky and formal.

Note that specifying recurring .9999 as the sum of the series 9/10 + 9/100 + 9/1000. . . is independent of the radix. It is equivalent to the hex series 9/A, 9/64, 9/3E8. . . The proof could be done using hex notation followed by showing that the infinite series is equivalent to recurring nines in decimal notation.

Similarly recurring .FFFF in hex can be proven to approach one as the limit independent of the radix used for numerical notation.

[marnitzg]

I never knew you got decimal points in hex. How do you represent it? My HP won't accept F.FFFFF (or any other hex fraction)

[Guv]

Marnitzg: My HP calculator will do Hex, octal, and binary arithmetic on integers only. I think there are special calculators which do general purpose radix arithmetic, but I do not know of any regular calulator which will do radix arithmetic on fractional values.
However, such values exist in the Mindscape of Mathematics.

Depending on the country you favor, a dot or a comma separates the integer and fractional parts of Hex (or other radix) numbers. In Hex: 3.2FA represents 3 + 2/16 + 15/256 + 10/4096

If you kept track of the radix point yourself, you could use your calculator to do hex arithmetic on fractional values. On my calculator, division would be a problem because it gives an integer quotient and no remainder. Perhaps you could do division if you added some trailing zeros to the dividend.

If you wanted to multiply 3C.F by 2.F5, you could use integer multiplication of 3CF by 2F5, getting B431B. Since there are 3 digits (total) to the right of the radix point in the multiplier and multiplicand, there must be 3 digits right of the radix point in the product. B4.31B is the product.

In this day and age, I would how many people know how to do multiplication and division by hand methods, keeping correct track of the decimal point.

[Sam Finch]

0.999999............. = 1


Remeber this axiom

For any 2 real numbers A and B exactly one of the folowing is true.

A = B

or

there exists a real number C s.t.

A < C < B

or

there exists a real number C s.t.

A > C > B



And as there is no real number between 0.9999999... and 1 it must be true that 0.999999.... = 1.


QED.

[Guv]

Sam: You do come up with interesting ideas. Remember this axiom

For any 2 real numbers A and B exactly one of the following is true.
A = B
or
there exists a real number C s.t.
A < C < B
or
there exists a real number C s.t.
A > C > BMy answer to the question is that I do not remember such an axiom. There could very well be such an axiom, but I do not remember it. It looks valid for any real number whose value has a finite expression.

I do wonder about the context of such an axiom. Is it really applicable to real numbers which cannot be expressed in a finite manner? I would be surprised if there are not some caveats with that axiom.

What about the following? Recurring .99999 is a representation of the geometric series: 9/10 + 9/100 + 9/1000..... In hex, recurring .FFFFF is a representation of 15/16 + 15/256 + 15/4096....For a given number of finite terms, the sum of the second series is greater than the sum of the first, and is also less than one. Hence there is always a finite real number between recurring .99999 and one. Hence, they are not equal. This argument could be dressed up with some picky-picky mathematical language, but I think you get the picture.

The sum of both series is limited by one. However, I still agree with Noble.0.9999 recurring forever equals 0.9999 recurring forever and 1 equals 1. 0.9999 recurring forever "approaches one and for simplicity can be represented as 1".

[Sam Finch]

I figure this line of argument can only continue with increacingly complicated nit picky maths, so I suggest we take a different route.


What is 1 - 0.99999999..... ?

What is the reciprocal of this number?

[Guv]

Sam: Variations on this argument have occurred several times now.

If you tell me how many nines we are talking about, I will tell you what 1 - .99999... is equal to, and I will estimate the reciprocal. I will almost always insist on being picky, because I believe in avoiding sloppy terminology and sloppy methods. In practice, I will use one as a replacement for .999999 and 1/3 as equivalent to .333333, but I will not accept statements like " recurring .33333 equals 1/3," except as convenient shorthand notation.

This thread started with a proof that recurring .999999 = 1.0000 using some questionable operations on "infinitely long" decimal numbers. I objected to the proof for various reasons and suggested viewing the recurring .99999 as a geometric series. It is fairly easy to prove that the limit of the series is one, without using questionable arithmetic operations on infinitely long decimal numbers.

I agree with Noble0.9999 recurring forever equals 0.9999 recurring forever and 1 equals 1. 0.9999 recurring forever "approaches one and for simplicity can be represented as 1".Limits are defined by terminology similar to the following.The number a will be called the limit of the sequence X1, X2, X3 . . . Xn . . . provided that, given any positive number h, no matter how small, there exists a corresponding term of the sequence Xn such that every succeeding term of the series lies between the numbers a-h and a+h.The wording is slightly different for the limit of a function and the limit of the sum of a series, but the idea is always the same. As some number or variable grows without bound, some value gets closer to the limiting value.

The formal definitions use terms like increases indefinitely or there exists or grows without bound. The formal definitions refer to an extremely small difference between the value and its limit. They avoid referring to infinity and avoid the statement that the value equals its limit.

The use of the limit as a value is viewed as a convenience which avoids lengthy terminology.

Every serious mathematical text I have ever read avoids claiming that a function equals its limit. When not dealing with transfinite numbers ala Cantor, they also avoid using the term infinity. I have a book (Fundamentals of Mathematics by Moses Richardson) which specifically warns against the use of such a term.The notation "Limit(Xn) = a as n —>lazyeight" is commonly used, but it often misleads the unwary student into believing that there is a peculiar number called infinity which n approaches. This is, of course, not true.In a later chapter of this book. . .It would be even more misleading to write 1/0 = infinity, although some books do. . . .they do not mean that infinity is a number which you obtain by dividing 1 by zero. As we have seen 1/0 is a meaningless symbol and infinity is not a number. "Infinity" in this sense is merely a way of describing the manner in which certain function behave.In a footnote it is mentioned that historically the statement 1/0 = infinity was taken literally before mathematicians were familiar with modern mathematical logic.

[jmcswain]

1 / 9 = .1111111...
2 / 9 = .2222222...
3 / 9 = .3333333...
4 / 9 = .4444444...
5 / 9 = .5555555...
6 / 9 = .6666666...
7 / 9 = .7777777...
8 / 9 = .8888888...
9 / 9 = 1

It does seem like a breach of the progression.

9 / 9 = .9999999...

would fit so much more naturally.

[jmcswain]

Isn't there some rule about all real numbers can be expressed as

a / b

where both a and b are real numbers? If so, then .999... is not a real number.

[parksie]

But then what about 1/9?

[jmcswain]

.111... is fine because it fits the "find the a / b" mold


10x = 1.111...
- x = .111...
-------------------------
9x = 1

x = 1 / 9 = .111...

[Guv]

JmcSwain: Rational numbers can be expressed as a / b, where both a and b are integers. I think numbers like SquareRoot(3) are called surds (I am not sure about this). Numbers which are the roots of polynomials are called algebraic numbers. Numbers like e (2.71828) and Pi (3.141592653589793) are called transcendental numbers.

[jmcswain]

Guv,

Ahhh, I knew there was some rule like that. It was starting to bug me what the rule actually was; thanks for the explanation.

[Ajai]

until now 2 wasnt 1

but now 1 and 1 which is 2 got married and became 1!!

[simonm]

Can you point out the flaw in the supposed proof that 1 = 0.9999... ?

[Guv]

Simonm: From a common sense or intuitive point of view, there is nothing wrong with the proof that recurring .99999 equals one. Almost every serious mathematician up to the end of the 19th and many in the early 20th century would have accepted it as valid. For all practical purposes it is valid. After all, there is a valid proof that recurring .99999 has one for a limit. How far off base could the proof's conclusion be?

A serious mathematician would not claim that the proof is erroneous. He would claim that it is invalid due to undefined operations. In particular, the multiplication of recurring .9999 by ten and the subsequent subtraction of recurring .9999 from the product. These steps require doing operations on decimal fractions of unbounded length, which are undefined operations.

For any finite number of recurring nines, the proof breaks down. As some posters have already mentioned, perhaps then is a nine missing or unaccounted for in the proof. For a finite number of nines, you merely prove that recurring nines equals recurring nines. It requires a leap of faith to assume that for infinitely many nines the proof would work.

Some time starting in the late 19th or early 20th century, mathematicians became very picky about what constituted a valid proof and what are acceptable operations. The new attitude came about because some extremely subtle errors crept into the mathematical discipline. Proofs which later turned out to be invalid (and contradictory to other proofs) were discovered.

The problems were attributed to the use of common notions rather than formal logic.

Concepts relating to infinity are primarily (perhaps exclusively) dealt with in Set Theory. In analysis, calculus, algebra, analytical geometry, et cetera only finite operations are defined. Terminology like increases without bound is used to avoid words like infinity. When dealing with limits, a concept like the following is used. Consider the difference between the sum of a series and its proposed limit. Choose any small finite amount which we will call Delta. If for any arbitrarily small value of Delta a numbers of terms can be specified which makes the difference smaller than Delta, then the series sum is said to have or to approach the proposed limit.The above definition does not say anything about an infinite number of terms. It does not refer to some value becoming infinite. It does not require any operations involving an infinite number of terms. It merely says "You choose a small finite Delta and I will tell you a finite numbers of terms that will result in the difference between the sum and the limit being less than Delta."

[nathantkd]

guys, that whole thing with .9999... is only true because you are treating infinity as a number. Infinity is not a number, rather, it is a mathematical animal in and of itself that describes an action. Of course regular mathematic breaks down when you stop using actual numbers. Dammit. I'm turning into math geek. i'm leaving before i start going to geek parties and asking for some Pi.

[bver]

1 - .9999~ = 0 is flawed logically. With an infinite set of 9's there would never be a "Last" 9 to borrow from.