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 ) ] 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