Simple mathematical problem (?)

[Zep]

I know this was meant as a joke, but you goofed up your order of operations. You can't go from 1-1+1 to 1-(1+1).

How strong are your Math skills, Zep?

It was indeed a joke - see the smilie?? And of course it was wrong - deliberately so.

My maths skills are quite rusty - it's been many years since I completed my science degree and had to do heavy-duty engineering maths and calculus and stuff. Mostly it's been computer-related stuff since then.

 

[Zep]

Xouper, perhaps a question to make this clear in my mind.

How far away from 1 must a number be for it NOT to be 1? What is that delta that "makes the difference"?

 

[angard]

"Xouper, perhaps a question to make this clear in my mind.

How far away from 1 must a number be for it NOT to be 1? What is that delta that "makes the difference"?"


Hi, I think he will answer more than 0.

 

[geni]

My take is that as long as we stick to finite numbers they are the same. If we go to infinty I can get them behave differenty (try squaring tham both an infinte number of times). Conclution : within finite limits they are the same.

 

[Colloden]

I’m still unconvinced that 0.9999etc. is 1.
After four pages I’m none the wiser.

 

[slimshady2357]

Zep, are you talking about some different concept of numbers? Or are you talking about the Reals?

The only way I can see 0.999.... not being equal to 1 is if you are working in some other number system.

If you are positive that 0.999.... is NOT equal to 1. Can you please answer this questions, so as to help clear up any confusion:

What is 1 - 0.999...?

What is 4*(1 - 0.999....)?

What is 0.999.... - (1 - 0.999....)?

The reason I ask more than one is that I don't want you to give "iota" as an answer. What is 4 times an iota? What is 0.999... minus an iota?

Just curious, because it seems to me either you're talking about a different number system, or you really have no idea what you're talking about.

Adam

 

[slimshady2357]

(1) Incorrect. After the "infinite-th" cut you will still have 10% of the previous length of the string left over. And so it will be for ANY number of iterations of that process.

Ludicrous! You contact me after you have made an infinite amount of cuts and I'll show you how there is no string left

You just let me know when you're done

(2) Yes, the number is indeed a "process". The very definition of it requires some calculation to even conceive of it. Mathematically that may be as shown by a previous poster: 9/10^1 + 9/10^2... Conceptually it may be the "one iota down from one" idea.

Both these ideas are really an attempt to represent a surreal number as accurately as possible. [/B]

A 'surreal' number? Since when is 0.999... a surreal number? It's always been a rational (sub-set of Reals) number for me.

I think 0.999.... is a notation that accurately describes the number. So is 1.

Again, out of curiosity, would you say then that 1/3 is not equal to 0.3333.....?

You must right? Otherwise, 1/3 + 1/3 + 1/3 would be equal to two different things for you (1 and 0.9999....)

Adam

 

[slimshady2357]

"Xouper, perhaps a question to make this clear in my mind.

How far away from 1 must a number be for it NOT to be 1? What is that delta that "makes the difference"?"


Hi, I think he will answer more than 0.

Great answer!

How much greater than 0 is an 'iota' anyway?

Adam

 

[Suggestologist]

This is total nonsense.

Wrong. Tell us what is tautalogical about the proof I gave in that other thread, the one that uses the geometric series theorem. That is in fact one of the standard proofs.

Nonsense.

You did no such thing.

Yes, that's true. So is the step I used. One divided into ten gives nine with a remainder of one. There is nothing false about that statement.

Well then you may as well get .888888888..... as a result of your division. Since 1 into 10 is 8 remainder 2. And then 1 into 20 is 8 remainder 12. And then 1 into 120 is 8 remainder 112. Etc.

 

[Suggestologist]

let x be any single digit number.

a/9 is what in decimal. If we want decimal we desire a denominator which is power of 10. b/10 is very easy to write in decimal form. So we try to solve:

a/9 = b/10
b = (10/9)a

so if multipl a by 10/9 we can get the fraction into a form that is easy to change to decimal notation.

10/9 = 1.111.... (sorry, my equation editor is acting funny, try the long division yourself if your not convinced)

b = a x 1.111 ...
b = a.aaa.... (assuming a is a single digit)

Therefore
a/9 = b/10 = a.aaa/10 = 0.aaa...

a/9 = 0.aaa... for any a which is a single digit

Walt

It doesn't matter what your equations seem to show. They are obviously not generalizable because a counterexample exists.

 

[LuxFerum]

1-0,9         =0,1
1-0,99        =0,01
1-0,999       =0,001
1-0,9999      =0,0001
1-0,99999     =0,00001
1-0,999999    =0,000001

when exactly the result will be zero??
never!!!!

 

[Suggestologist]

Pete has 3 apples. He gives 5 apples to Mary. How many apples does Pete have left?

Well, then obviously Pete stole at least two apples from John.

All I am suggesting is that people are taught mathematics with simple rules first, and that they should well understand those, before they are bothered with more complex mathematical theories. They are not going to understand the complex rules, if they don't understand the simpler underlying rules first.

I agree, in general. But when they express interest is the time to give them something interesting to learn.

 

[Suggestologist]

I "see" it as exactly equal to one. I "see" all the nines already there simultaneously. All infinity of them.

No.

Wrong. The definition of the notation is that all the nines are already there simultaneously. There is no process in that definition. I do not accept your attempt to redefine the meaning of that standard notation. It may be convenient in some contexts to think of it in terms of a process, but that is not how the notation is defined.

No. It is exactly equal to one.

Yeah, and the number "infinity" is not a process? Anytime you deal with infinity of anything, decimals or steps in a limit, or whatever -- it IS a process that theoretically does not stop.

 

[Mendor]

1-0,9         =0,1
1-0,99        =0,01
1-0,999       =0,001
1-0,9999      =0,0001
1-0,99999     =0,00001
1-0,999999    =0,000001

when exactly the result will be zero??
never!!!!

The result tends to zero as the thing you are subracting adds on more nines. The "limit" of the result is zero as the number of nines tends to infinity - which is what the "..." in 0.99... means.

0.99 != 1
0.999 != 1
0.999999999999 !=1

but 0.99... = 1

I put "limit" in quotes as I get the feeling this is not a terribly mathematically rigourous way to talk about it, but it works for me.

To anyone unconvinced, I'll echo slimshady's question: what do you get if you subtract 0.99... from 1?

 

[LuxFerum]

The result tends to zero as the thing you are subracting adds on more nines. The "limit" of the result is zero as the number of nines tends to infinity - which is what the "..." in 0.99... means.

The result tends to zero!!!
The result is never zero!!!
infinity is not a magical stuff that makes things dissapears.

 

[DanishDynamite]

For those still in doubt that 0.999... = 1, allow me to cut and paste xouper's proof from the link provided by boooeee on page 1 of this thread:

One of the standard proofs uses the geometric series theorem, which says that the sum of the infinite series a*(k^n) where n goes from 0 to infinity, is equal to a/(1-k).

Another way of expressing 9.999... is to use the above infinite series with a=9 and k=1/10. Since -1< k <1, this is a convergent series and according to the theorem the sum is equal to 9/(1-(1/10)) = 10.

Thus we have proved that 9.999... = 10. Now subtract 9 from both sides. End of proof.

This is airtight. There is no "taking the limit". There are just equal signs. In short:

0.999.... = 0.9 + 0.09 + 0.009 + ...
= sum from i =1 to infinity of 9*(0.1)^i
= (sum from i =0 to infinity of 9*(0.1)^i) - 9*(0.1)^0
= 9/(1-0.1) - 9
= 10 - 9
= 1

QED.

 

[LW]

Yeah, and the number "infinity" is not a process?

For starters, "infinity" is not a number. There are actually an infinite number of different orders of infinity.

But anyway, about the whole 0.99... = 1 thing.

I have two books about mathematics on the reach of my hand. Erwin Kreyszig's Advanced Engineering Mathematics (7th Ed) and Råde&Westergren's Mathematics Handbook for Scientists and Engineers. (If these are not enough I can walk to the other end of hallway and get few more).

They both contain the following formula for geometric series:

sum( a x^k ) = a / (1 - x)

when the summation goes from 0 <= k < infinity and -1 < x < 1.

I hope that you accept that this formula is valid. If you don't, you can refer to Kreyszig to see the proof. If you still don't accept it, then I'd like to make a blatant appeal to authority and note that Kreyszig is a professor of mathematics in Ohio State University and you are not (or at least he was in 1993 when the book was printed). This might suggest that he knows a lot more about mathematics than you. Also, as it is the seventh edition of the book one might thing that most blatant errors have already been found.

Next, when you examine the number 0.999... you notice that it is actually:

0.9 * (1/10)^0 + 0.9 * (1/10)^1 + 0.9 * (1/10)^2 + ...

If we replace 0.9 with 'a' and 1/10 with x, we get:

a * x^0 + a * x^1 + ...

Wow, this is actually the very same sum that was shown above.

So,since (1/10) < 1, the value of the sum is:

a / (1 - x)

If we put the numbers back, we have:

0.9 / (1 - (1/ 10)).

After performing the subtraction we have:

0.9 / (9/10)

This is equivalent to:

(0.9 * 10) / 9

Performing the multiplication we have:

9 / 9.

I don't know how your mathematical system works, but mine says that

9/9 = 1.

What part of this particular proof you disagree with?

 

[LW]

infinity is not a magical stuff that makes things dissapears.

It is .

 

[Mendor]

The result tends to zero!!!
The result is never zero!!!
infinity is not a magical stuff that makes things dissapears.

Yes, it is. That's what makes maths great

But seriously: both your reasoning and my answer to it are considering 0.99... as a "process" again, and as we've seen in this thread, that definitely isn't mathematically rigourous, so I'll stop.

 

[LuxFerum]

What part of this particular proof you disagree with?

with this one

sum( a x^k ) = a / (1 - x)

when the summation goes from 0 <= k < infinity and -1 < x < 1.

When k tend to infinity
the result tends to a / (1 - x)