Let's discuss Mathematics

[onejayhawk]

since 0,999...=1 it's not in [0, 1) you don't have a better arguement for why this isn't the case

I see, wave a magic wand.

You have stated one truth, 1 is not an element of [0,1). 0.999... is an element of the set {0.9, 0.99, 0.999...} which is a subset of [0, 1). Hence 0.999... is an element of [0,1). This is a contradiction and your assumption that 0,999...=1 is disproven.

J

 

[AdrienIer]

It's certainly NOT an element of the set {0.9, 0.99, 0.999...}. Or you could point out which one it is.

 

[Samson]

What about the proposition that 0.999... =/= 1.000...? The proof is trivial. 0.999... is an element of [0,1) and 1.000 is not.# QED

J

The proposition can be writen as when x = 0.999... then x <> 1.
If you put [0,1) into wolfram alpha it translates it to :

[EDIT] Wolfram alpha images are not permanent. It says:
Inequality: 0 <= x < 1
x < 1 therefore x<> 1 does not seem a very strong argument.

 

[AdrienIer]

J's argument can also be applied to the limit of 2^n when n-> infinity. All numbers in this set are in [0;+infinity[ (an interval that exludes +infinity), therefore the limit (ie +infinity) is also in [0;+infinity[

 

[onejayhawk]

It's certainly NOT an element of the set {0.9, 0.99, 0.999...}. Or you could point out which one it is.

Why? That's like saying you can point to an infinite number. You are just saying that the set is not finite, which we already knew.

J

 

[AdrienIer]

Ok so according to you +infinity is an element of the set {1, 2, 4, 8, 16, 32...). Good for you I guess, but you're the only person in the world who believes that.

 

[Lohrenswald]

I'll try to explain why 1=0,999... as best as I understand it
In

, the "number density is infinite". If two numbers a and b are different and b>a, then there's an infinite amount of numbers r that's a<r<b
so if 0,999... and 1 where different numbers, you should be able to produce at least one number that fits between them, like the average
however, you can't

I'm sure there's a much more vigurous "version" of this proof, but this is roughly how I remember learning about it

and speaking of me learning about it: I thought initialy that 0,999... and 1 were different, however I learned about it, saw it was true, accepted it and then moved on with my life

like, what do you want to achieve by this inanity, hawkboy?

J

 

[onejayhawk]

Ok so according to you +infinity is an element of the set {1, 2, 4, 8, 16, 32...). Good for you I guess, but you're the only person in the world who believes that.

No, and don't be absurdist.

What I claim is that your set contains all numbers of the form 2^n. You cannot point to the largest element of either set.

I'll try to explain why 1=0,999... as best as I understand it
In

, the "number density is infinite". If two numbers a and b are different and b>a, then there's an infinite amount of numbers r that's a<r<b
so if 0,999... and 1 where different numbers, you should be able to produce at least one number that fits between them, like the average
however, you can't

I'm sure there's a much more vigurous "version" of this proof, but this is roughly how I remember learning about it and speaking of me learning about it: I thought initialy that 0,999... and 1 were different, however I learned about it, saw it was true, accepted it and then moved on with my life like, what do you want to achieve by this inanity, hawkboy?

This is a difficulty with irrational numbers--you cannot write them out. If you were to try to define the highest number less than 1, 0.999... is a candidate. Instead, we absurdly state that these sets have no highest number and that open segments have no endpoints. Like Schroedinger's cat, things only resolve on inspection. Regardless, nothing in your post contradicts any part of my proof.

As to the value, abstract mathematics needs no defined purpose. They are thought experiments (also like Schroedinger's cat).

J

 

[AdrienIer]

What I claim is that your set contains all numbers of the form 2^n. You cannot point to the largest element of either set.

The set {0.9, 0.99, 0.999...} contains all the numbers in that form with a finite number of 9s. That's what you don't seem to understand.

As to the value, abstract mathematics needs no defined purpose. They are thought experiments (also like Schroedinger's cat).

Abstract mathematics are not a thought experiment, it's not just a mental game. Some things in abstract mathematics are true, others are false, and your assertion is 100% false. The proof for 0.999999... = 1 was given to you several times in the other thread about this.

 

[onejayhawk]

The set {0.9, 0.99, 0.999...} contains all the numbers in that form with a finite number of 9s. That's what you don't seem to understand.

False. You know better. It is a bounded, infinite set with an accumulation point that lies outside the set.

Abstract mathematics are not a thought experiment, it's not just a mental game. Some things in abstract mathematics are true, others are false, and your assertion is 100% false. The proof for 0.999999... = 1 was given to you several times in the other thread about this.

Say rather attempted proofs--failed attempts. Regardless, the discussion is profitable, or it can be. No one can make you stretch your mind.

J

 

[formerdc81]

Mathematics are indeed arbitrary in the sense that the axioms can be chosen however you wish. HOWEVER, once a set of axioms is chosen, all the logical consequences of it follow and it is not possible to then wave your hands and say "it's just a theoretical construct and I can believe whatever I like" towards results you personally disagree with.

IN your case, you are making a basic mistake. If S is a bounded sequence that can be contained in an open set X, then the sup and inf of S lie NOT in X, but in the closure of X.
Proof: (I'll just show it for the supremum since the proof is the same for the other extrema.)
Since S is a bounded sequence, sup S is not infinite. By definition if s is an element of S, s is less than or equal to sup S.
If s = sup S, this is trivial because sup S is in X, and thus, is in the closure of X.
If s < sup S, S is an infinite sequence. More importantly, by the definition of supremum, for any r > 0, we can always find an n such that s_n is the nth element of S and sup S - s_n < r. However, since the above is true for any r, sup S by definition exists in the closure of X.

Now back to your problem. The closure of the set X [0.9, 1) is [0.9, 1]. Thus, if S = (0.9, 0.99, 0.999, ....), then S is in X and sup S = lim n->infinity of s_n is in the set [0.9, 1]. Since clearly sup S is not in X, sup S must be in the set cl(X) - X = {1}. Hence, 0.999.... = 1.

 

[onejayhawk]

Mathematics are indeed arbitrary in the sense that the axioms can be chosen however you wish. HOWEVER, once a set of axioms is chosen, all the logical consequences of it follow and it is not possible to then wave your hands and say "it's just a theoretical construct and I can believe whatever I like" towards results you personally disagree with.

IN your case, you are making a basic mistake. If S is a bounded sequence that can be contained in an open set X, then the sup and inf of S lie NOT in X, but in the closure of X.
Proof: (I'll just show it for the supremum since the proof is the same for the other extrema.)
Since S is a bounded sequence, sup S is not infinite. By definition if s is an element of S, s is less than or equal to sup S.
If s = sup S, this is trivial because sup S is in X, and thus, is in the closure of X.
If s < sup S, S is an infinite sequence. More importantly, by the definition of supremum, for any r > 0, we can always find an n such that s_n is the nth element of S and sup S - s_n < r. However, since the above is true for any r, sup S by definition exists in the closure of X.

Now back to your problem. The closure of the set X [0.9, 1) is [0.9, 1]. Thus, if S = (0.9, 0.99, 0.999, ....), then S is in X and sup S = lim n->infinity of s_n is in the set [0.9, 1]. Since clearly sup S is not in X, sup S must be in the set cl(X) - X = {1}. Hence, 0.999.... = 1.

Interesting approach though you seem to assume your conclusion. I would like to see a more definitive explanation of why the closure of S is relevant, since C(S) = [0.9,1.0) union 1. This seems equivalent of lim (0.9+0.09+0.009+...) = 1.0 which is inadequate for reasons given previously.

On a silly level, I ran across this. Assume you have ordered a sausage and mushroom from your local Italian restaurant. If the radius of the pie = z and the thickness = a, then the volume of your repast is V = Pi*z*z*a.

J

 

[onejayhawk]

The closure of a set is always closed, no?

The closure of a set is defined as the set joined with its limit points. Since you simply added the point 1 to the set [0,1) you assumed the conclusion. However, I did rephrase to make my objection clearer.

Once again, the fact that lim (0.9+0.09+0.009+...) = 1 is insufficient to show that 0.999... = 1.

J

 

[Kaitzilla]

I've seen this one before.

1 divided by 3 is 0.333....
Multiply by 3 and the answer is 0.999...

1 / 3 * 3 = 0.999... = 1

So as long as you believe that multiplication and division are inverse operations, 0.999... = 1.



The solution is obviously to abolish the use of decimals and switch entirely to fractions.
When asked what each decimal place means, the teacher always uses fractions like 1/10th, 1/100th, 1/1000th etc. to explain anyway.
Fractions are far superior to decimals!

https://abcnews.go.com/Technology/story?id=4194473&page=1

A few years ago, Dennis DeTurck, an award-winning professor of mathematics at the University of Pennsylvania, stood at an outdoor podium on campus and proclaimed, "Down with fractions!"

"Fractions have had their day, being useful for by-hand calculation," DeTurck said as part of a 60-second lecture series. "But in this digital age, they're as obsolete as Roman numerals are."

Pistols at dawn sirrah
Math should always be precise.

 

[onejayhawk]

That's one approach. The problem is that 1/3 is not equal to 0.333... for the same reason that 1 is not equal to 0.999.... It works as a practical matter because Rational numbers are dense on the Real numbers.

J

 

[Kaitzilla]

The problem is that 1/3 is not equal to 0.333... for the same reason that 1 is not equal to 0.999..

1 is between [0,1) and (1,2] right?

If 0.999... is not equal to 1, and is in fact less than 1, then it must be part of [0,1)
By that logic, an infinitely small value must exist, and 1 - 0.999... is equal to the infinitely small value.
1 plus the infinitely small value must exist too, and would be written as 1.000...
This 1.000... with an infinite amount of zeros is part of (1,2]

I find it hard to imagine the 1.000... with infinite zeros is somehow greater than 1, but it must be if the infinitely small value exists since it was added to 1 to create 1.000...
In addition, the infinitely small value must exist, because how do we count from 0 to 1 without it?


I think I see your objection to 0.999... = 1
https://en.wikipedia.org/wiki/0.999...

It is something like this maybe?
https://en.wikipedia.org/wiki/Infinitesimal

Pioneering works based on Abraham Robinson's infinitesimals include texts by Stroyan (dating from 1972) and Howard Jerome Keisler (Elementary Calculus: An Infinitesimal Approach). Students easily relate to the intuitive notion of an infinitesimal difference 1-"0.999...", where "0.999..." differs from its standard meaning as the real number 1, and is reinterpreted as an infinite terminating extended decimal that is strictly less than 1.[14][15]

They tried to make math work where 0.999... did not equal 1 with Non-Standard Analysis I think?
https://en.wikipedia.org/wiki/Non-standard_analysis

Here is some talk about it I found.
http://math.coe.uga.edu/tme/Issues/v21n2/5-21.2_Norton & Baldwin.pdf

The Hyperreals
The argument that 0.999... only approximates 1 has grounding in formal mathematics.
In the 1960’s, a mathematician, Abraham Robinson, developed nonstandard analysis (Keisler, 1976).
In contrast to standard analysis, which is what we normally teach in K–16 classrooms, nonstandard analysis posits the existence of infinitely small numbers (infinitesimals) and has no need for limits.
In fact, until Balzano formalized the concept of limits, computing derivatives relied on the use of infinitesimals and related objects that Newton called “fluxions” (Burton, 2007).
These initially shaky foundations for Calculus prompted the following whimsical remark from fellow Englishman, Bishop George Berkeley: “And what are these fluxions? ... May we not call them ghosts of departed quantities?” (p. 525).
Robinson’s work provided a solid foundation for infinitesimals that Newton lacked, by extending the field of real numbers to include an uncountably infinite collection of infinitesimals (Keisler,1976).
This foundation (nonstandard analysis) requires that we treat infinite numbers like real numbers that can be added and multiplied.
Nonstandard analysis provides a sound basis for treating infinitesimals like real numbers and for rejecting the equality of 0.999...and 1(Katz & Katz, 2010).
However, we will see that it also contradicts accepted concepts, such as the Archimedean property.

Shouldn't we use the current math system with limits where 0.999... = 1 if the math is easier to understand and gives more useful results?
The other math system with infinitesimals where 0.999...≠ 1 requires learning the current math system first I think?

I know there should only be one truth for 0.999... = 1 or 0.999... ≠ 1, but non-math people should be spared too much pain!

 

[onejayhawk]

1 is between [0,1) and (1,2] right?

If 0.999... is not equal to 1, and is in fact less than 1, then it must be part of [0,1)
By that logic, an infinitely small value must exist, and 1 - 0.999... is equal to the infinitely small value.

You lost it here. For this to work the distance between [0,1) and 1 must be nonzero. It is not even defined.

You are correct that 0.9999... is in [0,1) and that 1 is not. This provides the contradiction of the assumption that 0.999... = 1.

J

 

[Lohrenswald]

Alright so I've got some like repetition style problems I evidentally absolutely need

so obviously the answer is f(x)=exp(bx)
but I don't remember the like formalism to reach that answer (haven't really solved any differential equations in at least a year, probably more)

so like, what is the formal way to do it, again?

sorry for annoying question

 

[Petek]

For simplicity, write y for f(x) and assume that y <> 0. Then your equation is

dy/dx = by

Rewrite as

dy/y = bdx

Integrate both sides to get

log (y) = b(x + C)

where C is a constant. Exponentiate both sides:

y = exp(b(x + C)) = k exp(bx)

where k = exp(C) is a constant. Therefore, the most general solution is

y = k exp(bx)

For some constant k.

 

[Lohrenswald]

Thanks a lot, it was very helpful!

(also stupid of me to forget the C constant so thanks a lot for that as well)