If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333...

 

 

 

If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number.

 

 

 

You can't--it's zero.

If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333...

 

 

 

If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number.

 

 

 

You can't--it's zero.

The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero.

I showed this to my math teacher, specifically the second point.

 

 

 

x = .999...

 

10x = 9.999...

 

10x - x = 9.999... - 999...

 

9x = 9

 

x = 1

 

 

 

He looked at it for a bit, and found an error in 9x = 9.

 

Since x (at least at that moment) is not equal to one, .999... x 9 isn't equal to 9.

 

 

 

So I guess there's something impossible that happens?

 

 

 

No. Your math teacher doesn't understand math very well, and that's why he's teaching high school instead of getting his PhD.

 

 

 

 

 

Care to explain why he's wrong, ye mighty mathematical genius?

Then frankly Laura, your talking about the wrong subject.

Then frankly Laura, your talking about the wrong subject.

Again, I detest.

Laura, I suspect the word your looking for is 'dissent'.

 

 

 

Insertnamehere; Pay attention to the third line as the sequence of 9's is infinite there is no end for their to be a difference at. As such all 9's after the decimal point vanish when the subtraction on the right hand side in the third line is performed, and on the left hand side you are left with 9x which leads you to line 4.

 

x = .999...

 

10x = 9.999...

 

10x - x = 9.999... - .999...

 

9x = 9

 

x = 1

If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333...

 

 

 

If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number.

 

 

 

You can't--it's zero.

The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero.

 

 

 

The number you are trying to visualize is, in fact, 0.

Laura, I suspect the word your looking for is 'dissent'.

No, because the majorities ruling does not matter to me; I meant detest, as in, I I disagree immensely with your idea that logic cannot be overturned. I agree, that may not be done here but that doesn't change my stance. I see and understand why you believe that .999... is equal to one and the theorems regarding rational and irrational numbers. I do not agree however, that .333 or .999 extending in any direction is equivalent to any fraction; the fraction being the more accurate of the two. This explain all of these proofs.

 

If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333...

 

 

 

If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number.

 

 

 

You can't--it's zero.

The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero.

 

 

 

The number you are trying to visualize is, in fact, 0.

The number at the end is greater than a value of zero.

Laura, I suspect the word your looking for is 'dissent'.

No, because the majorities ruling does not matter to me; I meant detest, as in, I I disagree immensely with your idea that logic cannot be overturned. I agree, that may not be done here but that doesn't change my stance. I see and understand why you believe that .999... is equal to one and the theorems regarding rational and irrational numbers. I do not agree however, that .333 or .999 extending in any direction is equivalent to any fraction; the fraction being the more accurate of the two. This explain all of these proofs.

 

If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333...

 

 

 

If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number.

 

 

 

You can't--it's zero.

The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero.

 

 

 

The number you are trying to visualize is, in fact, 0.

The number at the end is greater than a value of zero.

 

 

 

What end? There is no end to an infinite string of 0's.

Detest means to hate, to loathe. Dissent is to disagree.

 

 

 

Majorities have nothing to do with it, nor does belief.

To minorly add to the validity of the following statement (not that it needs any), I'm currently attending one of the best Engineering schools in the nation of the United States of America, one that is also a part of an Ivy League institution, so I feel I can say this with a respectable amount of backing.

 

 

 

Laura, hun, math isn't your strong suit.

Detest means to hate, to loathe. Dissent is to disagree.

 

 

 

Majorities have nothing to do with it, nor does belief.

 

Detest - dislike intensely

 

Dissent - to express an opinion at variance with those commonly stated

 

 

 

Laura, hun, math isn't your strong suit.

While I would agree, that certainly doesn't mean I was never, and am not strong in the field of mathematics. I don't agree with these proofs though. I believe that it approaches 1 infinitely, yet you can use infinity to contradict this. How does that work? For that matter, it's a concept, not a real number, so how is it plugged in as substitutes in these equations?

 

 

 

But the rest of the mathematics community apparently disagrees with me. I'll stop.

It is perfectly possible to dislike/loathe/hate something while accepting that it is true.

 

 

 

And to your question you would need to study limits, and the foundation of calculus to see (real analysis).

It is perfectly possible to dislike/loathe/hate something while accepting that it is true.

I was talking about your statement that logic cannot be overturned.

 

And to your question you would need to study limits, and the foundation of calculus to see (real analysis).

I'm in Calc II for what it matters, as my last year of mathematics, though I'm not doing nearly as well as in High School - nowhere near in fact. I was taught that no definitive answer could be discerned without the use of these concepts such as infinity.

Laura you may find this a helpful resource to help you understand why and how.

 

 

 

http://www.math.lsu.edu/~sengupta/4031f ... sNotes.pdf

Laura you may find this a helpful resource to help you understand why and how.

 

 

 

http://www.math.lsu.edu/~sengupta/4031f ... sNotes.pdf

 

I actually have his work bookmarked, as it was relating to my studies in theoretical physics during high school.

Try taking alook at the section on Cauchy sequences (page 49-52) and using that to prove .9... = 1.

I showed this to my math teacher, specifically the second point.

 

 

 

x = .999...

 

10x = 9.999...

 

10x - x = 9.999... - 999...

 

9x = 9

 

x = 1

 

 

 

He looked at it for a bit, and found an error in 9x = 9.

 

Since x (at least at that moment) is not equal to one, .999... x 9 isn't equal to 9.

 

 

 

So I guess there's something impossible that happens?

 

 

 

No. Your math teacher doesn't understand math very well, and that's why he's teaching high school instead of getting his PhD.

 

 

 

 

 

Care to explain why he's wrong, ye mighty mathematical genius?

 

 

 

lets just write the .999... as .99 here for the sake of simplicity

 

 

 

x=.99

 

10x=9.9

 

10(.99)-.99=9.9-.99

 

9(.99)=9

 

.99=1

 

 

 

since multiplying both sides by 10 and subtracting .99 are both legitimate operations (not multiplying by 0) this functions as a proof.

if all you can do is argue over how many decimal places I use to represent an endlessly recurring decimal (because its pretty difficult, and ultimately pointless, to use the actual symbol for recurring on this forum), then it is pretty clear you dont have much of a leg to stand on in regards to the actual debate going on.

 

 

 

My proof is right.

whoa there matimaticiann, calm down. i'm just enjoying browsing this thread. i don't have any "stand" to begin with, because frankly m'dear, i don't give a [cabbage]. i just got confused because i honestly didn't realize you were using repeating values.

 

lets just write the .999... as .99 here for the sake of simplicity

 

 

 

x=.99

 

10x=9.9

 

10(.99)-.99=9.9-.99

 

9(.99)=9

 

.99=1

 

 

 

since multiplying both sides by 10 and subtracting .99 are both legitimate operations (not multiplying by 0) this functions as a proof.

 

You've made a calculation error here: 10(.99) -.99 does not equal 9(.99) it equals 9.9-.99 You have to multiply before you subtract. Even if you could, it would be 9.11(.99) not 9(.99)

 

You proved that 1=1 not .99=1

 

 

 

If

 

x=2

 

10x=20

 

10(2)-1=20-1

 

20-1=19 not 9(2)=19

 

 

 

If I had calculated it the way you did, I would have proved that 18=19.

 

lets just write the .999... as .99 here for the sake of simplicity

 

 

 

x=.99

 

10x=9.9

 

10(.99)-.99=9.9-.99

 

9(.99)=9

 

.99=1

 

 

 

since multiplying both sides by 10 and subtracting .99 are both legitimate operations (not multiplying by 0) this functions as a proof.

 

You've made a calculation error here: 10(.99) -.99 does not equal 9(.99) it equals 9.9-.99 You have to multiply before you subtract. Even if you could, it would be 9.11(.99) not 9(.99)

 

You proved that 1=1 not .99=1

 

 

 

If

 

x=2

 

10x=20

 

10(2)-1=20-1

 

20-1=19 not 9(2)=19

 

 

 

If I had calculated it the way you did, I would have proved that 18=19.

 

 

 

I subtracted x from both sides not 1(well x equaled one but you get the point)

 

 

 

x=2

 

10(2)=20

 

10(2)-2=20-2

 

9(2)=18

 

 

 

the reason I change to 9(2) is the same property that gives 5x-2x=3x

[hide=]

 

lets just write the .999... as .99 here for the sake of simplicity

 

 

 

x=.99

 

10x=9.9

 

10(.99)-.99=9.9-.99

 

9(.99)=9

 

.99=1

 

 

 

since multiplying both sides by 10 and subtracting .99 are both legitimate operations (not multiplying by 0) this functions as a proof.

 

You've made a calculation error here: 10(.99) -.99 does not equal 9(.99) it equals 9.9-.99 You have to multiply before you subtract. Even if you could, it would be 9.11(.99) not 9(.99)

 

You proved that 1=1 not .99=1

 

 

 

If

 

x=2

 

10x=20

 

10(2)-1=20-1

 

20-1=19 not 9(2)=19

 

 

 

If I had calculated it the way you did, I would have proved that 18=19.

[/hide]

 

I subtracted x from both sides not 1(well x equaled one but you get the point)

 

 

 

x=2

 

10(2)=20

 

10(2)-2=20-2

 

9(2)=18

 

 

 

the reason I change to 9(2) is the same property that gives 5x-2x=3x

 

Okay, I see what you're saying. I missed that you had subtracted x from each side. #-o

 

Can you tell it's been many years since my last Calculus class? :lol:

Why use all these algebraic proofs when basic analysis of the reals tells us that it must be so?

[hide=]

[hide=]

 

lets just write the .999... as .99 here for the sake of simplicity

 

 

 

x=.99

 

10x=9.9

 

10(.99)-.99=9.9-.99

 

9(.99)=9

 

.99=1

 

 

 

since multiplying both sides by 10 and subtracting .99 are both legitimate operations (not multiplying by 0) this functions as a proof.

 

You've made a calculation error here: 10(.99) -.99 does not equal 9(.99) it equals 9.9-.99 You have to multiply before you subtract. Even if you could, it would be 9.11(.99) not 9(.99)

 

You proved that 1=1 not .99=1

 

 

 

If

 

x=2

 

10x=20

 

10(2)-1=20-1

 

20-1=19 not 9(2)=19

 

 

 

If I had calculated it the way you did, I would have proved that 18=19.

[/hide]

 

I subtracted x from both sides not 1(well x equaled one but you get the point)

 

 

 

x=2

 

10(2)=20

 

10(2)-2=20-2

 

9(2)=18

 

 

 

the reason I change to 9(2) is the same property that gives 5x-2x=3x

 

Okay, I see what you're saying. I missed that you had subtracted x from each side. #-o

 

Can you tell it's been many years since my last Calculus class? :lol:

[/hide]

 

 

 

no problem, it took me a minute or two to see what was being done that step the first time around.

Here is an easy explanation for those who are brain dead when it comes to math (hey, didn't everyone here have 130+ IQs? hmm).

 

 

 

Anyway, let's pretend that when the first mathematicians had to define a term to represent a number that increases forever (infinity), they defined that term in such a way that 0.9999... = 1. If they had done it any other way, our modern calculations could possibly have be different, but since "advanced" math is based on this, there's no escaping it. End of discussion.