Why it doesn't matter if .9 repeating = 1

[slopeslider]

.9r (I will use this to stand in for .999 repeating) gets closer and closer to 1 but never gets there, or so I've heard. But why do we even distinguish the two? It's an infinite number, and can't be distinguished from 1. How?
lets say you take the amount of all atoms in the known universe, x, and raise it to itself, then raise that product to .9r. Now we do the same except raise the product to 1.
Will there be any difference at all between the two answers? No. Why? Because in order for us to measure a difference, .9r would have to be finite, as in it ends at a fixed value. But it is not. We have no use for infinite numbers in a finite universe, as we can never measure the difference between .9r and 1.
If two lines start at point A on a flat plane, and line y leaves .9r degrees to the right from 0 and line z leaves at 1 degree to the right, no matter what finite number you give me as a distance until point b, the two lines will be EXACTLY on each other, down to the atomic level, when they reach point b. In order for there to be a difference, .9r would have to be a measurable, finite value, which it is not.

Am I right? Or close? Or severely flawed?
I have no math major, so feel free to flame me on how I used the wrong terms and stuff, I just want to start the discussion.

 

[Aardvark]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

 

[Shru1kan]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

 

[MazzaTheFirst]

May I direct you to here? http://qntm.org/?pointnine
(Don't worry, it's all text, no fancy flash or pictures to eat bandwidth.)

 

[slopeslider]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

 

[Trivun]

It's not the same thing. What is with all of the threads from people who simply don't understand math?

I remember those threads from a while back. As a Maths undergraduate I can prove, conclusively, that 0.999 recurring does indeed equal 1. And if anybody tries to prove me wrong, then those people will obviously have very little grasp of mathematics. Not to mention it's an extremely important concept in the dealings of infinity, which is vital for all sorts (such as basic Calculus, which is used in pretty much everything in the modern day...).

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[Shru1kan]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

Then with whatever you're working with, you'd cut off the 9's in that spot. for for thousandths it would be .990, rounding doesn't carry up the number until it stops, only at the cutoff point.

 

[slopeslider]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

Then with whatever you're working with, you'd cut off the 9's in that spot. for for thousandths it would be .990, rounding doesn't carry up the number until it stops, only at the cutoff point.

That's not .9r, that's .999 or .9990
It's like saying pi=3.14 and not pi~3.14

 

[Shru1kan]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

Then with whatever you're working with, you'd cut off the 9's in that spot. for for thousandths it would be .990, rounding doesn't carry up the number until it stops, only at the cutoff point.

That's not .9r, that's .999 or .9990
It's like saying pi=3.14 and not pi~3.14

Rounding to the accuracy of whatever you are measuring this number with is THE ACCEPTED method of the real world. Nice theory, but its that, a theory.

 

[slopeslider]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

Then with whatever you're working with, you'd cut off the 9's in that spot. for for thousandths it would be .990, rounding doesn't carry up the number until it stops, only at the cutoff point.

That's not .9r, that's .999 or .9990
It's like saying pi=3.14 and not pi~3.14

Rounding to the accuracy of whatever you are measuring this number with is THE ACCEPTED method of the real world. Nice theory, but its that, a theory.

What's your point? And what theory did I assert? Are you talking about the whole '.9r is indistinguishable from 1' thing?
You've made slopeslider a confused boy :/

 

[BuckminsterF]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Heisenburg's uncertainty theory says in practice, you can't know a value to an arbitrary (infinite) precision. In theory, however you can prove it equals one:

.9r = (9/10) + (9/100) + (9/1000)... so .9r- (1/10).9r = (9/10) = (9/10) + (9/100) + (9/1000)... - (9/100)+ (9/1000)...= (9/10) so (9/10).9r = (9/10) so .9r = 1 (divide by (9/10) on both sides)

 

[Shru1kan]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Thank you. These can go to millionths of an inch in precision machining.

Parts with tolerances of 3 millionths of an inch keep your planes in the sky. Why don't we round? Cause then its not PRECISE, and you will experience massive amounts of DEATH.

Rounding .9r to 1 will result in an unmeasurable difference in the parts, down to the atomic level.

Then with whatever you're working with, you'd cut off the 9's in that spot. for for thousandths it would be .990, rounding doesn't carry up the number until it stops, only at the cutoff point.

That's not .9r, that's .999 or .9990
It's like saying pi=3.14 and not pi~3.14

Rounding to the accuracy of whatever you are measuring this number with is THE ACCEPTED method of the real world. Nice theory, but its that, a theory.

What's your point? And what theory did I assert? Are you talking about the whole '.9r is indistinguishable from 1' thing?
You've made slopeslider a confused boy :/

It's a theory that .9r is indistinguishable until you bring proof from the math community, not your notebook. To be frank, thinking that you are the first to postulate on this is arrogant. No offense, but do you think that in the near couple thousand years we have had complex math that nobody voiced this before you?

 

[Ghonzor]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

You sir, win this thread.
Congratulations.

 

[MelasZepheos_v1legacy]

Of course, one could ask what the point of needing to know whether .9 recurring is equal to 1, or not?

One could also ponder a while as to whether it in fact means anything, given that numbers, and language, both of which we deem 'necessary' to talk about this 'important' issue, are simply human constructs, and as such, not being derived from any external point of reference divorced from our human conerns, in fact mean anything.

But since I'm a philosopher and a linguistics student in a mathematics thread, I suspect you will all now prove me wrong.

 

[benylor]

Many people here are making the mistake of assuming that in analysis you have tolerances. A number is precisely as it is labelled.

0.9r can be defined as the limit of the series 9(SIGMA)1/(10^-n) from n = 1 to infinity. The limit of the series is 1, therefore 0.9r = 1.

I am willing to accept that it's a bit clumsy here, but I don't want to bother going into a full proof of convergence when I can't draw symbols into the text box and in any case if you understand what a limit is you probably don't need me to argue this to you.

 

[BeeRye]

My maths lecturer covered this day one, first lecture. People hummed and hawed, objected and made counter-arguments, however, none of this changed the fact that they were wrong.

Get someone to show you why .9 recurring equals 1, and when they have done so shelve all arguments to the contrary, because they are all wrong no matter how clever they seem. There is nothing more pointless than someone arguing to the death that .9r does not equal one even when they have been shown it does.

 

[Steveh15]

I love doing this proof so even though it's completely unneseccary here it is!

x = 0.9999r
10x = 9.9999r
9x = 9 <-------- 10x-x = 9.9999r - 0.9999r
x = 1
0.9999r = 1

woo hoo! I paid attention in math(s)!

 

[Daze]

If you're talking theory, you can round if you're lazy. In practice, you round to the precision of the instruments you're working with.

Heisenburg's uncertainty theory says in practice, you can't know a value to an arbitrary (infinite) precision. In theory, however you can prove it equals one:

.9r = (9/10) + (9/100) + (9/1000)... so .9r- (1/10).9r = (9/10) = (9/10) + (9/100) + (9/1000)... - (9/100)+ (9/1000)...= (9/10) so (9/10).9r = (9/10) so .9r = 1 (divide by (9/10) on both sides)

This is a way of showing it that I haven't seen before, but I like it. I like to show it using a slightly simpler method.

3(.3)=.9
3(.33)=.99
3(.3333)=.9999

By inference, we can say that 3 times .3 repeating equals .9 repeating.

3(.3r)=.9r
1/3=.3r
3(1/3)=3(.3r)
1=.9r

 

[Aardvark]

Heisenburg's uncertainty theory says in practice, you can't know a value to an arbitrary (infinite) precision. In theory, however you can prove it equals one:

.9r = (9/10) + (9/100) + (9/1000)... so .9r- (1/10).9r = (9/10) = (9/10) + (9/100) + (9/1000)... - (9/100)+ (9/1000)...= (9/10) so (9/10).9r = (9/10) so .9r = 1 (divide by (9/10) on both sides)

.5r = (5/10) + (5/100) + (5/1000)... so .5r- (1/10).5r = (5/10) = (5/10) + (5/100) + (5/1000)... - (5/100)+ (5/1000)...= (5/10) so (5/10).5r = (5/10) so .5r = 1 (divide by (5/10) on both sides)

Would somebody kindly point out where I'm going wrong, because it seems to me that I just proved that just over half = 1.