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Math problem

Dec 11, 2012#1

WasabiPoptart

Last night I took a comprehensive quiz for my math course. I got a 95! WOO! But the one problem I got wrong is really bugging me:
There are 4 senators on a committee: Amy, Brian, Carol, and Dennis. Any or none of them may be chosen for other subcommittees. How many possible subcommittees are there?

I thought the solution was n^n (n to the nth power) which gave me (4^4) 256. But that isn't right. I can't review the correct answers until after the final is over. However, this is driving me crazy. Did I have the right idea or was I completely off?

Dec 11, 2012#2

Dei

Uh, if there can be subcommittees with none of them in it, that kind of makes the number infinite... unless I am reading that wording way off.

Dec 11, 2012#3

Allen who is Quiet

I think you had the right idea, but there's the possibility none of them are on the subcommittee, one of them is, two of them are, three of them are, or all of them are. So the number of possible ones should be:

(number of possible subcommittees containing none) + (containing one) + (containing two) + (containing three) + (containing all of them)

Dec 11, 2012#4

Dave

I get 16 total.

Dec 11, 2012#5

tegid

I just did the combinations on a piece of paper and got 16 as the result. Now, that's not useful without the math behind it, right? So I looked up on the internet the various formulas to help me explain and here it is:

The formula you used, n^n, is for permutations with repetition, i.e. the order matters and you can have the same senator any number of times, and you are always choosing exactly 4 of them. Maybe there's some other case the formula is good for...

In this case, what you need to use is the formula for combinations (the order doesn't matter) without repetition, which is n!/(r!(n-r)!), where n is the number of elements you choose from (in this case 4) and r the number of elements you pick (in this case it ranges from 0 to 4). So you'd get:
# of combinations of 0 senators + # of combinations of 1 senator + #of combinations of 2 senators + ... 3 senators + ... 4 senators =
4!/(0! 4!) + 4!/(1! 3!) + 4!/(2! 2!) + 4!/(3! 1!) + 4!(4! 0!) =
1 + 4 + 6 + 4 +1 = 16

(n!= n * n-1 * n-2 * ... * 2 * 1, and 0! = 1)

EDIT:

The trick to laying out all the combinations by hand is fixing the elements in order: you fix the first one and cycle through the rest:
all the possibilities with A
AB,AC,AD
now all the possibilites with B. Since we already accounted for A, we leave it out.
BC,BD
now all the possibilities with C. We leave out A and B:
CD
And that's it.

It may sound ridiculous but I'm not good with combinatorics and having some sort of system helps in more complicated cases.

Dec 11, 2012#6

Bubble181

Dei said:
Uh, if there can be subcommittees with none of them in it, that kind of makes the number infinite... unless I am reading that wording way off.

This would be mu answer too.

Dec 11, 2012#7

Ravenpoe

Allen who is Quiet said:
I think you had the right idea, but there's the possibility none of them are on the subcommittee, one of them is, two of them are, three of them are, or all of them are. So the number of possible ones should be:

(number of possible subcommittees containing none) + (containing one) + (containing two) + (containing three) + (containing all of them)

For the more visual:

Amy, Brian, Carol, and Dennis
Amy, Brian, Carol
Amy, Brian, Dennis
Amy, Carol, Dennis
Brian, Carol, Dennis
Amy, Brian
Amy, Carol
Amy, Dennis
Brian, Carol
Brian, Dennis
Carol, Dennis
Amy
Brian
Carol
Dennis
None of them

(1)+(4)+(6)+(4)+(1)

Dec 11, 2012#8

evilmike

Wouldn't the solution be 2^n? 4 binary digits -- either they are on the committee or they aren't.

Dec 11, 2012#9

tegid

evilmike said:
Wouldn't the solution be 2^n? 4 binary digits -- either they are on the committee or they aren't.

I like that answer. Way cleaner than mine, mathematically.

Dec 11, 2012#10

WasabiPoptart

Thanks everyone! I was going to write it all out like Poe posted (visual is definitely the way to go for me on this) and decided not to since I left this problem for last and time was ticking away. I should have gone with it.

Dec 12, 2012#11

PatrThom

The binary method is the most visual, but you can use Tegid's method to calculate your own lottery odds.

--Patrick

Dec 12, 2012#12

Krisken

I'm sorry, I'm still trying to wrap my head around .9999(infinity) being equal to 1.

Dec 12, 2012#13

Chad Sexington

Krisken said:
I'm sorry, I'm still trying to wrap my head around .9999(infinity) being equal to 1.

.99999...=x
9.9999...=10x
9.9999... - .9999...=9 and 10x - x=9x
9 = 9x
1 = x
Therefore .9999...=1

Dec 12, 2012#14

MindDetective

Something seems off about that. Let's try swapping in some different values...

.33333...=x
9.33333...=10x
9.33333... - .33333...=9 and 10x - x=9x
9 = 9x
1 = x
Therefore .33333...=1[DOUBLEPOST=1355337448][/DOUBLEPOST]

Ah, you just reassert the assumption in step two is all.

Dec 12, 2012#15

Dave

That would be 3.33333...=10x

3.33333...-.33333...=3 and 10x-x=9x

3=9x

1/3 = x

Dec 12, 2012#16

tegid

MindDetective said:
snip

Actually with what you did (minus the incorrect first step!) you'd end up with 1/3=0.3333... Which is really the same result! (Just dividing by 3 on each side)

EDIT: What Dave said

Dec 12, 2012#17

MindDetective

Dave said:
That would be 3.33333...=10x

3.33333...-.33333...=3 and 10x-x=9x

3=9x

1/3 = x

I was intentionally trying to break it, which is why I only changed the decimals. I wanted to create an inconsistency to see where things were off so that x = both 1 and 0.33333...

Dec 12, 2012#18

Chad Sexington

The 1/3 thing does illustrate the point perfectly, actually.

.3333... = 1/3 and 1/3 + 1/3 + 1/3 = 3/3 or 1. (while adding the decimals gives you .99999999...)

Dec 12, 2012#19

MindDetective

Chad Sexington said:
The 1/3 thing does illustrate the point perfectly, actually.

.3333... = 1/3 and 1/3 + 1/3 + 1/3 = 3/3 or 1. (while adding the decimals gives you .99999999...)

Well, that was my point. It just is another version of the same problem. I wanted to see where the equality really occurred, so I broke it on purpose. It turns out all you are doing in that progression is reasserting that 1 = 0.99999... in step 2. Thus it doesn't actually demonstrate equality, it just appears to.

Dec 12, 2012#20

Chad Sexington

MindDetective said:
Well, that was my point. It just is another version of the same problem. I wanted to see where the equality really occurred, so I broke it on purpose. It turns out all you are doing in that progression is reasserting that 1 = 0.99999... in step 2. Thus it doesn't actually demonstrate equality, it just appears to.

I'm not sure I follow? I set x=.99... and in step two I multiply both sides by 10. Solve for x and you get 1. I am not sure how this doesn't demonstrate equality?

Dec 12, 2012#21

MindDetective

Chad Sexington said:
I'm not sure I follow? I set x=.99... and in step two I multiply both sides by 10. Solve for x and you get 1. I am not sure how this doesn't demonstrate equality?

Because step two is where you are essentially asserting that 9.99999... = 10. It isn't apparent because of the number of decimals you are dealing with. But when you break it on purpose like I did earlier:

.33333...=x
9.33333...=10x
9.33333... - .33333...=9 and 10x - x=9x
9 = 9x
1 = x
Therefore .33333...=1

You can see that in step two 9.33333...=10x is essentially false, given our x initial x value. So at this step of the proof I have to reassert that these are in fact equal (thus ignoring my initial given value) and continue on from there. You are doing the same thing in your version as well, but it is cleverly hidden in the millionth decimal place somewhere.

Dec 12, 2012#22

Chad Sexington

No, I'm only saying that if you multiply .9999... by ten, then the decimal moves one place over (if you multiply 10.0 by 10, you get 100). I'm not stating that 9.999...=10 I'm stating 9.999... is equal to 10 times x which is .9999...

It does so happen that because .999... is equal to 1 that 9.999... is equal to 10, but the math isn't flawed.

Your setup makes no sense. In step two you don't multiply both sides by 10. You add 9 to the left and multiply the right by 10. That's not equivalent.

Dec 12, 2012#23

MindDetective

Chad Sexington said:
In step two you don't multiply both sides by 10. You add 9 to the left and multiply the right by 10. That's not equivalent.

This is it exactly. It is the exact same problem with your set up.

Dec 12, 2012#24

Chad Sexington

The actual setup with .333... is

.333...=x
3.333...=10x
3.333...-.333...=3 and 10x-x=9x
3=9x
1/3 = x

Dec 12, 2012#25

MindDetective

Chad Sexington said:
The actual setup with .333... is

.333...=x
3.333...=10x
3.333...-.333...=3 and 10x-x=9x
3=9x
1/3 = x

Only if you want to hide the flaw again, yes.

Dec 12, 2012#26

Chad Sexington

We are in agreement that .999... does in fact equal 1, right? You're just disputing my methodology?

When you do your method the way I just posted, to get 1/3 = x that also proves .9999... = 1. I also disagree that my previous method is wrong, however. I am not adding nine. I'm multiplying both sides by the same value.

Dec 12, 2012#27

MindDetective

Chad Sexington said:
We are in agreement that .999... does in fact equal 1, right? You're just disputing my methodology?

No, I'm not convinced of that.

Chad Sexington said:
When you do your method the way I just posted, to get 1/3 = x that also proves .9999... = 1. I also disagree that my previous method is wrong, however. I am not adding nine. I'm multiplying both sides by the same value.

Not really. You are also adding 9 to one side and multiplying the other by 10.

Dec 12, 2012#28

Chad Sexington

This section agrees with me that it demonstrates it.

http://en.wikipedia.org/wiki/0.999...#Digit_manipulation

Vihart:
[DOUBLEPOST=1355340280][/DOUBLEPOST]

MindDetective said:
No, I'm not convinced of that.

Okay, but then, 1/3 which is .333....

1/3 + 1/3 + 1/3 = 1 Right?

And .333... + .333... + .333... = .999...

So these must be equivalent values.

Dec 12, 2012#29

MindDetective

Chad Sexington said:
This section agrees with me that it demonstrates it.

http://en.wikipedia.org/wiki/0.999...#Digit_manipulation

Vihart:

Did you read the discussion bit on Wikipedia? That's what we're fundamentally going round and round about here.

Dec 12, 2012#30

strawman

What is 0.9999... * 10?

Dec 12, 2012#31

MindDetective

stienman said:
What is 0.9999... * 10?

What is 0.9999... + 9?

Dec 12, 2012#32

strawman

Does 0.9999... * 10 == 0.9999... + 9?

Dec 12, 2012#33

Chad Sexington

MindDetective said:
Did you read the discussion bit on Wikipedia? That's what we're fundamentally going round and round about here.

I agree that it isn't a good explanation of why .999...=1 but the discussion section does state that it does demonstrate it. I am surprised you don't agree with the idea that .999...=1(in some number systems you can have infinitely small repeating decimals but not in the real number system).

I will try to come up with better sources/explanation if I can but I'm at work and should probably, you know, do some of that.

Dec 12, 2012#34

MindDetective

Well, it definitely approximates it. In fact, it is the best approximate of it after the number 1 itself (since we don't have a way to easily represent 1.000...1). We simply don't differentiate at such a fine level, so we round up.[DOUBLEPOST=1355341132][/DOUBLEPOST]

stienman said:
Does 0.9999... * 10 == 0.9999... + 9?

That's the fundamental question, isn't it?

Dec 12, 2012#35

strawman

MindDetective said:
That's the fundamental question, isn't it?

I thought the fundamental question was: do we have free will, or are we merely meat robots?

Dec 12, 2012#36

MindDetective

In fact I will go 1 step further and offer my opinion. 0.9999... + 9 definitely preserves all of the infinite 9s in the decimal places. 0.9999... * 10 may not preserve them all, effectively losing the 9 in the infinity place. In Chad Sexington 's proof, the assumption is that 9.9999... has preserved all of infinity of the 9s in the decimal place, which to me implies it is more accurate to say that 9 has been added rather than 10 has been multiplied.

*edit* Argh, I should have said I would go 0.9999... steps further. Opportunity missed.
[DOUBLEPOST=1355341504][/DOUBLEPOST]

stienman said:
I thought the fundamental question was: do we have free will, or are we merely meat robots?

No, the answer to this question informs us to the answer on that one.

Dec 12, 2012#37

strawman

MindDetective said:
the answer to this question informs us to the answer on that one.

I don't know about anyone else, but I bought chickens, then I got eggs

Dec 12, 2012#38

PatrThom

stienman said:
Does 0.9999... * 10 == 0.9999... + 9?

I see where you're going here…

--Patrick

Dec 12, 2012#39

Krisken

It really is as simple as 1/3 of 1 is .333(repeating), .333(repeating) times 3 is .999(repeating), 1=.999(repeating). Not that my brain accepts this.

Dec 12, 2012#40

MindDetective

Krisken said:
It really is as simple as 1/3 of 1 is .333(repeating), .333(repeating) times 3 is .999(repeating), 1=.999(repeating). Not that my brain accepts this.

That is because there is some rounding going on here as well. It is again occurring in the infinity place, so is excused (or goes unnoticed).

Dec 12, 2012#41

strawman

MindDetective said:
That is because there is some rounding going on here as well.

Only if you don't accept that 0.999... = 1.

If you do not accept this, however, then calculus stops working.

Dec 12, 2012#42

WasabiPoptart

Dec 12, 2012#43

MindDetective

stienman said:
Only if you don't accept that 0.999... = 1.

If you do not accept this, however, then calculus stops working.

Well, that's an oversimplification. You don't have to accept it as true, only as close enough.

Dec 12, 2012#44

tegid

MindDetective said:
In fact I will go 1 step further and offer my opinion. 0.9999... + 9 definitely preserves all of the infinite 9s in the decimal places. 0.9999... * 10 may not preserve them all, effectively losing the 9 in the infinity place. In Chad Sexington 's proof, the assumption is that 9.9999... has preserved all of infinity of the 9s in the decimal place, which to me implies it is more accurate to say that 9 has been added rather than 10 has been multiplied.

*edit* Argh, I should have said I would go 0.9999... steps further. Opportunity missed.
[DOUBLEPOST=1355341504][/DOUBLEPOST]

No, the answer to this question informs us to the answer on that one.

Wait, then where you have the problem is with the infinity of decimal places. 0.999... x 10 =9.999... because, since there's an infinity of 9's after the coma, you don't lose any. There isn't a last 9 that gets shifted out of place. You can accept that without assuming that 0.99...=1 (except it's an immediate consequence so it's normal to equate the two).

Let's try it with a different periodicity in the decimals: Would you accept that 0.898989... x 10 = 8.989898... and 0.898989... x 100 = 89.8989... preserving an infinity of 89's in the decimal places? (although in the case where we multiply by 10 they get shifted in position, if you identify them by that)?

Dec 12, 2012#45

Chad Sexington

If you have a pattern of 110110110... that repeats infinitely, are there enough zeroes to pair up one 0 to each 1? (yes.) Infinity, man. You can't lose it.

Dec 12, 2012#46

MindDetective

tegid said:
Wait, then where you have the problem is with the infinity of decimal places. 0.999... x 10 =9.999... because, since there's an infinity of 9's after the coma, you don't lose any. There isn't a last 9 that gets shifted out of place. You can accept that without assuming that 0.99...=1 (except it's an immediate consequence so it's normal to equate the two).

Let's try it with a different periodicity in the decimals: Would you accept that 0.898989... x 10 = 8.989898... and 0.898989... x 100 = 89.8989... preserving an infinity of 89's in the decimal places? (although in the case where we multiply by 10 they get shifted in position, if you identify them by that)?

Sure you do. You lose it in the infinity place! That's like saying infinity minus 1 is infinity. It's true but we can't do much with that information. You've effectively shifted infinity over by 1 decimal place. But we don't have a notation for that, so we round it off and say good enough.

Dec 12, 2012#47

strawman

MindDetective said:
Well, that's an oversimplification. You don't have to accept it as true, only as close enough.

[DOUBLEPOST=1355348500][/DOUBLEPOST]

MindDetective said:
Sure you do. You lose it in the infinity place! That's like saying infinity minus 1 is infinity. It's true but we can't do much with that information. You've effectively shifted infinity over by 1 decimal place. But we don't have a notation for that, so we round it off and say good enough.

You keep using that word. I do not think it means what you think it means.

Dec 12, 2012#48

MindDetective

stienman said:
You keep using that word. I do not think it means what you think it means.

The word "that"?

Dec 12, 2012#49

Krisken

Thank you, fellas. You've given me an opportunity to post this.

Dec 12, 2012#50

strawman

MindDetective said:
The word "that"?

The word infinity. The following provides several proofs, but more interestingly a few posts point out that this depends on what 0.999... actually means, what real numbers are (and aren't), and the fact that you can go both ways depending on your assumptions. This touches on one of the foundations of math:

http://math.stackexchange.com/questions/11/does-99999-1

Dec 12, 2012#51

Krisken

Snarky image aside, I really am enjoying your discussion.

Dec 12, 2012#52

MindDetective

stienman said:
The word infinity. The following provides several proofs, but more interestingly a few posts point out that this depends on what 0.999... actually means, what real numbers are (and aren't), and the fact that you can go both ways depending on your assumptions. This touches on one of the foundations of math:

http://math.stackexchange.com/questions/11/does-99999-1

I honestly am not sure what you're getting at. I thought you were arguing that they were equal but now it seems like you are saying "it depends on how concepts like infinity are defined", which really doesn't put you in much of a place to tease me about how I'm using the word myself.

Dec 12, 2012#53

Krisken

How does math define infinity?

Dec 12, 2012#54

strawman

I'm arguing that we've reached the point in the discussion where we should realize we're arguing semantics.

This all depends on whether 0.999... is actually unending. If it never ends, then there is no "infinity place" and your solution doesn't work. It must equal 1 simply because there is no number between 0.999... and 1, therefore they are the same number.

If, however, your version of infinity allows for an "infinity place" and is thus countably infinite then there is a number between 0.999... and 1, thus they are separate and distinct numbers.[DOUBLEPOST=1355349944][/DOUBLEPOST]

Krisken said:
How does math define infinity?

Which version of math?

:awesome:

Dec 12, 2012#55

Krisken

So do mathematicians who focus on different fields punch each over over this kind of thing?

Dec 12, 2012#56

Chad Sexington

Krisken said:
So do mathematicians who focus on different fields punch each over over this kind of thing?

Well, no, not really. It's just that sometimes you do things with complex numbers, sometimes with real numbers, natural numbers... Someone doing calculus would know that .999...=1 in the real number system, while also knowing that there are systems wherein you can have an infinite infinitesimal number .999... that is not 1.

Dec 12, 2012#57

strawman

Krisken said:
So do mathematicians who focus on different fields punch each over over this kind of thing?

It's really fun to watch mathematicians fight.

Dec 12, 2012#58

Krisken

Dec 12, 2012#59

tegid

So, I(we)'m using infinity as neverending and Mind Detective is using it as very, very large, so large you can't wrap your head around it?[DOUBLEPOST=1355354121][/DOUBLEPOST]

stienman said:
If, however, your version of infinity allows for an "infinity place" and is thus countably infinite then there is a number between 0.999... and 1, thus they are separate and distinct numbers.

Is that what countably infinite means? I'd have said that the 'regular' 0.999... has a countable infinite amount of 9's. I don't even know if it applies, since it's not a set.

Dec 12, 2012#60

strawman

Well, I'm not sure infinity actually applies here. You can count the number of nines after the decimal place, but it's actually not useful. Sure, there are an infinite number of them, but who cares?

Along the same lines is wondering whether 0.999... is rational. Can it be represented by an integer fraction?

Dec 12, 2012#61

Chad Sexington

MindDetective I've thought about this and I think I've come to the conclusion I lack the expertise to give a simple enough explanation. I was originally caught off guard that you didn't agree with .999...=1, and thought you just didn't like my example (which I continue to defend as not flawed!

). I suppose the only argument I have left to make is that if you agree 1/3 + 1/3 + 1/3 = 3/3, do you then think 3/3 is only approximately 1? Or do you think .333... is only approximately 1/3? If you accept that 1/3 = .333... then surely you must accept that 3/3 = .999... and therefore .999... = 1, not approximately but absolutely.

But I think that is essentially covered in this thread and I suspect you feel 1/3 is only aproximately .333... that somewhere in the infinite digits there is a change that accounts for the missing ....1 at the end. I feel like it's something I understand clearly and yet I'm reminded of Einstein's "If you can't explain something simply, you don't understand it well enough." so I maybe just have to admit I'm not the man to explain this to you.

But I'm right, dammit! I'M RIIIIIIIIIIIGHT.

Dec 12, 2012#62

PatrThom

stienman said:
Which version of math?

Why…New Math, of course.

stienman said:
Along the same lines is wondering whether 0.999... is rational. Can it be represented by an integer fraction?

Not in base ten, no.

--Patrick

Dec 12, 2012#63

strawman

If not, then this is an irrational discussion.

Dec 12, 2012#64

Krisken

I wish I could choose more than one rating, Stienman.

Dec 12, 2012#65

bhamv3

Krisken said:
I wish I could choose more than one rating, Stienman.

Tell me what other rating you wanna give him and I'll do it for you. In your name.

Dec 12, 2012#66

MindDetective

stienman said:
It must equal 1 simply because there is no number between 0.999... and 1, therefore they are the same number.

See, this part bugs me. It doesn't seem necessary to me that a number has to be in between them for them to be different. A question that popped into my head as I was driving around doing errands this evening was spurred by seeing this argument on the link you gave. What is the closest number to 1 smaller than 1 that is not 1?[DOUBLEPOST=1355365630][/DOUBLEPOST]

Chad Sexington said:
MindDetective I've thought about this and I think I've come to the conclusion I lack the expertise to give a simple enough explanation. I was originally caught off guard that you didn't agree with .999...=1, and thought you just didn't like my example (which I continue to defend as not flawed! ). I suppose the only argument I have left to make is that if you agree 1/3 + 1/3 + 1/3 = 3/3, do you then think 3/3 is only approximately 1? Or do you think .333... is only approximately 1/3? If you accept that 1/3 = .333... then surely you must accept that 3/3 = .999... and therefore .999... = 1, not approximately but absolutely.

But I think that is essentially covered in this thread and I suspect you feel 1/3 is only aproximately .333... that somewhere in the infinite digits there is a change that accounts for the missing ....1 at the end. I feel like it's something I understand clearly and yet I'm reminded of Einstein's "If you can't explain something simply, you don't understand it well enough." so I maybe just have to admit I'm not the man to explain this to you.

But I'm right, dammit! I'M RIIIIIIIIIIIGHT.

It only approximates 1/3 because it only ever CAN approximate it. When you take 1 and divide it by 3, you get a 3 with a remainder of 1, which is then divided by 3, etc. After each 3 is a remainder of 1 that still needs dividing. The remainder is essentially dropped when we say "it is 3s for infinity", partly because the way we use infinity is a short hand for rounding. In my mind it is 3 for infinity with a remainder of 1, even at the "end" of infinity (I know, I know, there is no "end"...but that is why the remainder should still be considered as hanging around, I think!)

Dec 12, 2012#67

papachronos

Chad Sexington said:
Your setup makes no sense. In step two you don't multiply both sides by 10. You add 9 to the left and multiply the right by 10. That's not equivalent.

MindDetective said:
This is it exactly. It is the exact same problem with your set up.

MindDetective , the method you used to "break" the proof is incorrect. The initial setup doesn't multiply one side by ten and add nine to the other side and call it equivalent:

if x = a then 10x = 10a is true for every value of a. More generally:

if x = a then (base)*x = (base)*a.

For any rational number (i.e. one that can be expressed as a fraction, be that in numerator/denominator format or by using a radix point), multiplying by that number by its representational number base is equivalent to shifting the radix point one space to the right.

Therefore, x = 0.999... <=> 10x = 9.999... is a valid statement. The only assumption implicit in the second statement is that everything is happening in base 10.

Dec 12, 2012#68

MindDetective

papachronos said:
MindDetective , the method you used to "break" the proof is incorrect. The initial setup doesn't multiply one side by ten and add nine to the other side and call it equivalent:

if x = a then 10x = 10a is true for every value of a. More generally:

if x = a then (base)*x = (base)*a.

For any rational number (i.e. one that can be expressed as a fraction, be that in numerator/denominator format or by using a radix point), multiplying by that number by its representational number base is equivalent to shifting the radix point one space to the right.

Therefore, x = 0.999... <=> 10x = 9.999... is a valid statement. The only assumption implicit in the second statement is that everything is happening in base 10.

You should read the rest of the thread because I've been arguing that the "working" proof does exactly the same thing I did. It only adds 9 to one side and multiples the other side by 10, not multiple both sides by 10.[DOUBLEPOST=1355366735][/DOUBLEPOST]My argument is thus:
.99999...=x
9.99999...=9+x
9.99999... - .99999...=9 and 9+x - x=9
9 = 9
x is still = .99999...

Dec 12, 2012#69

papachronos

MindDetective said:
You should read the rest of the thread because I've been arguing that the "working" proof does exactly the same thing I did. It only adds 9 to one side and multiples the other side by 10, not multiple both sides by 10.

That's just it, though - the working proof doesn't do this. Well, okay, it does, but only as a side effect. It takes the more general proof and applies it to a specific value of a.

Working proof: x = a <=> 10x = radixpointshift(a)
Your "broken" proof: x = a <=> 10x = 9 + a

The two methods are incompatible. The second is only valid for a = 1 (which is what you're saying), otherwise algebraic law is broken. The first is applicable to any value of a, and algebraic law is never broken.

Dec 12, 2012#70

papachronos

MindDetective said:
My argument is thus:
.99999...=x
9.99999...=9+x
9.99999... - .99999...=9 and 9+x - x=9
9 = 9
x is still = .99999...

Problem with that method is that you remove x from the proof entirely in the third step, so you end up with an identity statement which doesn't prove anything.

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Dec 12, 2012#71

MindDetective

papachronos said:
That's just it, though - the working proof doesn't do this. Well, okay, it does, but only as a side effect. It takes the more general proof and applies it to a specific value of a.

Working proof: x = a <=> 10x = radixpointshift(a)
Your "broken" proof: x = a <=> 10x = 9 + a

The two methods are incompatible. The second is only valid for a = 1 (which is what you're saying), otherwise algebraic law is broken. The first is applicable to any value of a, and algebraic law is never broken.

papachronos said:
Problem with that method is that you remove x from the proof entirely in the third step, so you end up with an identity statement which doesn't prove anything.

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That's basically my point.

Dec 12, 2012#72

papachronos

Like this:

x = a
9 + x = 9 + a
9 + x - x = 9 + a - a
9 = 9
Therefore, x still equals a... which was never in doubt in the first place.

Dec 12, 2012#73

MindDetective

You haven't said anything I don't already know, nor argues against my point. I was using those proofs illustratively to demonstrate the very things you've pointed out more explicitly. As said earlier in the thread, it basically boils down to: is 9 + 0.99999... == 10*0.99999...

My contention revolves around the number of decimal places in each one. In the first, there is infinite decimal places. In the second, there is infinite-1 decimal places...which we treat as infinite decimal places.

I contend that 9+0.99999... is a less ambiguous interpretation of the "working" proof than 10*0.99999... because it doesn't in any way adjust the number of decimal places. That's where I think some implicit rounding occurs.

Dec 12, 2012#74

papachronos

I see. It's an issue of real number construction and the presence or absence of non-zero infinitesimals.

But we've gone quite far afield now from the original topic, and I feel bad enough having inflicted Halforums to this discussion without adding number theory on top of it.

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Dec 12, 2012#75

Allen who is Quiet

What else are we going to read? Wholesome literature?

Dec 13, 2012#76

PatrThom

MindDetective said:
I contend that 9+0.99999... is a less ambiguous interpretation of the "working" proof than 10*0.99999... because it doesn't in any way adjust the number of decimal places. That's where I think some implicit rounding occurs.

If you want to approach it quasi-arithmatically, then remember that multiplication is just a shorthand form of addition, so "10x" is exactly equivalent to writing "x+x+x+x+x+x+x+x+x+x," and therefore:

Code:

10 * 0.999...

is in fact exactly the same as stating:

Code:

  0.999...
  0.999...
  0.999...
  0.999...
  0.999...
  0.999...
  0.999...
  0.999...
  0.999...
+ 0.999...
--------
  9.999...

which is in fact completely different from stating:

Code:

  0.999...
+ 9.000...
--------
  9.999...

...which can be verified simply by inspection. If anything, you accidentally show that 9.999... is the same as 9.999... (both methods produce the same result), which would actually prove that 9x0.999...=9 and therefore that 0.999...=1.

Now each of the ten terms in that tall addition problem above all have infinite significant digits. Thus it follows that there is no "lost" digit on the end, because every digit always has another 9 to the right of it to keep it from "rolling over" to zero. Your argument that you can make an equivalent by saying, "We are not multiplying by 10, we are adding a 9 in front of it" makes no sense for the same reason I can't arbitrarily say that "we are not multiplying by ten, we are adding a zero on the end." If that were true, then 10 x 0.999... would instead be 0.999...0 which is impossible (that rule has a hidden restriction, which is that the number you are multiplying by 10 can't have any significant digits after the decimal point). Even if you try and rationalize it by moving the decimal point and then sticking a zero into the spot which "opened up" (10x0.999... = 9.999...0) this "rule" still fails because there is no end, no terminus where you could stick that zero. The hard part to visualize is that the series of nines is not dynamically growing and pushing out into infinity faster and faster as you race to put your zero on the "end." Instead it is static. Its foreverness does not change since it was already forever when you started. You could count off a dozen googolplexes of 9's as you try to reach the end, but once you had done so, you would literally be no closer to the "end" than when you started. Not "figuratively" no closer, but literally. Even half of forever is still forever. This is what "Infinite" means.

You can convert it to other bases if you really want, but the result ends up being the same.
0.999... (base 10) is exactly equal to 0.222... (base 3) or 0.FFF... (base 16). It is easiest to see in (base 2) because 0.111... perfectly illustrates the concept of "there is no room for any number to exist between 0.111... and 1.000..."

If you want to look at this algorithmically instead of algebraically, the concept being stated is that a zero followed by an infinite series of digits-which-are-all-one-unit-less-than-the-base will always be the same as (or simplify to) one unit.

Infinity is a hard concept to grasp, and transcends such concepts as, "how many." In Mathematics, you say something is "Infinite" not because it would be impractical to determine (like how many atoms exist in the Universe), but because it would be impossible to quantify (like how many real numbers exist between 0 and 1).

--Patrick

Dec 13, 2012#77

Ravenpoe

My best friend is a mathematics professor with a PhD in mathematics. I asked him about this, and he told me that 0.999... is 1 and you guys are dumb.

That's about all I have to contribute to this.

*I didn't actually ask him specifically about this, but we've had the exact same conversation before. I added the you guys are dumb part.

Dec 13, 2012#78

tegid

MindDetective said:
You haven't said anything I don't already know, nor argues against my point. I was using those proofs illustratively to demonstrate the very things you've pointed out more explicitly. As said earlier in the thread, it basically boils down to: is 9 + 0.99999... == 10*0.99999...

My contention revolves around the number of decimal places in each one. In the first, there is infinite decimal places. In the second, there is infinite-1 decimal places...which we treat as infinite decimal places.

It's really a matter of defining infinity. What we're all saying, I think, is that infinity-1=infinity, due to how infinity works. In your explanation of 1/3 where you keep adding decimal 3's, you say that we approximate the division by 'an infinity of 3's' because we're not able to get to the last decimal place, but there's really an infinity of 3's because you know every division begets the same result and you'll keep getting 3's. There's no last decimal to the division unless you decide to stop it... if you do the 'exact' division, you actually get the infinity of 3's!
Anyway, I don't think we're going to agree here.

I contend that 9+0.99999... is a less ambiguous interpretation of the "working" proof than 10*0.99999... because it doesn't in any way adjust the number of decimal places. That's where I think some implicit rounding occurs.

Even if you interpret it that way, your step would yield
9.9999...=9+x instead of
9.9999...=10x, right? So your breaking it doesn't really work.
(Please, if you answer prioritize this second part. I think it may be more productive.)

Dec 13, 2012#79

MindDetective

tegid said:
9.9999...=9+x instead of
9.9999...=10x, right?

No, thats not true.[DOUBLEPOST=1355413692][/DOUBLEPOST]

PatrThom said:
Infinity is a hard concept to grasp, and transcends such concepts as, "how many." In Mathematics, you say something is "Infinite" not because it would be impractical to determine (like how many atoms exist in the Universe), but because it would be impossible to quantify (like how many real numbers exist between 0 and 1).

--Patrick

I think therein lies the problem.[DOUBLEPOST=1355414056][/DOUBLEPOST]

Ravenpoe said:
you guys are dumb.

I may very well be. I like playing devil's advocate with my math friends, which might be what makes me dumb.

Dec 17, 2012#80

fade

Do you remember the scene in Police Academy where Tackleberry shows up after the gunfight and realizes he missed it? That's me right now.

Dec 17, 2012#81

PatrThom

fade said:
Do you remember the scene in Police Academy where Tackleberry shows up after the gunfight and realizes he missed it? That's me right now.

Feel free to correct/comment any misconceptions or inconsistencies you see. That way we'll all learn. Just remember to show your work.

--Patrick

Dec 17, 2012#82

Krisken

fade said:
Do you remember the scene in Police Academy where Tackleberry shows up after the gunfight and realizes he missed it? That's me right now.

Yes, but you can rekindle it. You have the power.

Dec 17, 2012#83

strawman

It's only over if no one starts shooting again.

Take up your weapon and smite your enemies!

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