The Instigator
abard124
Pro (for)
Losing
23 Points
The Contender
dvhoose
Con (against)
Winning
29 Points

.99999...(repeating) is equal to 1

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Post Voting Period
The voting period for this debate has ended.
after 12 votes the winner is...
dvhoose
Voting Style: Open Point System: 7 Point
Started: 6/1/2009 Category: Miscellaneous
Updated: 7 years ago Status: Post Voting Period
Viewed: 4,135 times Debate No: 8506
Debate Rounds (3)
Comments (21)
Votes (12)

 

abard124

Pro

Hi!

I usually like debating serious political issues, but I though tit would be fun to debate this little thing I figured out because i have no life...

.99999...=1

I'll prove it after someone accepts.

Thanks!
dvhoose

Con

Despite all the "proofs" I've seen for this so called "fact" I still remain skeptical and believe otherwise.

To keep everything fair, I'll allow you the first argument.

Somebody (me) has accepted, and now I believe it's time for you to prove it ;)

Hope to see a great debate, and good luck!
Debate Round No. 1
abard124

Pro

Thank you for accepting!

I have a few different ways to look at it, but I think the most straightforward is this:

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

Let's walk through that.

we are trying to prove that .9(r)=1
If you divide them by 3 using the division property of equality, you get the same answer. By the same property, that proves their equality.

Here's another (using inductive reasoning)

1/9=.1(r)
2/9=.2(r)
3/9=.3(r)
4/9=.4(r)
5/9=.5(r)
6/9=.6(r)
7/9=.7(r)
8/9=.8(r)
9/9=.9(r)--Or does it?

if each time you add 1/9, or .1(r), what happens when you add .1(r) to .8(r)? You get .9(r), but you also get 1. Well, you learned in third grade that anything divided by itself equals 1, so 9/9 obviously is worth 1, but using inductive reasoning, you'd find that it is worth .9(r). This leads to the only conclusion that .9(r)=1

Looking forward to your response!
dvhoose

Con

The proofs my opponent has provided are fairly common ones at making the case that .9(r)=1

To counter these proofs, I have a couple of proofs myself, but before I get there, I want to talk about the number .9(r).

.9(r) relies upon the concept of infinity; that the numbers never end. Ever! This concept is impossible for the human mind to grasp. We're not able to understand that things go on forever, after all, everything in our world ends. It's all finite. How far from here to there? How many jellybeans in the jar? How much money does it cost? It's all finite. Even when we take things suck as How far from here to the edge of the universe? How many jellybeans in a grain silo? How much money does Warren Buffet have? While astronomical, these numbers are all finite.

This relates to the debate, because though .9(r) gets infinitely closer to 1, it never reaches 1, which brings me to my first proof.

PROOF 1: PROOF BY DECIMAL

The number 1 can be written as 1.0, or in the case of this debate 1.0(r) [the 0's would continue forever]. When we compare this to the number .9(r), we forget that there is a "0" in the ones place, making the number 0.9(r).
Clearly, just a 1.0 > .9 and 1>.99 and 1>.999 etc., 1.0>0.9(r)

PROOF 2: PROOF BY SUBTRACTION

Because infinity is such an impossible concept to grasp, lets use an ever increasing sting of nines to help with this proof.

1-.9=.1
1-.99=.01
1-.999=.001
1-.9999=.0001
and the pattern progresses until we reach
1-.9(r)=.0(r)1

PROOF 3: PROOF BY ADDITION

When saying that .9(r)=1, we should be able to add them together in any combination and get the same result.

1+1=2
1+.9(r)=1.9(r)
.9(r)+1=1.9(r)
.9(r)+.9(r)=? <-- This is the one I'd like to focus on.

Lets use the ever-increasing string of 9's to help with this as well.

.9+.9=1.8 [notice, no 9]
.99+.99=1.98 [only one 9]
.999+.999=1.998 [two 9's]
.9999+.9999=1.9998 [three 9's]
We can expect this pattern to continue ever onward, eventually leaving us with
.9(r)+.9(r)=1.9(r)8

2 =/= 1.9(r) =/= 1.9(r)8

These three proofs should be enough to prove that 1 =/= .9(r)

With that, I'll turn the debate back over to my opponent. I look forward to another great response.
Debate Round No. 2
abard124

Pro

Proof by decimal
_________________
Inductive reasoning works almost all the time. Clearly, my inductive reasoning contradicts with yours, so for the purposes of this debate, I propose that any and all uses of inductive reasoning are null and void, regardless of who said them.

Proof by subtraction
_________________
.0(r)1? Is that legal? I don't think so... The only way to attain it is subtracting .9(r) from 1. This is a moot point anyway, because if I am correct, .0(r)1 (which I don't think exists anyway) is equivalent to 0. And once again, your inductive reasoning could be wrong.

Let me delve into why you can't have .0(r)1

As you said, infinity is a very confusing and deceptive concept. You can't put a digit after #infinity, because there is no # infinity. Because if you put the one after the repeating 0's, you lose the infinite quality of the zeroes, in that there is nothing after them.

Proof by addition
_________________
Once again, it's not infinite if you out something after it, so 1.9(r)8 is impossible. and once again, you used inductive reasoning. I know I did, but I also had a solid deductive proof, which you do not.

Also, you claim that 1.9(r)=/=1.9(r)8, which raises an interesting point dealing with infinity, and goes yet again to show why there can't be another digit.

1.9<1.98, but 1.99>1.98. When the nines are infinite, though, which is it? There's always one more nine on each one than the other, because infinity does not act like a number. Maybe they are equal. Then it does not go against my point, because if 1.9(r)=2, so would the elusive 1.9(r)8. But that can't be the case, therefore, you can't repeat and have it end with something else.

In my argument, I used inductive reasoning to supplement my mathematical proof. You used only empirical observations which do not necessarily hold up to infinity. In your argument, you need to PROVE it mathematically, not just observe what happens before you insert infinity.

I have one more algebraic proof, then I will let you close it up!

x=.9(r) given
10x=9.9(r) mult. prop. of eq.
10x-x=9.9(r)-x sub. prop. of eq.
9x=9 simp.
x=1 div. prop. of eq.
x=x reflexive
.9(r)=1 substitution

This has been a very interesting debate, and you made some very good (but empirical) points. I look forward to your response, and may the better mathematician (regrettably probably neither of us :-/) win!
dvhoose

Con

A very well written response by my opponent.

For this final speech, I want to take a look at all of my opponents proofs, and explain why they aren't valid, explaining why 1=/=.9(r) after each. Then I'll look to my opponents 3rd round and comment as necessary.

[Division by 3]

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

This proof says that the two must be equal because you can divide by 3 and get the same answer. This entire conflict is caused by the precision infinity allows.

With each new '9' added to .9(r), it gets closer to 1. Just look at my proof by subtraction above. Whether you believe in the end result is irrelevant for now, just note that with each '9 after the decimal point, the number gets closer and closer to one.
Now, what happens when the difference becomes so small that it's impossible to tell a difference between them? A great example of this is dividing by 0. It can't be done, but you can get really close. With 1/x, as x approaches 0, 1/x gets progressively bigger. Can x ever be 0? No, it'd give us an invalid answer. Same concept here. Eventually, the numbers become so close together that a difference is neigh impossible to tell, but exists nonetheless.

[The Mysterious Ninths]

The same fallacy is seen in the 1/9's proof. 9/9, we're taught, equals one. but 1/9 * 9 = .9(r). Surely they're the same thing? No. The numbers are just so close together that for all intensive purposes, they act the same. But a difference is there. I'll get to how we know this in a little (actually, I've already covered it, but I'll reaffirm it later)

[Solving for 'x']

And to cover the final proof, found at the end of my opponents final speech.
Infinity makes this proof an extremely hard one to grasp. With .9(r), when you multiply by 10, it becomes 9.9(r). But according to common math ideas, that would shift the decimal point. It'd be .9(r) with one less nine. But it's infinity! There is no end; there is no "one less 9." Nevertheless, the two numbers are once again so close together that humans cannot tell a difference, but one exists.

BUT HOW DO WE KNOW?

Simple, really. Look to my Proof by Decimal from Rd.2. Both PRO and CON show examples of how .9(r)=1 and how .9(r)=/=1. We can't both be right! And we aren't. Prefer CON's argument because I've shown how it's easily mistaken that .9(r)=1 (Incredibly unnoticeable gap) yet proven that a gap exists.

Just look at the ones digit of both numbers.
1.0
0.9(r)

1>0 thus 1>0.9(r)

Don't discard all evidence used in this debate, instead adapt to fit the truth I've shown and the easily made misconception due to the accuracy of infinity.

Proof By Subtraction
-----------------------

We're dealing with infinity. .0(r)1 is legal, but hard to understand. In the real world, no it doesn't exist. It's simply a concept used to experiment with infinity, or representing an infinitely small number. .0(r)1 is as close to 0 as .9(r) is close to 1.

Proof By Addition
--------------------

You raise a good point on which is greater, .9(r) or .9(r)8. Infinity is certainly a tricky concept and I honestly don't know the answer to it. Even with this concession, however, the first argument in this speech still stands. It looks like this:

Units digit is all you need to look at here.
0.9(r)
1.0

Because 1>0, 1.0>0.9(r).
The misconception [that 1=.9(r)] occurs because infinity is so precise, so accurate, that the difference is so small, so insignificant, that the two numbers ACT as one, yet are different. You might say that .9(r) is 1's stunt double. (Hollywood would have a fit with this one, no?)

A miniscule difference exists, one that allows the two DIFFERENT numbers to ACT as the SAME.

A very well fought debate on my opponents side, I wish him luck :)

Resolution Negated, Vote CON
Debate Round No. 3
21 comments have been posted on this debate. Showing 1 through 10 records.
Posted by abard124 7 years ago
abard124
Damn... I thought I had this one in the bag... Oh well, I suppose I can't win everything... Congrats to my opponent, and pj, whatever happened to that automatic win that you claimed was nearly inevitable??? I suppose I beat you at your own game because I lost this one... So it's like some awful backhanded victory for me...
Posted by abard124 7 years ago
abard124
If it was "maliciously weighted" and "factual," would (a) anyone have taken it, and (b) would I actually be losing if that were the case? Think about it.
Posted by PoeJoe 7 years ago
PoeJoe
Conduct goes to CON. PRO started a maliciously weighted, factual debate.

Argument is a tie, because PRO gets penalized.
Posted by s0m31john 7 years ago
s0m31john
So an infinite amount of men walk into a bar. The first guy says, "I'll have a beer". The second guy says, "I'll have half a beer". The third guy says, "I'll have a quarter of a beer". And so on.

The bartender says "f*ck you guys" and pours 2 beers.

That bartender knows his math. And so should you guys before opening your mouth.

.999~ = 1.
Posted by tBoonePickens 7 years ago
tBoonePickens
One cannot divide an infinite amount of times and get a result because the result can only be attained once the calculation is completed and a calculation that has an infinite number of steps will never be completed and thus never have a result. Therefore, .9(r) itself must be rounded in order to use it in any calculation as it would require an infinite number of steps to complete the calculation otherwise and thus not be able to yield a result.
Posted by dvhoose 7 years ago
dvhoose
You just said, and I agree, that one can take any number and continuiously divide it by two. By doing this, one would NEVER reach 0. But when we do it an infinite number of times, we will? How is that possible? It's not. It's IMpossible.
Posted by abard124 7 years ago
abard124
True, but infinity does not work the same as finite numbers.

Example: Take any number and divide it by 2. Then repeat it any number of times. You will never got to zero. You will get really close, but you will never get there. Now if you divide it an infinite amount of times, you will reach zero.

As you said, no matter how many 9's come after the decimal point, you won't reach one, but with infinity 9's, you will.
Posted by dvhoose 7 years ago
dvhoose
The Lysol analogy wasn't intended to replicate the situation. NOTHING in our world is infinate, and thus, no analogies exist. My analogy was merely intended to show how humans tend to hear a fact, and assume another fact that's relatively close. Such is the case with 1 and .9(r)
Posted by abard124 7 years ago
abard124
Exactly what you said... Humans can't comprehend infinity... It is such a strange concept that it is actually very hard to grasp. Your Lysol analogy doesn't work at all, because it didn't deal with infinity. 99.99%=/=99.9(r)%.
Posted by dvhoose 7 years ago
dvhoose
"Just want to clarify...
In your last argument, when you mention the inability for humans to detect a difference...
How does that work? I mean, math acts the same with infinity... Patterns don't, but math does...
Can you explain that? I'm interested..."

Correct me if I'm wrong, but you're asking why human perception has anything to do with this?

Basically, because humans are unable to comprehend infinity, the two numbers appear to be the same, through all the proofs you showed and a couple others I've seen. Go to youtube and search for 1= -1 or 2+2=5 and watch all the youtube videos where people have seemingly done the impossible. It's all just tricks and mis-manipulations of numbers, and this is no different. Because it all deals with precision (.99 is closer than .9, .999 is closer than .99, etc) by the time we reach an infinite amount of digits, the difference is so small that humans can't tell a difference, thus, our minds make the simple jump that the two are equal. It's like Lysol, claiming that their products kill 99.99% of bacteria. Does it kill all of them? No. Does it kill enough that we think it's killing all of them? Yes. So we make the jump and assume Lysol kills all germs.
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