The Instigator
Commondebator
Pro (for)
The Contender
Con (against)
Anonymous

.999...(repeating) =1

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Voting Style: Open Point System: 7 Point
Started: 8/16/2017 Category: Science
Updated: 3 years ago Status: Debating Period
Viewed: 1,655 times Debate No: 103562
Debate Rounds (4)
Comments (32)
Votes (0)

 

Commondebator

Pro

This is a debate on whether or not .999...(repeating)=1.

First round is acceptance.

I'm looking forward/curious to see any opposing arguments on how .999...cannot equal 1.

No semantics, no trolling.

Kewl.

Con

no, it never reaches 1
Debate Round No. 1
Commondebator

Pro

Introduction:

In order to prove that 0.999...=1, I will be showing this via 3 different mathematical methods. While the numerical portion of this argument is the main argument, I would briefly try to explain using words how exactly .999...=1. 0.999 repeating is a number that can only exist within the realm of mathematics. 0.999 repeating is a number with infinite accuracy, so it's not something that can possibly exist in real life. However, using mathematics, a number that has an infinite number of "9s" after the decimal place would evaluate to 1. This is simply because the limit of 0.999 repeating written as an infinite geometric series would evaulate to 1, something I will show later in this argument. 0.99... may not seem like it ever reaches 1, but if we evaluate the limit of a geometric series that represents 0.999 with an infinite number of 9s, it would indeed be calculated as 1.

C.1 Fractions

This is the simplest argument to show that 0.999...=1.

1/3=0.333....
Adding 1/3=Adding 0.333...
2/3=0.666...
Adding 1/3=Adding 0.333...
3/3=0.999....

And as it turns out, 3/3 which simplifies to 1, does indeed equal 0.999....

C.2 Algebra


x=0.999...
10x=9.999....(Multiply by 10 to both sides)
10x=9+0.999....(We broke up 9.999...to 9+0.9999)
10x=9+x (0.999...is equal to x, as seen in step 1)
9x=9 (Subtract both sides by x)
x=1 (Divide both sides by 9)

C.3 Limits and geometric series

Let's try writing 0.99...as a fraction, and see what we get. Since this number isn't irrational, it can be written as a fraction. In order to do that, we would have to rewrite this number as an infinite geometric series, and then find the sum of that series. The limit of that series as the number of terms tend to infinity, would be 1. While the limit isn't telling you the value of the function at that number, it is telling you what that function is approaching as you get closer and closer to some number (in this case, it's infinity). Since 0.999...as a geometric series has an infinite number of terms, the limit would tell us what the value if 0.999...had an infinite number of "9s". Which it does.

Sum of finite geometric series:

Sn=a1((1-r^n)/(1-r)

Where

Sn=Sum of n terms
a1=First number in series
n=Number of terms
r=common ratio

If con requests, I can derive this formula.

However, if we assume that r is greater than 0 and less than 1, we can take the limit of the function as n tends to infinity. Therefore, the sum of an infinite number of terms would be...

Sinf.= a1/(1-r).

(r^n=0, as n tends to infinity. Therefore, we're left with a1/(1-r))

If con requests, I can do a more rigourous proof, though this is the formula for the sum of an infinite geometric series.

So, we can now rewrite 0.9999... as...

0.9+0.09+0.009+....

Which is equal to

9/10+9/100+9/100+....

In this series,

a1=9/10
r=1/10

Since |r|<1, we can use the formula of Sinf.

Sinf.=9/10/(1-1/10)
=9/10/(9/10)
=1

And that's it. Using the definition of the sum of an infinite geometric series, we figured out that 0.999...as a fraction does indeed evaluate to be 1. Since we can rewrite 0.999 as an infinite geometric series, we can evalaute it's sum in order to figure out it's fractional form. And that form turns out to be 1/1.

This debate doesn't really need sources, since it's just math. However, if Con likes, I can show sources of the formula for the sum of a finite and infinite geometric series.

Con

but it dosnt equal true, if thats what you are saying, it never gets to equal as something true like 1 soda, what is a 0,9999 soda?

dividing it into 3 parts and saying those 3 which neither equates to 1 is 1 all togeather, as if you have changed or added something... then you are simply wrong

its like all you are arguing is that 3/3 is a FULL fraction.. that dosnt make it valid, you cant just change around the value of numbers, they have actual usage.. what does, 0,3 point to in real life? a stone? which is one?

3=1+1+1

your faker 3(0,33..+0,33..+0,33..)=not 3

i assume the rest of it is non sense as well
Debate Round No. 2
Commondebator

Pro

Rebuttal:

"it never gets to equal as something true like 1 soda, what is a 0,9999 soda?"

Con's example fails here since Con is trying to use a real world example/object to describe a concept such as infinity. It literally cannot be done since we can never have infinite accuracy and everything in our universe is bounded by an finite number of "stuff". If you could theoretically measure 0.999... of a soda, then you would have 1 full soda. The reasoning behind this is something I have already posted, something that Con fails to adress.


"its like all you are arguing is that 3/3 is a FULL fraction.. that dosnt make it valid, you cant just change around the value of numbers, they have actual usage"

Again, this seems to be an argument from incredulity. Just because 0.999... doesn't look like 1, doesn't mean it's not 1. 0.999... and 1 have the same exact same value, just like how 0.5 and 1/2 have the exact same value. 3/3 and 0.999... are equivalent statements.

It's also worth mentioning to point out that 0.999...is a decimal number, it's a real number. A number that gets infinitesimally close to 1, but not 1, is also a number, but it's not 0.999... A number that would somehow get infinitesimally close to 1, would not be real number, whereas 0.999...is.

Con seems to be getting confused between this statement

lim(x->1+) f(x)=x

(where a number get's infinitesimally close to 1)

and this statement

lim(n->inf.) 9/10((1-(1/10)^n))/(1-9/10)

(where a repeating decimal number IS 1)

"your faker 3(0,33..+0,33..+0,33..)=not 3"

Again, Con states that this cannot be 3, yet provides no mathematical reasoning or evidence.

"i assume the rest of it is non sense as well"

Extend

Con

0,0000000000.. infnitiy

you are adding things unnecessarily

a soda is 1, not 0,999

0,999 is not 0,999? ...
Debate Round No. 3
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Debate Round No. 4
32 comments have been posted on this debate. Showing 11 through 20 records.
Posted by Anonymous 3 years ago
DeletedUser
wauw you still dont get it, read the rounds dude.. obviusly 0.999 dosnt equal 1.. no matter how you try to use fractions to trick yourself
Posted by Anonymous 3 years ago
DeletedUser
right, something is nothing, i get it
Posted by Anonymous 3 years ago
DeletedUser
hahah
Posted by Anonymous 3 years ago
Commondebator
@vi_spex , at this point you're just trolling.
Posted by Anonymous 3 years ago
DeletedUser
if you add up all the numbers that make up 0.999.. you get 0.999...
Posted by Anonymous 3 years ago
Commondebator
I'm really sorry about that, NDECD1441. I'll try to explain a little more in depth.

The simplest "mathematical" way of proving .999...=1 is just by doing the algebra. Or, just know that 1/3=0.3... 2/3=0.6..., and notice how we're adding 0.3.... every time to get the next fraction? Meaning 3/3=0.9... Or another words, 1=0.9....

0.9... may seem to get really close to 1, but the reality is that with an INFINITE number of "9s", the VALUE of 0.9...=1. A lot of people seem to think that once you're "at infinity", there's still gonna be a 9. That's not true, because you're never going to reach infinity. Here's a more mathematical way of looking at it.

We can rewrite 0.9... as a sequence of terms. Meaning

0.9+0.09+0.009+0.0009....and so on. If you add up all those terms, it would be 0.9999...correct? So we can rewrite this sequence as...

9/10+9/100+9/1000+9/10000.... and so on. We now know that the first term of this sequence is 9/10. The common multiplier is 1/10. If you ADD UP all those numbers for the infinite sequence (there's formula for that, which you can derive), the number turns out to be 1.

Meaning, if you add up all the numbers that make up 0.999..., it all adds up to be 1.
Posted by Anonymous 3 years ago
DeletedUser
you might as well argue that 0 is 1
Posted by Anonymous 3 years ago
NDECD1441
Oh my god, you just said a bunch of equations i dunno how to understand. Could you explain a little more?
Posted by Anonymous 3 years ago
Commondebator
This statement

lim(x->1+) f(x)=x

(where a number get's infinitesimally close to 1)

IS NOT THE SAME as this statement:

lim(n->inf.) 9/10((1-(1/10)^n))/(1-9/10)

(where a repeating decimal number IS 1)
Posted by Anonymous 3 years ago
Commondebator
Masterful, NDECD1441

it's not about whether or not it "reaches" 1. It IS 1. It's just another way of writing 1. The series converges into a single number, which is 1. This is mathematically provable. You're working with infinite accuracy here, so again, it's not something that can be that can be worked with in real life. .9 repeating only exists within the realm of mathematics.
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