# .9 Repeating = 1 YES -or- NO

## Does .9 Repeating equal to 1?   116 members have voted

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.Atlantis
Uhh, explain why I can't subtract 0.9 recurring from 9.9 recurring?

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OPaul

All these proof are all well and dandy but I still don't see how something that approaches 1 can be 1. It never reaches 1, it isn't a real number.

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Dazzla
you can do that. that part is not invalid. your 9x part is.

Hang on hang on, if you get that then what's the problem?

x = 0.9 recurring right?

so 10x = 9.9 recurring right?

With me?

You're fine with 9.9 recurring - 0.9 recurring = 9 right?

So what's the problem?

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Samoa
you can do that. that part is not invalid. your 9x part is.

10x-x=9x

10(.9999) - (.9999) = 9 * (.9999)

9.9999 -.9999 = 8.991

9 = 8.991    that is the flaw

this is the problem DAZZLA

9 times 9 is 81

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bangbang023

ok I'm going to try to break this down but i have to hurry:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

1) now he sets x = 0.9999...

2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999...

3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9

4) the rest is simple

eidt: I think thats what he meant. I started and put on music and lost my train of thought lol.

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Samoa
ok I'm going to try to break this down but i have to hurry:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

1) now he sets x = 0.9999...

2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999...

3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9

4) the rest is simple

your math leaves of 9 times .9999

what does the 9x part just disappear?

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Dazzla
ok I'm going to try to break this down but i have to hurry:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

1) now he sets x = 0.9999...

2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999...

3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9

4) the rest is simple

^^

Bingo.

Do you also disagree with this Samoa?

0.3 recurring = 1/3

0.3 recurring * 3 = 0.9 recurring

1/3 *3 = 1

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OPaul
All these proofs are all well and dandy but I still don't see how something that approaches 1 can be 1. It never reaches 1, it isn't a real number.

Can someone respond to this without giving a proof. Dazzel or who ever is says 1 = .99... tell me how that can be if the number never gets there. If it got there it would be a real number and would no longer approach 1.

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bangbang023
ok I'm going to try to break this down but i have to hurry:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

1) now he sets x = 0.9999...

2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999...

3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9

4) the rest is simple

your math leaves of 9 times .9999

what does the 9x part just disappear?

ok so 10x - x

if you plugin the value for x yo uhave 10(0.999...) - 0.9999 qhich equals 9

but 10(x) - x = 10(0.9999....) - 0.999....

9x = 9

x = 1

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Samoa
bangbang, you're out of your depth, don't bother.

Proof 1:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

Proof 2:

0.111... = 1/9

0.111...*9 = 0.999...

since 0.111.... = 1/9 -> 1/9 * 9 = 0.999...

1/9 * 9 = 9/9 = 1

Proof 3:

0.3 recurring = 1/3

0.3 recurring * 3 = 0.9 recurring

1/3 *3 = 1

Disprove...

:sleep:

http://forums.philosophyforums.com/showthr...38&page=1&pp=25

I'm off to bed now, time to unsubscribe and leave you talking to a brick wall...

Proof 2:

0.111... = 1/9

0.111...*9 = 0.999...

since 0.111.... = 1/9 -> 1/9 * 9 = 0.999...

1/9 * 9 = 9/9 = 1

0.1111.... does equal 1/9 [ 0.111... = 1/9]

.1111..... times 9 equals 1 not [0.111...*9 = 0.999...]

you cannot divide something then multiply by the same number and arrive at a different original number.

once again a mistake.

however this is correct, but is not proving .9999 is equal to 1

Proof 3:

0.3 recurring = 1/3

0.3 recurring * 3 = 0.9 recurring

1/3 *3 = 1

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un4given1
10x = 9.9999...

10x - x = 9x

Where in the hell did you get this? that's not proper math.

x=0.9999...

10x=9.9999...

10x/x=9.9999.../x (I don't know where you came up with the rule that you can subtract x from 10*x)

10=9/x

10=9/.9999...

10=10.0000...9 This is a FALSE PROOF!

let's try this with a known variable...

x=5

10*5=10*5

10/5=10/5

10=10

and Dazzla's method with the same variable

x=5

10*5=50

10*5-5=45

45=10/5

50=50

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Dazzla

How is it not proving .999 = 1?

0.3 recurring = 1/3

0.3 recurring * 3 = 0.9 recurring

1/3 * 3 = 1

^^^^That you 100% agree with, can also read (seeing as 0.3 recurring = 1/3) as:

1/3 * 3 = 0.9 recurring

But on the final line we've shown that 1/3 * 3 = 1, and above we've shown 1/3 * 3 = 0.9 recurring:

1/3 * 3 = 0.9 recurring

1/3 * 3 = 1

Wow, what do you think that means?

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applelover

Here is the way I would:

0.99999999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

This is a geometric series where the nth term can be represented by a(n) = a(0)*0.1^(-n) for n=1,2,3,4..., where a(0) is the first term of the series.

As such, the sum of the terms is S = a(0)/(1-r) where r=0.1 and a(0)=0.9

So,

0.9/(1-0.1) = 0.9/0.9 = 1

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bangbang023

you guys are missing the point!

x = 0.999... right?

ok we got that. now slow down a bit

10x - x = 9.999... - 0.9999

now what is 9.999... - 0.999....? that's right, 9.

10x - x = 9

9x = 9

x = 1

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Dazzla
Here is the way I would:

0.99999999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

This is a geometric series where the nth term can be represented by a(n) = a(0)*0.1^(-n) for n=1,2,3,4..., where a(0) is the first term of the series.

As such, the sum of the terms is S = a(0)/(1-r) where r=0.1 and a(0)=0.9

So,

0.9/(1-0.1) = 0.9/0.9 = 1

That's the way I was taught it at college by my Math professor.

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Samoa
ok I'm going to try to break this down but i have to hurry:

x = 0.9999...

10x = 9.9999...

10x - x = 9x

9.9999... - 0.9999... = 9

9 = 9x

1 = x = 0.9999...

1) now he sets x = 0.9999...

2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999...

3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9

4) the rest is simple

your math leaves of 9 times .9999

what does the 9x part just disappear?

ok so 10x - x

if you plugin the value for x yo uhave 10(0.999...) - 0.9999 qhich equals 9

but 10(x) - x = 10(0.9999....) - 0.999....

9x = 9

x = 1

9x = 9

x = 1 yes that would be correct if X=1, and only if x=1

we already said it did not with x=.9999

but let's use another interger [y] to show the point

y=.9999

x=1

9xy = 9y?

9 (1) (.9999) = 9 (.9999)

8.991=8.991 this holds true but in no way says 1 is equal to .9999

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un4given1

x=1

y=.9999...

1000000x=1000000y

1000000 = 999999.9999...

if x=1 and y=.9999... but x should = y, then the statement above should be correct, right? Looks a little off to me

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Dazzla

Sammy, you choose to ignore this:

0.3 recurring = 1/3

0.3 recurring * 3 = 0.9 recurring

1/3 * 3 = 1

^^^^That you 100% agree with, can also read (seeing as 0.3 recurring = 1/3) as:

1/3 * 3 = 0.9 recurring

But on the final line we've shown that 1/3 * 3 = 1, and above we've shown 1/3 * 3 = 0.9 recurring:

1/3 * 3 = 0.9 recurring

1/3 * 3 = 1

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John

in regards to the title: no.

0.999... is 1 - an infinitely small number (ie. less than 0.000...1, which doesn't exist anyway, cause you can't have a digit AFTER an infinite number of digits)

someone can explain this with limits, but i don't really want to right now...

i'm 100% sure the answer is no.

edit: AFAIK, you can't solve this with just algebra, so dazzla's method, although it looks correct, isn't correct...

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OPaul
Can someone respond to this without giving a proof. Dazzel or who ever is says 1 = .99... tell me how that can be if the number never gets there. If it got there it would be a real number and would no longer approach 1.

Alright, enough proofs. Proofs are all well and dandy and I see what there doing and such. My problem is this;

0.999999.... is approaching 1 indefinitely, it is not a real number, it never reaches 1. So how does it equal 1 if it never gets there? Can someone explain that without giving another proof.

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Samoa

anyone who says .9999 is equal to one cannot be correct

in dazzla's equation 9x is repeated left out. When you can show me 9x is not 8.991, when x=.9999 then the world will fall apart.

you can show me over an over that 1 is equal to 1 all you want. Cause I know it is true. 10x-x when x=1 is going to be 9. There is no argument. But when X=.9999, then 10x-x also equal 9

10x-x

10(.9999)-.9999

9.9999-.9999

9

10(1)-(1) = 9 = 10(.9999)-(.9999)

that is no way proves .9999 equals 1, started to fall for that flawed reasoning as well for a moment but...

Edited by Samoa
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Dazzla

Chosen to ignore applelovers post and mine 3 posts up?

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OPaul

I think the problem is you guys are trying to solve a calculus problem with plain algebra, your ignoring the fact the limit of something doesn't equal what it approaches. That's just simple Calculus I.

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.Atlantis

I'm gonna have to agree on both sides of the argument :| I really need to ask our Math proffessor tommorow...

-Grant

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Samoa
Chosen to ignore applelovers post and mine 3 posts up?

no I am showing where an how your calculation is flawed, Please consider my post again.

10x-x

10(.9999)-.9999

9.9999-.9999

9

9 = 10(.9999)-(.9999)

10(1)-(1) = 9

then,

10(1)-(1) = 9 = 10(.9999)-(.9999)

10(1)-(1)=9=10(.9999)-(.9999)

if x=1, z=.9999 then

10(x)-(x)=9=10(z)-(z)

10(x)-(x)/10=9/10=10(z)-(z)/10

(x)-(x)=.9=(z)-(z)

plug in the values for x and z it don't work.

(1)-(1)=.9=(.9999)-(.9999)

0=.9=0

Edited by Samoa