As penance for liberally beating a dead goat in the other brainteaser thread, here is a fresh one that should be easy for many of you:

Does .9 repeated = 1?

Please prove mathematically and explain in English. SB for the first to do both.

A Geometric series is probably easier.

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just construct a cauchy sequence

You can explain this without any fancy math, using the definition of numbers alone.

x = 0.9
10x = 9.9
10x - x = 9

GBS

.9 repeated has a limit value of 1, but is never actually 1.

TechBanking:

.9 repeated has a limit value of 1, but is never actually 1.

If abs (x) <1, the infinite series sum(x^n) converges to 1/(1-x)
let x = (1/10)

finite sum = 1/(1-.1)

this gives you 1 + .1 + .01 + .001 ....
subtracting 1 and multiplying by 9 gives you .9 + .09 + .090 ....

solution = [1/(1-.1) -1] *9 = [10/9 - 9/9] * 9 = 1

TechBanking:

.9 repeated has a limit value of 1, but is never actually 1.

This statement makes no sense at it stands. If you mean to say constructing a sequence where each additional term is made by appending another 9, then yes, this sequence has a limit of 1. If you mean the number .9 repeating, well, it's a constant, so its "limit value" must be itself. By the way, that is most definitely 1.

x = 0.9

10x = 9.9

10x - x = 9

9x = 9

x = 1

illiniPride:
Cola Coca:

x = 0.9

10x = 9.9

10x - x = 9

9x = 9

x = 1

That's the math, now why does it work?

Because an infinite string of 0.9 will always equal 1? There is no number between 0.9 repeating and 1; they are the same number, represented differently.

0.25 = 0.24(9) repeating

Cola Coca:

x = 0.9

10x = 9.9

10x - x = 9

9x = 9

x = 1

10*.9=9

not 9.9, so in your equation x does not equal .9 like you said.

the short answer is no it does not equal 1

Cola Coca:

x = 0.9

10x = 9.9

10x - x = 9

9x = 9

x = 1

Wtf is this shit...

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You just saw this on reddit didn't you

Yes - my simplest 'explain in english' is that .3 repeating is generally known to equal one third, and 3 times one third equals 1

JDimon:

Yes - my simplest 'explain in english' is that .3 repeating is generally known to equal one third, and 3 times one third equals 1

That's a good way to explain it. Also if
1/3 = .3 repeating
2/3= .6 repeating

Add them together and you get

3/3=.9 repeating
1=.9 repeating

JDimon:

Yes - my simplest 'explain in english' is that .3 repeating is generally known to equal one third, and 3 times one third equals 1

Give it to this guy.

Let me also do it with math:

0.999 repeating equals x.
10x equals 9.999 repeating equals 9 + x
Subtract x from both sides
9
x = 9
therefore x=1

JDimon:

Let me also do it with math:

10*x equals 9.999

Assuming you mean 10* .999 repeating = 9.999 repeating, this step is only true if the series converges (i.e it equals 1). In order to finish your proof you essentially assumed that what you were proving was true.

Are you retarded?

FreezePops:
JDimon:

Let me also do it with math:

10*x equals 9.999

Assuming you mean 10* .999 repeating = 9.999 repeating, this step is only true if the series converges (i.e it equals 1). In order to finish your proof you essentially assumed that what you were proving was true.

Are you retarded?

All I'm doing is using that fact that the 9s go on infinitely. If they go on forever, then multiplying everything by 10 doesn't reduce the amount of 9s after the decimal point....they still go on forever (not forever - 1, because that value still equals 'forever')

JDimon:
FreezePops:
JDimon:

Let me also do it with math:

10*x equals 9.999

Assuming you mean 10* .999 repeating = 9.999 repeating, this step is only true if the series converges (i.e it equals 1). In order to finish your proof you essentially assumed that what you were proving was true.

Are you retarded?

All I'm doing is using that fact that the 9s go on infinitely. If they go on forever, then multiplying everything by 10 doesn't reduce the amount of 9s after the decimal point....they still go on forever (not forever - 1, because that value still equals 'forever')

JDimon is right. This is a legitimate proof. He's not assuming anything about the series convergence- all he assumes is the presence of infinite 9s (follows by the assumption of repeated 9s). Multiplication by 10 is a well defined operation- even for infinite decimal places- as a "place shifter".

Oh Cola Coca got the answer first. Everyone who's disagreeing with him isn't taking the time to think about what it means for the decimal to be repeating indefinitely

Yes. When you want a repeated decimal, you just take that number and divided it by 9.
0.7777.....=7/9
0.5555.....=5/9

Thus, 0.999999.....should theoretically equal 9/9, and 9/9 = 1.

Do I get the SB?

Gave it to Dimon, wasnt thinking of it that way but it makes sense.

You can prove this another way using negation

how can one number equal another number? even if a limit, it still doesn't technically equal it...

plz splain

This was how I thought of it:

If a <> b,
x exists where a < x < b or a > x > b.
If not, a=b

The answer is yes. There is a long mathematical proof that you learn in calc 2.

But the easy straightforward answer is...

1/3 = .3 Repeated
2/3 = .6 Repeated

3/3 = .9 Repeated

Wikipedia has a pretty cool proof with (Dedekind) cuts too! I didn't know that one since my analysis class skipped the construction of real numbers, but it was nice. I liked it!

Assume that 9.999.../=10

Then you can list a number that is greater than 9.999... and less than 10.

Since you can't, 9.999...=10

1/3 + 1/3 + 1/3 = 3/3 = 1 and 1/3 is 0.33... so 0.3333..... + 0.3333..... + 0.3333.... = 0.9999 = 1

Weird corollary imo: every terminating number has 2 equivalent decimal representations

juked07:

Weird corollary imo: every terminating number has 2 equivalent decimal representations

One terminating and one non-terminating?

Models and bottles!

Did I get it right?

I know why our financial system failed.
Because there are too many guys who think 0.9 is 1.

There are several ways of showing this. An alternate version is using infinite series.

Let S(i) = i_Sum_k=1 .9*10^(-k)

And so .99 repeated is just Lim_(i--> infiniti) S(i)

But S(i) = .9* (1-10^(-i-1))/(1-1/10)

So Lim_(i--> infiniti) = .9*(1/(1-1/10)) = 1

-MBP

Numbers are concepts. If the number isn't the same number, then it's not the same number. The idea of .9 repeating is that its as close as possible to 1 without actually being 1. Otherwise it would be 1. So I'm gonna ruin all the fun and say it's not possible based on the idea of what numbers are.

I hate victims who respect their executioners

BlackHat:

Numbers are concepts. If the number isn't the same number, then it's not the same number. The idea of .9 repeating is that its as close as possible to 1 without actually being 1. Otherwise it would be 1. So I'm gonna ruin all the fun and say it's not possible based on the idea of what numbers are.

No it's a fact that .9 repeating is another representation of 1. There are many proofs for it. It's not counterintuitive if you look at how the real numbers are constructed.

-MBP

manbearpig:
BlackHat:

Numbers are concepts. If the number isn't the same number, then it's not the same number. The idea of .9 repeating is that its as close as possible to 1 without actually being 1. Otherwise it would be 1. So I'm gonna ruin all the fun and say it's not possible based on the idea of what numbers are.

No it's a fact that .9 repeating is another representation of 1. There are many proofs for it. It's not counterintuitive if you look at how the real numbers are constructed.

MBP has it right. Any number has an infinite number of representations: for example, 1/2 can be written as 2/4, 3/6, etc..

.9 repeating is simply another representation of 1.

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