278 replies to this topic

[WI_Rifleman]

My brother posed me this proof that says that .9999999 repeating = 1. Doesn't this violate the law of identity?

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9X

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

[The Wrath]

No...there's a similar proof that proves that 0=1. However, there's a step that screws up, b/c it requires division by zero.

[Nate T.]

Nah, no violation of identity-- this is an example of the same thing being given two different labels. It's no more contradictory than saying that 1/2 = 2/4; it's just that the concepts involved (limits) are more complicated.

The real resolution to this is to recognize that while 1 is just that same unit that everyone knows and loves (?), .999... is actually the sum of a geometric series, and that when you sum the series you get 1, which is what .999... = 1 is really saying.

I'm not sure how much math you've had, but I can probably give a quick proof if you're really interested.

[Bryan]

My brother posed me this proof that says that .9999999 repeating = 1. Doesn't this violate the law of identity?

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9X

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

<{POST_SNAPBACK}>

9X does not equal 9, its 8.99999999999999.

[Nate T.]

Moose,

No...there's a similar proof that proves that 0=1. However, there's a step that screws up, b/c it requires division by zero.

That's something different. That .999... = 1 isn't a fallacy, it's just counterintuitive, kind of like some of Xeno's paradoxes.

[EC]

It's a limit, which is a concept that is used in calculus all the time. As the 9's after the decimal point approach infinity the value of the decimal aproaches 1. The limit therefore equals 1.

[GoodOrigamiMan]

You can use the same method with any repeating decimal – but you will end up with the fractions that represent them (1/9 for .111111, 2/9 for .222222, etcetera - see below). But just like .99999 these repeating decimals do not equal the fractions or in your case 1. This is because as Rational One said, they are limits, so a repeating decimal is always approaching it’s limit (the fraction or 1) but technically it will never get there. What is happening in the subtraction is you are erasing the indefinite series by subtracting a lower power from a higher one; you are then left with the result and the factor of the power (9). So the answer becomes a fraction – representing the limit (becuase in the process you erased the reapeating) of the repeating decimal.

X=.33333333333 repeating
10X = 3.3333333333 repeating
10X - X = 3X
9X = 3
X = 3/9
Therefore
.33333333333 repeating = 3/9

[Free Capitalist]

Yep, Bryan caught the error:

In the problem above, this is wrong: 9X = 9

[Nate T.]

Free Capitalist,

Yep, Bryan caught the error:

In the problem above, this is wrong: 9X = 9

In what way is this an error?

[dwwoelfel]

See if you can figure out this one:

Let a = b
a² = ab
a² - b² = ab -b²
(a-b)(a+b)=b(a-b)
a+b=b
2b=b
2 = 1

Lifted from absolutereason.com

[Bryan]

Free Capitalist,

Yep, Bryan caught the error:

In the problem above, this is wrong: 9X = 9

In what way is this an error?

<{POST_SNAPBACK}>

Initially X is set to .999999, so 9X = 8.99999999.

Only after that step is X set equal to 1.

[Nate T.]

Initially X is set to .999999, so 9X = 8.99999999.

How do you know that 9(.999...) = 8.999...? That seems just as dubious a step as saying that 9.999... - .999... = 9.

[Bryan]

See if you can figure out this one:

Let a = b 
a² = ab 
a² - b² = ab -b² 
(a-(a+=b(a-
a+b=b 
2b=b
2 = 1

Lifted from absolutereason.com

<{POST_SNAPBACK}>

The only way that a + b = b is if a = b = 0.

If a = b = 0, then 2b = b reduces to 0 = 0 not 2 = 1.

[Bryan]

How do you know that 9(.999...) = 8.999...?  That seems just as dubious a step as saying that 9.999... - .999... = 9.

<{POST_SNAPBACK}>

Multiplying two numbers together is dubious?

Get a calculator and multiply 9*.9999 and see what you get.

[Hal]

it isnt 0.99999, it is 0.9999~ where the sequence of 9's is infinite. If you use a calculator (or do it by hand) youre not going to get the right answer because you'll have to terminate the sequence at some point - you need to manipulate the (non-truncated) infinite series to get the correct result.

Edited by Hal, 22 February 2005 - 10:56 AM.

[Bryan]

it isnt 0.99999, it is 0.9999~ where the sequence of 9's is infinite. If you use a calculator (or do it by hand) youre not going to get the right answer because you'll have to terminate the sequence at some point - you need to manipulate the (non-truncated) infinite series to get the correct result.

<{POST_SNAPBACK}>

No matter how many 9s you put at the end of the "infinite" decimal if you multiply it by 9 you will get a number less than 9.

.999~ < 1, therefore 9*.999~ < 9.

[Hal]

No matter how many 9s you put at the end of the "infinite" decimal if you multiply it by 9 you will get a number less than 9.

Not if you put infinite 9's there.

Edited by Hal, 22 February 2005 - 11:05 AM.

[Bryan]

There is no largest number less than 1, as I said the real numbers are infinitely dense. I'm not sure what you mean by 0.99~ doesnt exist in reality.

It makes more sense (to me at least) if you think of 0.999~ as being an infinite series (0.9 + 0.09 + 0.009 + ...), which is how it's defined mathematically, rather than being a big long list of 9s

<{POST_SNAPBACK}>

This is a quote from the other thread, I took upon myself to combine them because this thread contains the actual topic.

The infinite series (0.9 + 0.09 + 0.009 + ...) is just a big long list of 9s if you actually calculate the sum of the series.
.999~ is theoretically the largest number less than one, but it doesn't actually exist in reality.

Let's pretend you have 1 cup of coffee. You take the smallest sip of it that you possibly can. You now have less coffee in the cup than you had before. We'll say that you now have .999~ cups of coffee. But in reality, no matter how small of a sip you took, you still removed a measurable amount of coffee from the cup. You can't take an infinitely small amount of coffee out of the cup, which is why .999~ doesn't actually exist.

[Hal]

This is a quote from the other thread, I took upon myself to combine them because this thread contains the actual topic.

Good call, having 2 threads on the same subject seemed a bit silly.

The infinite series (0.9 + 0.09 + 0.009 + ...) is just a big long list of 9s if you actually calculate the sum of the series.

No, it's 1. I gave the formula for calculating the sum of the series in the other thread.

Let's pretend you have 1 cup of coffee.  You take the smallest sip of it that you possibly can.  You now have less coffee in the cup than you had before.  We'll say that you now have .999~ cups of coffee.  But in reality, no matter how small of a sip you took, you still removed a measurable amount of coffee from the cup.  You can't take an infinitely small amount of coffee out of the cup, which is why .999~ doesn't actually exist.]

I agree, but we arent talking about cups of coffee, we are talking about numbers in the abstract sense. The square-root of -1 doesnt exist in reality, but it's still a perfectly valid number. There is perhaps a smallest number that has significance in reality (Planck), but this doesnt mean that there is a smallest real number, mathematically speaking.

Edited by Hal, 22 February 2005 - 11:26 AM.

[dougclayton]

I think an easier way to see that 0.99999999999999 = 1 is the following:

1/3 = 0.33333333333333333 (repeating)

3 * (1/3) = 0.99999999999999999 (repeating)

but 3 * (1/3) can be rearranged to (3*1)/3 and therefore 3/3, which must equal 1 (multiplication is associative and commutative over the real numbers), thus

1 = 0.99999999999999999 (repeating)

It seems counter-intuitive, I know, but so did using a number for zero to the Greeks. By the way, this is not a flawed proof like those hide-the-division-by-zero proofs. If you can find a mistake I'd love to hear it.

[Bryan]

No, it's 1. I gave the formula for calculating the sum of the series in the other thread.

That formula is just a way to approximate inifinite sums. The infinite sum does approximately equal 1, but its is actually slightly smaller.

I agree, but we arent talking about cups of coffee, we are talking about numbers in the abstract sense. The square-root of -1 doesnt exist in reality, but it's still a perfectly valid number. There is perhaps a smallest number that has significance in reality (Planck), but this doesnt mean that there is a smallest real number, mathematically speaking.

<{POST_SNAPBACK}>

In the abstract sense, .999~ is the largest number that is less than 1.

The square root of negative one is a mathematical tool that does represent things in reality. In fact, I used complex numbers this morning to represent the power in a transformer in my electrical machinary lab.

[dougclayton]

My brother posed me this proof that says that .9999999 repeating = 1. Doesn't this violate the law of identity?

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9X

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

<{POST_SNAPBACK}>

Part of the reason this is supposed to contain a fallacy is that it is misstated. It should be:

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9 (corrected)

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

This doesn't rely on the fact that 0.999~=1, which is what you are trying to prove. Instead it says that 9.999~ - 0.999~ = 9, which is different.

[dougclayton]

That formula is just a way to approximate inifinite sums.  The infinite sum does approximately equal 1, but its is actually slightly smaller.

No, the infinite sum is exactly equal to 1. This is the answer to Zeno's paradox that the Greeks could not appreciate. His paradox says that 1/2 + 1/4 + 1/8 + .... both must be equal to 1 and cannot be equal to 1, and hence the paradox, whereas in truth the "cannot be equal to 1" part is wrong.

Any finite sum is approximate, of course, but we are talking about infinite sums.

[Bryan]

I think an easier way to see that 0.99999999999999 = 1 is the following:

1/3 = 0.33333333333333333 (repeating)

3 * (1/3) = 0.99999999999999999 (repeating)

but 3 * (1/3) can be rearranged to (3*1)/3 and therefore 3/3, which must equal 1 (multiplication is associative and commutative over the real numbers), thus

1 = 0.99999999999999999 (repeating)

It seems counter-intuitive, I know, but so did using a number for zero to the Greeks.  By the way, this is not a flawed proof like those hide-the-division-by-zero proofs.  If you can find a mistake I'd love to hear it.

<{POST_SNAPBACK}>

Your mistake is the that you say:
3 * (1/3) = 0.99999999999999999 (repeating)

3 multiplied by 1/3 is 1. Not anything other than 1!

1 divided by 3 is not a number that can be accurately represented by a decimal. You can put as many 3s at the end of .333 until you drop dead, you still won't get exactly 1/3.

By the same method you can put 9s at the end of .999 until you drop dead, you will never get exactly 1.

[dougclayton]

Ummm... you started with X = .9999999 (repeating) and ended with X = 1.

If that doesn't prove that .9999999 (repeating) = 1, then the mathematical Law of Identity (if a=b and b=c, then a=c) isn't true.

<{POST_SNAPBACK}>

It does prove that 0.999~ = 1. My point is that it doesn't rely on that fact before it proves it, as the original proof appears to.