0.9 recurring = 1

[agurkie]

I know this a topic which has been discussed to infinity, but I have a problem with this theory and would like other peoples' opinion on it.

 

The 'proof' states that:

 

x = 0.9999

10x = 9.9999

9x = 9

x = 1

 

I believe the problem lies in the second line already. Is it possible to do an arithmetic operation on a number with an infinitely repeating fraction?

 

Let's take a finite number. x = 0.9999, then 10x = 9.9990. We have to know there are 4 digits after the decimal point, so each one moves one to the left, and the fourth digit after the decimal is replaced by a 0.

 

I suppose the argument is with regard to the theoritical interpretation, but in practice it is not possible to multiply a number with a repeating fraction.

 

Looking also at the irrational number PI, any computer or calculator actually stores PI to about 40 significant digits. So to any computer or calculator, PI is a number with a finite number of digits after the decimal point. If you think about it, how could you expect a computer to use PI if it continued forever?

 

So my opinion is, yes I agree that 0.9 recurring approaches 1, but we do not have the mathematics to prove that it is in fact 1 and until then it just...well....isnt.

[ajb]

You could also think about 1/3 + 1/3 +1/3 =?

 

You can also prove it a lot more rigorously using series and limits.

[agurkie]

But 1/3 is not equal to 0.3 recurring.

0.3 recurring approaches 1/3, but never gets there.

 

same situation:

 

x=0.3333

10x=3.3333

9x=3

x=1/3

 

and in my opinion the same problem in line 2.

the series proof proves convergence. but just because something converges to something else does not mean it is actually equal to that something does it?

[Bignose]

So my opinion is, yes I agree that 0.9 recurring approaches 1, but we do not have the mathematics to prove that it is in fact 1 and until then it just...well....isnt.

Ok, so if two numbers aren't the same number, there is always a number between them.

 

If 0.9999... isn't the same as 1, please post the number that comes between them.

[HalfWit]

I know this a topic which has been discussed to infinity, but I have a problem with this theory and would like other peoples' opinion on it.

 

The 'proof' states that:

 

x = 0.9999

10x = 9.9999

9x = 9

x = 1

 

I believe the problem lies in the second line already. Is it possible to do an arithmetic operation on a number with an infinitely repeating fraction?

 

Let's take a finite number. x = 0.9999, then 10x = 9.9990. We have to know there are 4 digits after the decimal point, so each one moves one to the left, and the fourth digit after the decimal is replaced by a 0.

 

I suppose the argument is with regard to the theoritical interpretation, but in practice it is not possible to multiply a number with a repeating fraction.

 

Looking also at the irrational number PI, any computer or calculator actually stores PI to about 40 significant digits. So to any computer or calculator, PI is a number with a finite number of digits after the decimal point. If you think about it, how could you expect a computer to use PI if it continued forever?

 

So my opinion is, yes I agree that 0.9 recurring approaches 1, but we do not have the mathematics to prove that it is in fact 1 and until then it just...well....isnt.

You are correct that this commonly seen "proof" is a bit bogus. The field axioms for the real numbers only allow you to multiply a constant times a finite sum. In order to multiply 10 * .999... term by term, you need to develop the theory of infinite sequences and series ... by which time, the fact that .999... = 1 has already been proved. So this "10x" proof is really just a heuristic hand-wavy pseudo-proof for the benefit of high school students.

 

However, it's still true that .999... = 1. For one thing, can you name a number that's strictly larger than .999... and strictly less than 1? No, you can't.

 

Secondly, in freshman calculus they teach you about geometric series. .999... is a shorthand for 9/10 + 9/100 + 9/1000 + ..., which is a geometric series whose sum is 1.

 

To sum up: .999... = 1, and that's a provable mathematical fact. However, the common "10x" proof is flawed, since it assumes facts that are already more mathematically sophisticated than the fact that .999... = 1. Namely, the "10x" proof assumes that you can do term-by-term multiplication of an infinite sum. However, that operation is not allowed by the axioms for real numbers; and can only be proven after you have developed a rigorous theory of the real numbers and then a rigorous theory of infinite sequences and series.

[John Cuthber]

I'm not sure about this "Secondly, in freshman calculus they teach you about geometric series. .999... is a shorthand for 9/10 + 9/100 + 9/1000 + ..., which is a geometric series whose sum is 1." solving the problem

The proof (at least, the one I have seen) for the sum of the series

http://en.wikipedia.org/wiki/Geometric_series#Formula

looks pretty much the same as the 9.99999...... -0.9999999 =9 proof that .999999...... =1

It seems not to tally with "The field axioms for the real numbers only allow you to multiply a constant times a finite sum."

 

Of course, that doesn't detract from the fact that .9 recurring is actually 1.

Edited September 2, 2013 by John Cuthber

[ajb]

If 0.9999... isn't the same as 1, please post the number that comes between them.

Not that I have actually seen it done, I imagine one can use a kind of epsilon delta proof here (or rather a proof by contradiction)?

[agurkie]

The sum of finite geometric series is a(r-r^n)/(1-r).

 

Now, to derive the sum of infinite series from the sum of finite series above, we are using the fact that r^n => 0 for n => infinity, if |r| < 1.

Then the formula becomes ar/(1-r). Great, and this is then how you prove using the sum of infinite geometric series, that 0.9 recurring = 1.

 

But wait a moment! r^n will never ever be 0 unless r=0. We just threw it away and used the result to prove what we wanted it to be.

That r^n that we threw away is in fact the difference between 0.9 recurring and 1, (if you divide it by (1-r)). This actually proves that we can never accurately determine the sum of that series unless we know what n is, which also highlights the error in the 10x = 9.9 recurring theory.

 

So once again, it gets very close, but it just isnt.

 

I think where the cofusion comes in, is around whether or not 0.9 recurring is a constant number or a number that converges.

Lets take (x-1)/(x+1). We could easily prove using L'Hospitals rule that this converges to 1 as x => infinity.

The reason is because the value of this expression actually changes as x gets larger and larger.

 

But what happens to 0.9 recurring as x goes to infinity. Well, nothing. 0.9 recurring is not dependant on any variable.

So, in fact you could argue that it does not converge to anything at all. The limit of 0.9 recurring is 0.9 recurring.

 

It just feels to me that saying 0.9 recurring = 1 is turning mathematics into some form of fancy trickery.

Look I can prove that this hat contains a rabbit, as long as nobody checks my sleeve.

Edited September 3, 2013 by agurkie

[ajb]

It sounds like you are not happy with the notion of limits, I understrand that it is not always very clear.

 

One quite direct way to understand that 0.999... =1 is to use the fact that the real numbers are dense and try to find a p (strictly positive) such that 0.999...+p =1, where we assume explicitly that 0.999... is not equal to 1. That is, as Bignose suggest, tell us what p could be? You can see that there is no such number p.

 

Turning this around, can you find a proof that 0.999... is not equal to 1? This is a different task to trying to find holes in proof that they are equal.

[agurkie]

It sounds like you are not happy with the notion of limits, I understrand that it is not always very clear.

 

Don't you think this statement is a bit condescending? Anyway, moving on....

 

The definition of a limit states that if lim x->c f(x) = L, then for any epsilom > 0 there exists a delta > 0 such that 0 < |x-c| < delta and |f(x) - L| < epsilom.

Meaning, give me any arbritrary value, and i will find a value for x which will make the distance between f(x) and L closer than that arbritrary value without it being 0.

 

But it never says f( c ) = L. f( c ) does not even have to be defined.

 

Of course this works for a function of x. So if you are looking to use the limit argument on 0.9 recurring, then you are saying 0.9 recurring is some variable function. In which case, by the very definition of a limit I can prove to you that there is a value closer to 1.

 

Plus i thought i gave a pretty decent proof of why 0.9 recurring is not equal to 1 in my previous post.

Partial sum of an infinite geometric series: a(r-r^n)/(1-r), in this case 9( 1/10 - (1/10)^n)/ (1- 1/10)

To stretch this to the sum for an infinite series, we must assume that (1/10)^n = 0 as n => infinity, so that the formula becomes 9(1/10) / (9/10), which is equal to 1.

 

But for which n is (1/10)^n ever going to be 0?

 

I dont quite understand how the sum of geometric series proves to some people that 0.9 recurring = 1, while to me it proves exactly the opposite.

 

 

 

 

 

Edited September 3, 2013 by agurkie

[Bignose]

I dont quite understand how the sum of geometric series proves to some people that 0.9 recurring = 1, while to me it proves exactly the opposite.

Then please use what you think is conclusive about the geometric series to present the number that is between 0.9999.... and 1.

 

This notion of a number between two number cuts both ways. If there is a number between a and b, then a does not equal b. But vice versa, if there are no numbers between a and b, then a must equal b.

 

So, in order for a=0.99999 to not equal b=1, then there must be some number between them. What is that number?

[ajb]

Don't you think this statement is a bit condescending?

Sorry if you took it that way. Limits are not easy things to think about.

 

So if you are looking to use the limit argument on 0.9 recurring, then you are saying 0.9 recurring is some variable function. In which case, by the very definition of a limit I can prove to you that there is a value closer to 1.

No you can't argue that 0.999.. approaches anything as it is a fix real number. You can of course show that a series approaches both 1 and 0.999..., but as a convergent series has only one limit these must be equal.

 

But for which n is (1/10)^n ever going to be 0?

Don't you end up showing that

 

[math]\frac{1}{10^{n}} > 1- 0.999... \geq 0[/math]?

 

Now you can make the LHS as small as you like by increasing n. You don't actually need it to be zero. The idea is that you can show that 1-0.999... is smaller than any positive real number; hence it is zero.

[agurkie]

So, in order for a=0.99999 to not equal b=1, then there must be some number between them. What is that number?

 

There will always be another 9.

The whole idea with 0.9 recurring is that it approaches 1 but never gets there.

This is also the whole idea behind a limit.

Why do people want to make it equal to 1? What is the mathematical use of making 0.9 recurring = 1?

[ajb]

There will always be another 9.

The whole idea with 0.9 recurring is that it approaches 1 but never gets there.

This is just not right.

 

0.999... does not approach anything, it is a fixed real number. As it turns out, but you doubt, it is really just another way of writing 1.

 

 

Why do people want to make it equal to 1? What is the mathematical use of making 0.9 recurring = 1?

People don't want to, but rather it follows from the properties of the real numbers.

[Olinguito]

x = 0.9999

10x = 9.9999

9x = 9

x = 1

 

I believe the problem lies in the second line already. Is it possible to do an arithmetic operation on a number with an infinitely repeating fraction?

As long as the series is absolutely convergent (as this one is) then such an operation is perfectly valid.

[HalfWit]

As long as the series is absolutely convergent (as this one is) then such an operation is perfectly valid.

If you already know the series is convergent; and you already know what convergence means; then there is nothing to prove, you're already done. So you see, that step is not valid. Because it assumes what you are trying to prove.

 

By the way, you don't need absolute convergence to multiply a series term by term. Plain old convergence suffices. But you have to prove the theorem on term-by-term convergence of an infinite series, since it's not directly allowed by the field axioms. But if you can prove that theorem, you certainly have no confusion about .999... = 1.

[Bignose]

There will always be another 9.

There is not 'another' 9. There is no finite number of nines. It is infinite. There is no such thing as an 'infinite number of 9s and another 9'. It is infinite in the first place.

 

That is what the ellipses notion means, an infinite number.

 

Because there is not 'another' 9, there is no number between 0.9999... and 1.

[Olinguito]

If you already know the series is convergent; and you already know what convergence means; then there is nothing to prove, you're already done. So you see, that step is not valid. Because it assumes what you are trying to prove.

 

By the way, you don't need absolute convergence to multiply a series term by term. Plain old convergence suffices. But you have to prove the theorem on term-by-term convergence of an infinite series, since it's not directly allowed by the field axioms. But if you can prove that theorem, you certainly have no confusion about .999... = 1.

I don’t see why that step is not valid. Let me make be more clear.

 

To say that a series [latex]\sum_n a_n[/latex] converges means the sequence of partial sums [latex]\sum_{i=1}^n a_i[/latex] tends to a limit as [latex]n[/latex] tends to infinity. In fact I’m splitting hairs here: a series is rigorously defined as nothing more or less than the sequence of its partial sums. The limit of a convergent series is denoted [latex]\sum_{n=1}^\infty a_n[/latex]. This notation is often used to denote the series itself but for this post I’ll write [latex]\sum_n a_n[/latex] for a series and [latex]\sum_{n=1}^\infty a_n[/latex] for the limit of a (convergent) series.

 

My contention is that if the series [latex]\sum_n a_n[/latex] is (absolutely) convergent and [latex]k[/latex] is a fixed real number, then the series [latex]\sum_n ka_n[/latex] is (absolutely) onvergent and [latex]\sum_{n=1}^\infty ka_n=k\sum_{n=1}^\infty a_n[/latex].

[HalfWit]

I don’t see why that step is not valid. Let me make be more clear.

 

(snipped a lot of serious-looking math, far more sophisticated than the fact that .999.. = 1)

OF COURSE!!! But if you can do that, then you have already no doubts that .999... = 1.

 

Of course you CAN multiply a convergent series term by term. But you have to PROVE that you can do it, since infinitary operations are not mentioned in the field axioms. By the time you prove the theorem on term-by-term multiplication, you've already developed the theory of real numbers, limits, and infinite series. And if the sum of a series is the limit of the sequence of partial sums, then what is a sequence, and what is a limit? A sequence is a function whose domain is the natural numbers. And what's a function? It's a subset of the Cartesian product of two sets, and so on down the chain. In order to prove term-by-term multiplication, you need to start from the axioms of set theory and work your way up through half a semester of Real Analysis. It takes undergrad math majors three years to get to this point.

 

By the time you've done all that, you have no doubt that .999... = 1. The 10x proof is circular. It assumes facts far more powerful than .999... = 1. It's not that the 10x proof is wrong. It's that it already assumes a huge amount of technical machinery far more powerful than the mere fact that .999... = 1. That's why the 10x argument is only an informal justification, and not a proof.

Edited September 3, 2013 by HalfWit

[agurkie]

how many mathematicians does it take to change a lightbulb?

 

0.999...