Math Project Help: Year 8 Student Needs Advice

[chickenguy]

hi everyone, i am a year eight student and i need to do a project for maths. could you first of all please tell me what you think about this topic and do you believe that 0.99999999999999.....=1. thanks a lot! :rofl:

 

[chroot]

There are already many threads on this topic; please do a search. And yes, 0.999... equals 1, without any question.

- Warren

 

[Oggy]

If you don't believe it, look at this.
$$x=0.99999...$$
$$x=0.9+0.09999...$$ Multiply by 10
$$10x=9+0.99999...=9+x$$
$$9x=9,\ \Rightarrow x=1$$ Therefore
$$1=0.99999...$$

 

[Owen Holden]

hi everyone, i am a year eight student and i need to do a project for maths. could you first of all please tell me what you think about this topic and do you believe that 0.99999999999999.....=1. thanks a lot! :rofl:

It is the case by definition alone!

 

[jcsd]

An infinite decimal expansion 0.a₁a₂a₃...a_n... is defined as the limit of the sequence of partial sums of the series

$$\sum^{\infty}_{n = 1} a_n10^{-n}$$

which in the case of 0.999... is equal to 1.

 

[Icebreaker]

$$\frac{1}{3} = 0.333...$$

$$\frac{1}{3} + \frac{1}{3} = \frac{2}{3} = 0.666...$$

$$\frac{2}{3} + \frac{1}{3} = 0.666... + 0.333... = 0.999... = \frac{3}{3} = 1$$

 

[HallsofIvy]

Icebreaker and Oggy are perfectly correct except that it assumes that you can do arithmetic on the digits like that. That is true but is precisely the kind of thing a person who is asking about 0.999... would not accept without detailed proof.

jcsd's suggestion- using that fact that 0.999... MEANS the sum of the infinite series
.9+ 0.09+ 0.009+ ... and showing that that is a geometric series whose sum is 1 is, in my opinion, best.

By the way, this is "chickenguy"s second post on this. The first time he was asking for a poll on whether or not 0.9999... was equal to 1!
I guess this is an improvement.

 

[No Skills]

Well I have a problem with the 1 = .999... thing. Because if this is true it implies that 1/n = 0 when n is "at infinity". Please follow my thinking...

Consider the infinite series:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... which EQUALS 1. This follows from the fact that
1 = .9999... because the infinite series have a finite sum. Note that the partial sums are less than 1 and that equality is only achieved when the set of terms is infinite. I.E. "at infinity"

Now consider the infinite series:

[1/2 + 1/4 + 1/8 + 1/16 + ... + 1/(2^n)] + [1/(2^n)]

Please note that the first part of the series (in first brackets) equals 1 at infinity.
Please note that the partial sums of all terms always equal 1.
Please note that the second part of the series (in second brackets) is receeding to zero as n increases.
THEREFORE, the second part of the series must equal zero "at infinity" which means that 1/(2^oo) = 0.

I am not saying that the Limit is zero I am saying that it is necessary that EQUALITY to zero must true.

 

[matt grime]

good writers borrow, great writers steal outright, so to steal from red dwarf unashamedly: firstly there is no n "at infinity" and secondly there is no n "at infinity". I realize this is only one reason, but I thought it so important I had to say it twice.

 

[No Skills]

As you can see this was my 1st post and I am new to this forum. Sorry I stole from nobody. That said. If there is no "at infinity" then how does
1 = .9999...(not infinity??). Either the set of terms is finite (in that case the series is less than one) or the set of terms is infinite and we have achieved "at infinity".

P.S. Saying something two times does not make it so (its kind of weak argument, sorry).

 

[HallsofIvy]

Either the set of terms is finite (in that case the series is less than one) or the set of terms is infinite and we have achieved "at infinity".

No, that doesn't follow. There certainly exist "infinite" sets. The set of all positive integers is one. However, what was being objected to was "n at infinity". All n (by which I assume you mean a positive integer) are, by definition, finite.

"Now consider the infinite series:

[1/2 + 1/4 + 1/8 + 1/16 + ... + 1/(2^n)] + [1/(2^n)]

Please note that the first part of the series (in first brackets) equals 1 at infinity.
Please note that the partial sums of all terms always equal 1.
Please note that the second part of the series (in second brackets) is receeding to zero as n increases.
THEREFORE, the second part of the series must equal zero "at infinity" which means that 1/(2^oo) = 0.

I am not saying that the Limit is zero I am saying that it is necessary that EQUALITY to zero must true."

And that's the nub of the problem. Infinite sums are only defined AS LIMITS. There exist infinite sets, but there is no such thing as 'at infinity'.

 

[Hurkyl]

1 = .9999...(not infinity??). Either the set of terms is finite (in that case the series is less than one) or the set of terms is infinite and we have achieved "at infinity".

There is no "set of terms" in 0.9999...; it is not a series, it is a number.

 

[No Skills]

Sorry Hurky, but .9999... is a polynomial of the form
a*(1/10^n) for example 9*(10^-1) + 9(10^-2) + 9(10^-3) + ... to infinity

Sorry Halls of Ivy you are saying then
1 is NOT EQUAL to this Infinite sum .9999...
The Infinite sum converges on 1 and never equals it?!

 

[jcsd]

I don't believe polynomials can have infinite number of terms.

0.999.. = 1 by definition i.e. 0.999.. doesn't represnt a polynomial or a series it represents the limit of the sequence of partial sums of a series.

 

[No Skills]

SUM N FROM 1 TO INFINITY [9*(1/10^N)] = .99999...
What is so hard for you to see about this infinite series?

 

[Hurkyl]

Saying "Sorry, but you're wrong" is also fairly weak proof.

The thing you call a "polynomial" is not a polynomial, because it's an infinite series. :tongue:

0.999... is called a decimal representation of a number. It is, by definition, a sequence of digits.

And, it's true that $0.\bar{9} = \sum_{n=1}^{\infty} 9 / 10^n$ is an equality.

Note the wording -- it's an equality, not a definition. Both sides of the equation are real numbers. The left is one of the two decimal representations of the number 1, and the right is the value of the limit of a sequence of partial sums.

 

[Icebreaker]

SUM N FROM 1 TO INFINITY [9*(1/10^N)] = .99999...
What is so hard for you to see about this infinite series?

$$\sum^{\infty}_{n = 1} \frac{9}{10^{n}}$$

Is this what you're trying to say?

 

[jcsd]

SUM N FROM 1 TO INFINITY [9*(1/10^N)] = .99999...
What is so hard for you to see about this infinite series?

An infinite series should not be regarded as a number (even a finite series should be not be regarded as just a number, though it will always be associated with a number, as it has more structure), it should be regarded as a sequence of terms together with an additon operations which defines a sequence of partial sums, to illustrate this different inifnite series can converge to the same number. 0.999.. is not an infte series it is a number - the number which the sequence of partial sums of an infinite series converges to.

 

[HallsofIvy]

Sorry Hurky, but .9999... is a polynomial of the form
a*(1/10^n) for example 9*(10^-1) + 9(10^-2) + 9(10^-3) + ... to infinity

Sorry Halls of Ivy you are saying then
1 is NOT EQUAL to this Infinite sum .9999...
The Infinite sum converges on 1 and never equals it?!

An infinite sum EQUALS the number its sequence of partial sums converges to- that's the definition. There is no distinction between the convergence to a number and the number the infinite sum EQUALS.

 

[No Skills]

Hurkyl: It is all the same. A decimal representation of a number is a polynomial if it terminates and an infinite series if it doesn't. Of course .25 is a polynomial. This isn't worth discussing, I am glad you now have and enriched view of the numbering system we all use. Thank you.

IceBreaker: Yes that is what I am saying. Sorry I do not have the same command of this forums tools that you do. Would you agree that .9999... is an infinite series?

HallsOfIvy: Ok if there is no distiction why do you make it with all of your talk of convergence and limits. Why don't we just agree that the infinite series .99999... EQUALS 1? Further why don't we also agree that the series generates an infinite set of terms. And that when considering the sum of this infinite set, it EQUALS 1. Or in other words when the set has the actual value of 1 "at infinity".

jscd: .9999... is a generator of an infinite series terms in my mind. I am regarding the terms generated (digits) as an infinite set of terms and considering the sum of all the terms in this set.

 

[Hurkyl]

This isn't worth discussing

As you wish.

 

[HallsofIvy]

Why don't we just agree that the infinite series .99999... EQUALS 1?

? You were the one who started this by saying:
"Well I have a problem with the 1 = .999... thing. "

Okay, now you are willing to agree with what everyone else was saying all along?