.9 Repeating = 1 YES -or- NO

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

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ericnkatz

:no: Not I

I posted that i havent taken cal or pre cal yet

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John

i haven't read all 10 previous pages :(

i bet when you learn calc/pre-calc, you'll change your mind ;)

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El Bourricot

its internet... the truth isnt that important ;)

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CtotheJ
i'm curious. who here has taken a calculus or pre-calculus class?

Both.

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ike
i understand your concept, but i'm saying you can't multiply 10 times a repeating decimal... not without losing SOME accuracy. (i'm referring to a NONREAL number, so you won't be able to see a numerical difference between the number and 10 * the number)

the .whatever repeats forever, does it not?

so if you multiply .7777repeating by 10, then you get 7.7777repeating...

7/9 * 10 = 70/9

right?

i'm not seeing where accuracy is lost, as the .7 repeats forever, unless you're limiting it in your mind.

http://mathworld.wolfram.com/RepeatingDecimal.html

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Coolme

My TI-83+ said that 0.99999999.... was 1

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Tim Dorr

Ah yes, the wonders of an imperfect textual representation of mathematics...

All of this is due to our inability to correct represent math, really. Logically, the two don't represent the same number, so it is our incorrect representation of each number (or differing representation of the same number) that causes the problem.

So, you can't solve this with words :)

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kryptonik

No, it does not equal one because it is just repeating. However, if you were to round the .9 then yes, it would equal 1. :geek:

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elfangor

I vote yes.

And for the record, I'm taking linear algebra and multi-variable calculus.

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Samoa

I was just thinking..

where .9999...is infinately repeating...

.9999... Will constantly move away from 1 not towards it. It will not be constantly moving towards it. Just plot it on a graph. .9999... will never be equal to 1.

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yzero

nothing has been proven. I'd say your little algebra theories show the faults in our math system. plain and simple 0.99 is NOT 1.. wtf are you smoking?

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deathknight

I Said Yes... But Ah It s A Headache... Pllplplplp

I feel Serious... :wacko:

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Got3n

.99999999999999999999999999999999999999999999999999 does in fact not = 1. just look

1 and .99999999999999999999999999999999999999 do they look the same, NO, are they mathmatically equal, NO. so in conclusions 1 not = .999999999999999999999999999999999999

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Toxikk

yes.

.99 rounds it up to 1

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altezza

It equals to 1 for sure.

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johngalt

Having been through damn near ever [EDIT - should be every] math there is when I was in HS and College (gotta love engineering) you have to look at the big picture.

mathematically speaking, for all intents and purposes, an infinite series has no end - so when you get the the 1 millionth digit of the sequence, it is a lot closer to one than when you arrive at, say, the 1 thousandth digit.

Fundamentally, they are *not* equal - the series is *not* a representation of a true number, however, since there is no way to represent an infinite series using finite representations.

That being said, let's go back to the original calculations.

In the first one, where x was assigned the value of 0.9999... (notice the ellipses - that means it is repeating ad nauseum):

x = 0.99999...

there fore 10X = 9.99999...

Now, this part is separate from the beginning - this is where we make our *own* equations:

10X - x = 9x (this is not an equation from before, however, we are making one up to prove the equality or inequality for the sake of argument. Therefore, it is correct to create the equation this way and run it forth.)

now, we substitute the values for the X into the equation and get another equation:

9.99999... - 0.99999... = ?

There is the big question - if you have an infinite series, and you subtract another infinite series from it, can you really say that the results are going to be entirely accurate?

at first glance, it looks like the infinite series of 9s following the decimal place will in fact negate, leaving you with a value of 9. However, again, a representation of infinite series with finite representation system. So, it is our inability to accurately represent .99999... as a true number that is the flaw in the system in the first place.

Now on to the second proof:

1 / 3 = .33333... (again, the ellipses)

this is an equation, and you can manipulate both sides of the equation if you do so identically. So...

(1/3) *3 = 0.33333... * 3

Oops - here it comes.

3 thirds makes a whole - and 3 * 0.33333... again seems to result in 0.99999... - but it is the (again) inability of our system to represent infinite series that poses the problem. This is actually the proof that is used to show the *acceptance* of 0.99999... infinite series as being *equivalent* to 1.

However, here is the true logical proof:

If the series is continuous and repeating, then show me a point where it is separated from 1. For every point that you choose, there is another 9 after that - and so on - in other words, you can never come up with a number between 0.99999... and 1 - b/c the 9s never end.

Therefore, it is *accepted* that 0.99999... is the equivalent of 1.

Whew.

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Tran

Simply put .99999... = 1.

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Samoa

You also neglected in 10x-x=9x that 9 times x when x eqauls .99999... that 9 times 9 equals 81, leaving that side of the equation at 8.9999... with the impending 1. Which does not equal 10 times x. Therefore the equation is flawed.

and .9999... does not equal one

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Tran

Did you do this on calc.exe or did you actually write out all the numbers?

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figgy

I voted no.

But after reading Dazzla's proof. I know I was wrong.

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John
mathematically speaking, for all intents and purposes, an infinite series has no end - so when you get the the 1 millionth digit of the sequence, it is a lot closer to one than when you arrive at, say, the 1 thousandth digit.

Fundamentally, they are *not* equal - the series is *not* a representation of a true number, however, since there is no way to represent an infinite series using finite representations.

it is NOT 1. it is just soooooooo close to 1 that we count it as 1, that's basically what he said. 0.999... is not really 1, it's just so incredibly close (infinitely close, but not equal) to 1, that we say it's just 1.

btw, johngalt, nice explanation (Y)

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johngalt
You also neglected in 10x-x=9x that 9 times x when x equals .99999... that 9 times 9 equals 81, leaving that side of the equation at 8.9999... with the impending 1. Which does not equal 10 times x. Therefore the equation is flawed.

and .9999... does not equal one

Negative on that. the equation was made using the following proof:

if a-b=c, then ax-bx=cx

Thus, since 10-1=9, 10x - 1x = 9x. And thus it *is* correct.

If you multiply it out, you get:

9 *.9 = 8.1, and 9*.09 = .81, etc. ad nauseum - which means the inevitable result is 8.99999... (again, the ellipses, since there is no ending point where the "final" digit would result in a terminating 1.)

8.1

0.81

0.081

0.0081

0.00081...

_________

8.9999...

which, based upon the first supposition, that .99999... = 1 leads to the final result here being ... 9.

Again, though, in fact this is an unprovable inequality, since there is no way to *disprove* it other than the logic that it never *truly* approaches 1, for all intents and purposes it is *accepted* as being the equivalent of 1.

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johngalt
Simply put .99999... = 1.

No, it does *not* = 1 - but is accepted as being the equivalent of 1 since we cannot represent .99999... any better than that.

And no, I did not do any of it on a calculator (if you were asking me). I applied logic - that and some pre-calc that I learned 20 years ago (a lot of calc has to do with finite and infinite series and limits approaching a particular miniscule unit of measurement - hence the derivative.

[Edit - acknowledgment of kudos from gameguy]

Thanks gameguy. It basically comes down to how you want to define the equivalency - are you speaking for all intents and purposes, or are you speaking in the realm of pure mathematics.

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Samoa
You also neglected in 10x-x=9x that 9 times x when x equals .99999... that 9 times 9 equals 81, leaving that side of the equation at 8.9999... with the impending 1. Which does not equal 10 times x. Therefore the equation is flawed.

and .9999... does not equal one

Negative on that. the equation was made using the following proof:

if a-b=c, then ax-bx=cx

Thus, since 10-1=9, 10x - 1x = 9x. And thus it *is* correct.

If you multiply it out, you get:

9 *.9 = 8.1, and 9*.09 = .81, etc. ad nauseum - which means the inevitable result is 8.99999... (again, the ellipses, since there is no ending point where the "final" digit would result in a terminating 1.)

8.1

0.81

0.081

0.0081

0.00081...

_________

8.9999...

which, based upon the first supposition, that .99999... = 1 leads to the final result here being ... 9.

Again, though, in fact this is an unprovable inequality, since there is no way to *disprove* it other than the logic that it never *truly* approaches 1, for all intents and purposes it is *accepted* as being the equivalent of 1.

Please do not dictate math to me.

if a-b=c

then

ax-bx=cx

normally this is true.

the point being missed cosistantly..is that

x=.9999... (repeating infinately or not)

Then

10x-x=9x is not a true statement

10(.9999)-(.9999)=9(.9999) work it out yourself

oh what the heck

9.9999-.9999=8.9999(1) note that this 1 is merely that carry over until completed

9=8.9999(1) note that this 1 is merely that carry over until completed

so it is not true equation if x=.9999

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johngalt

Do not try to dictate math to me either, especially when you do not realize that what you have said I already proved as being inaccurately stated by the original poster in the first place.

And I already said *why* to - b/c of our inability to represent an infinite series using our method of representation of mathematical concepts in a finite system.

Sheesh.