Is 1 = 0.9999......

[silverpig]

Well, at least ATOT is smarter than Tom's Off Topic

My thread in the polls section from waaaay back... still going actually too

Results thus far:

Tom's: 37% equal, 63% not equal
ATOT: 44% equal, 50% not equal

 

[Darien]

Originally posted by: silverpig
Well, at least ATOT is smarter than Tom's Off Topic

My thread in the polls section from waaaay back... still going actually too

Results thus far:

Tom's: 37% equal, 63% not equal
ATOT: 44% equal, 50% not equal

Hahaha. I wonder if all those people that voted "no" took calculus up to series

 

[halik]

.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

 

[FrustratedUser]

Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.

 

[FuzzyBee]

Originally posted by: Skoorb
if .99999999999 = 1 then it would be 1 not, .99999999

 

[FuzzyBee]

dammit, i say .9 = 1

when i ring up something that's $.90, I'm not giving any change back. if anybody has an issue, i'm going to direct them to the scholars here.

 

[Darien]

Originally posted by: fuzzy bee
dammit, i say .9 = 1

when i ring up something that's $.90, I'm not giving any change back. if anybody has an issue, i'm going to direct them to the scholars here.

Have you taken calculus?

 

[TuxDave]

Originally posted by: FrustratedUser

Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.

So where's the 1 from? After an infinite string of zeros? It's like saying at the end of a never ending road there's a girl waiting for you.

 

[PowerEngineer]

Originally posted by: FrustratedUser

Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.

The key here is "infinitely long". Following your own line of reasoning, the difference you postulate between 0.9999... and 1.0 is a 1 preceded by an infinite number of zeros (to the right of the decimal point). Any number preceded by an infinite number of zeros is zero by definition!

 

[FrustratedUser]

Originally posted by: PowerEngineer

Originally posted by: FrustratedUser

Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.

The key here is "infinitely long". Following your own line of reasoning, the difference you postulate between 0.9999... and 1.0 is a 1 preceded by an infinite number of zeros (to the right of the decimal point). Any number preceded by an infinite number of zeros is zero by definition!

Who's definition? I say it's not.

 

[alphatarget1]

it sounds like an asymptotic (sp?) number to me... well for a function that is but i can find the similarities

graph f(x)=1/(x-1). x approaches 1 but it never reaches 1

apply the same logic to this question, .999999.... isn't 1

 

[Darien]

Originally posted by: kenleung
it sounds like an asymptotic (sp?) number to me... well for a function that is but i can find the similarities

graph f(x)=1/(x-1). x approaches 1 but it never reaches 1

apply the same logic to this question, .999999.... isn't 1

lim x-> infinity of f(x) = 1/(x-1) is 0.

 

[datalink7]

This thread has proven how I suck at math

 

[bleeb]

So has the issue been resolved beyond a doubt yet?

 

[kmac1914]

Ok, ok, ok people.....How bout this:


Who likes pie?

 

[silverpig]

Originally posted by: kenleung
it sounds like an asymptotic (sp?) number to me... well for a function that is but i can find the similarities

graph f(x)=1/(x-1). x approaches 1 but it never reaches 1

apply the same logic to this question, .999999.... isn't 1

I'll say it again for the 4th time this thread.

0.999... is NOT a progression. It doesn't approach anything. It's a number. That's like saying when 2 approaches 3...

 

[Haircut]

Holy crap, this thread is still going?
It's obvious this is getting like the HT thread on the same subject, two sides debating the point and neither refusing to back down.

One side, the side that knows math, put up lots of proofs showing that it is true.
The other side saying that it just can't be or bleating on about numbers such as 0.00....01 which cannot exist but never actually disproving anything that has been said by the other side.

I think we should just give up now, let the people who voted no carry on believing what they want to believe, but please can we let this thread die.

 

[SilentRunning]

Silverpig what is the fraction which is equal to 0.999999........ and don't say 1/1 because dividing 1 by 1 equals 1 with a remainder 0.


If you can represent 0.999999..... with a fractional equivalent then I will concede that 0.99999.... is not an irrational number.

 

[Woodchuck2000]

Originally posted by: SilentRunning
Silverpig what is the fraction which is equal to 0.999999........ and don't say 1/1 because dividing 1 by 1 equals 1 with a remainder 0.


If you can represent 0.999999..... with a fractional equivalent then I will concede that 0.99999.... is not an irrational number.

0.99999.... = 0.33333...x3 = 1/3 x 3
Repost, but it was worth it

 

[SilentRunning]

Originally posted by: Woodchuck2000

Originally posted by: SilentRunning
Silverpig what is the fraction which is equal to 0.999999........ and don't say 1/1 because dividing 1 by 1 equals 1 with a remainder 0.


If you can represent 0.999999..... with a fractional equivalent then I will concede that 0.99999.... is not an irrational number.

0.99999.... = 0.33333...x3 = 1/3 x 3
Repost, but it was worth it

Sorry but 1/3 x 3 is an equation, not a fractional equivalent.

 

[Woodchuck2000]

Do you dispute that
0.333333... x 3 = 0.999999
?

I left the last part (=1/1) implicit...

 

[SilentRunning]

Originally posted by: Woodchuck2000
Do you dispute that
0.333333... x 3 = 0.999999
?

I left the last part (=1/1) implicit...

No I do not dispute that. But I dispute that 0.3333333...... = 1/3. You still have a remainder when you divide 1/3. Lets say you have a decimal precision of 10.

Then 1/3 = 0.3333333333 with a remainder of 0.0000000001.

So 3 x 0.3333333333 = 0.9999999999
+ remainder 0.0000000001
--------------------------------------------------
equals 1.0000000000

 

[Alphathree33]

Originally posted by: SilentRunning

Originally posted by: Woodchuck2000 Do you dispute that 0.333333... x 3 = 0.999999 ? I left the last part (=1/1) implicit...

No I do not dispute that. But I dispute that 0.3333333...... = 1/3. You still have a remainder when you divide 1/3. Lets say you have a decimal precision of 10. Then 1/3 = 0.3333333333 with a remainder of 0.0000000001. So 3 x 0.3333333333 = 0.9999999999 + remainder 0.0000000001 -------------------------------------------------- equals 1.0000000000

Very good, you've just shown for yourself that the remainder is getting smaller and smaller.

Now, go do that infinitely many times and let me know how much of a remainder you have left.

You know how in calculus class when they were first teaching you limits and they made you punch in higher and higher values of n for some function and you went hmm 2.5, 2.75, 2.9, 2.95, 2.999, 2.999999999, 2.99999999999999999999999999

HEY!!! THERE'S A F*CKING PATTERN HERE! HOLY SHIT!

And then your calculus teacher told you that Newton already figured it out five hundred years ago?

That whole situation?

Yeah, it applies here, chief.

 

[SilentRunning]

Originally posted by: Alphathree33

Originally posted by: SilentRunning

Originally posted by: Woodchuck2000 Do you dispute that 0.333333... x 3 = 0.999999 ? I left the last part (=1/1) implicit...

No I do not dispute that. But I dispute that 0.3333333...... = 1/3. You still have a remainder when you divide 1/3. Lets say you have a decimal precision of 10. Then 1/3 = 0.3333333333 with a remainder of 0.0000000001. So 3 x 0.3333333333 = 0.9999999999 + remainder 0.0000000001 -------------------------------------------------- equals 1.0000000000

Very good, you've just shown for yourself that the remainder is getting <STRONG>smaller and smaller.

</STRONG>Now, go do that infinitely many times and let me know how much of a remainder you have left.

You know how in calculus class when they were first teaching you limits and they made you punch in higher and higher values of n for some function and you went hmm 2.5, 2.75, 2.9, 2.95, 2.999, 2.999999999, 2.99999999999999999999999999

HEY!!! THERE'S A F*CKING PATTERN HERE! HOLY SHIT!

And then your calculus teacher told you that Newton already figured it out five hundred years ago?

That whole situation?

Yeah, it applies here, chief.

Let me guess, your teacher taught you that e = mc^2 too.

You do realize that calculus is all about approximations don't you?

 

[Alphathree33]

Sorry... continuing my thought in a new post.

It's also important to realize that if f(x) converges on a value as x grows beyond all bounds, it never does reach the limit for any finite value of x. You are correct that for every single value of x you choose (i.e. number of 9s you choose), there will always be a difference between f(x) and lim f(x). That's the whole point. For finite values, you never get there.

When you do it infinitely many times, you do.

Take a look at the expansion series for e. Are you going to argue with me that, because if I only take the first ten terms of e, e ~= e? or e != e? That's a really nice argument you've got there.

You cannot find e by taking the summation of any n terms of its expansion series when n is a finite number. And yet e is still identically equal to e. Imagine that.

Just as 0.9 repeating is identically equal to 1, 1.9 repeating is identically equal to 2, 2.9 repeating is identically equal to 3, and so on.

Things like 0.8 repeating are not equal to 0.9 because there is a number between 0.888... and 0.9. There are infinitely many such numbers. For example 0.89 is one such number.

Between 0.9 repeating and 1 there is not another number. This can't happen in the real number system. For any two numbers x and y, you can always find another number between x and y by finding x + y / 2 (lets say for positive x and y).

When you do this for x= 0.9 repeating and y= 1, you get 1.9 rep. / 2. That's going to be either 1 or 0.9 rep. So in this case, x + y / 2 gives you either x or y back. Find me two other positive numbers in the real system that will do this for x != y. The idea here is this: if you find the midpoint between two points, you will ALWAYS get a NEW point unless the two original points were identically equal (they were on top of one another).