Mathematics to reconsider

[Kaervas]

Think of it like a cake though, with ten even wedged shape portions joining at the center.
The pointy bits get infinitely small (the recurring 9's) so they can, in theory, join up to be effectively 99.9999999999999999999999999999 recurring
However, if you remove one slice the other slices are still the same shape (the same value).

If you assign the value of "a" as 9.9999999999999999999999999999 recurring, and deduct that from "10a" you can only ever be left with "9a" which isn't 90.

To say that 9a = 90 would be to also say that 9x 9.999999999 recurring is also 90, which it is not.


The same should apply for value of a=0.999999 recurring.



Edit:
Or maybe I'm looking at things too literally Math was never really my strong point.

[+Colibri]

No actually this is solid, at least from what they taught me in school.

1) You say that a = 0.99999... Now, a is 0.9999..., it's not 1.
2) Then you take the 10 multiple of a. Which is 9.99999... (you shift the point).
3) then you substract a from 10a, so: 9.9999... - 0.99999.... The digits after the point cancel eachother out, and you get that 9a = 9. Divide both sides by 9 and you get a = 1. Hmm...

[Kaervas]

No actually this is solid, at least from what they taught me in school.

1) You say that a = 0.99999... Now, a is 0.9999..., it's not 1.
2) Then you take the 10 multiple of a. Which is 9.99999... (you shift the point).
3) then you substract a from 10a, so: 9.9999... - 0.99999.... The digits after the point cancel eachother out, and you get that 9a = 9. Divide both sides by 9 and you get a = 1. Hmm...

The more I think about it the more confused I get, I'm really not cut out for this kind of thing and actually I don't even know why I'm so fervently defending one side over the other, haha
You can take this as my withdrawal, I'm utterly not a maths-minded person so I'm probably making some kind of error of thought somewhere.
This is where i curl up in a fetal position with my dunce hat, over which "ignorance is bliss" is messily scrawled, obscuring the big D. I can't help but feel that 0.99999 recurring will never be equal to 1, but if maths says it and I disagree, maths is going to barrage me with numbers until I give in.


Edit:
Out of curiosity I did some Googling which I absolutely don't understand, but I suppose someone else would be able to make sense of this.

Don't worry, I thought it was wrong too at one time. Let me start off with a more general principle called Geometric progression (GP for short). for example,

1, 2, 4, 8, 16... and
3, 6, 12, 24... etc.

are GP's. We shall call the first term of a GP 'a' and the number we always multiply by 'r' - the common ratio, so in the first series, it is a GP of a = 1, r = 2 and the second one is a GP of a = 3, r = 2.

Now the problem is this... What is the sum of the first n terms? Let us denote it by S(n)

S(n) = a + a × r + a × r2 + a × r3 + .. + a × r(n-1)

but now,

r×S(n) = a × r + a × r2 + ....... + a × rn

Calculating

S(n) - r × S(n)= a - a × rn

S(n) = (a - a×rn) ÷ (1 - r) = a × (1 - rn) ÷ (1 - r)

Now let's consider a special case. If the absolute value of r is less than 1 (exclusively). eg. -1 < r < 1. you can see that rn ® 0 as n ® infinity.

So solving for S(n) and let n ® infinity, we get

S(inf) = a / (1 - r)

Now recall that in 0.9 recurring, it is just an infinite sum of

0.9 + 0.09 + 0.009 + 0.0009 +....

So above formula holds with a = 0.9, r = 0.1 and S(inf) = 1 from the formula, so 0.9 rec = 1

PS the above formula does not hold for r > 1 or r < -1, since the term rn would grow as big as we like, and hence the series diverges.

[Gumbi]

yeah lol i didn't make up this equation. I saw it online and i remembered this thread and decided to post it. Based off of what I have learned, the math is correct.

[Gumbi]

o yeah whoever posted above me, that is way to much info and I have no idea what it means

[+Colibri]

Well i'm glad when i can say that, yes i understand what that calculation is about, and i'm glad it took so many years of school so that i can understand something that is actually pointless

[Kaervas]

o yeah whoever posted above me, that is way to much info and I have no idea what it means

Hehe, that chunk of stuff I quoted was some mathematician's reply to a child asking exactly the same thing that I was asking (basically being firmly convinced that 0.9999999999999999 recurring was never equal to 1, and questioning the validity of that proof you and colibri used). I wonder if the poor kid was even more confused after trying to unravel the response, just like I was

I can't even begin to understand it, though mathematics has not been a subject of study for me for something like eight years.

[Bruce]

The ONLY time it DOES equal one is in accounting. Otherwise it's NO.

[Kimitsu]

0.9999999...(infinitely) is qual to 1, and those who deny it just plainly do not know the definition of infinite periodical numbers, which is normal for people who never use those numbers (everyone except mathemeticians). the infinite periodical number is qual to a limit of a sequence Ni (i->infinity), where Ni is the "cropped" number, consisting of the first i digits of an actual number. The limit of such sequence for 0.999... is 1, because lim 1-0.999...9(i) = lim 0.000...01(i) = lim 1/10^i = 0
As another example, 0.3333.... equals to 1/3, which can be proven: lim (1/3 - 0.333...3(i)) = lim (1-0.999...9(i))/3 = 0

Bryce is wrong. The only time it does equal to 1 is in mathematics. Otherwise, they do not use infinite periodical numbers, even in accounting. The numbers are always cropped and rounded, normally. So 0.9999999 (finite) is 0.999999. But can be rounded to 1 or 0.98 if neccessary.

[Penny]

1/9 = 0.1111111....

9*1/9 = 9*0.1111111..... = 0.999999..... = 1

But we are talking about infinitely many decimal places, and for me infinity is just "something that is large enough so that I can ignore it". So for example, if your comp could do only the numbers ... 0.7 , 0.8 , 0.9 , 1.0 , ... then 0.97 would be 1

Edit:
Infinity is just mathematical nonsense, or a symbol as some other people call it That is, it's not a number
that we can imagine, for example let's cut a cake into infinite number of slices, eat 3 slices, you still have
an infinite number of slices left, now just put the cake back together somehow and your mom/wife/GF will
never notice anything

[anarchy]

hahahaha we did something like this in highschool on apple 2e. lol yeah i'm that old. ok this is what we had to do. had to program it to multiply 7 x 7= 49 right, but if you put in the square root of 49 you get 6.99999999999. so that would mean to a computer even programed right would say that 49=48.99999999999. same as with the question on 1.0=0.999999. so to a computer they would be equol. to us i believe they are not equol. let's look at it as a math aspect. a mathematition would NEVER admit that there would be that much of a estimation of mathematics so math wise they are not equol. have you ever actually listened to a really dedicated mathematition? this ice cream taste good i don't want to hear why it takes good in a mathematical equations. jeeze. ;P also to a mathematition... there is a hugh infinate possiblilities between the two identities.

[Tectum]

S(inf) = 1 from the formula, so 0.9 rec = 1

I interperate the equals sign as "tends to" i.e. 0.9 recurring "tends to" 1 or 0.9 recurring ---> 1. A number so close to 1 but never actually 1.

If i remember correctly, in maths it would be denoted as " as (S) ---> (inf), (0.9rec) ---> (1)" You see it all the time in algebra proofs.

[Kimitsu]

Current poll results clearly show:
10 - mathematicians, or the ones who learned mathematics some day
37 - can't even understand the question, because they don't know what the periodical numbers are, but have their opinion on that for some reason.

[Penny]

0.9999.. with infinite recurring is 1.

Now, how about 0.9999...9 with finite recurring

I'd say that with enough 9s, it doesn't matter anymore.. it's 1

[Striker]

As a machine tool designer i'll say that if your car's motor used that kind of tolerances it wouldnt run.