Why .9r = 1
 

Yeah, so this was written on the chalkboard in math class. I'm using an r to represent the "repeating bar" in an irrational term.

Any thoughts? It makes sense, although I don't know if it's mathematically polite to do what he/she did in line three -- subtracting an x from one side and a constant from the other ven though they are equal.

You most likely right about the line 3 deal. x isn't really x, it's .9r. He/She is actually
subtracting 10x - .9r, which is definitaly not 9x. It's a hoax... or whatever... Probably just trying to get you guys to think...

If this somehow did workout (it doesn't, just hypothetical) you're teacher would be immediately awarded
a position at Fine Hall at Princeton to figure out how to fix Math. :P

So goto school Monday and tell you're teacher that you're not as stupid as you look! :D

no, what you're teacher's telling you is true; one of my teachers showed me the same thing a few years ago. think about it another way:
.3 repeating is equal to 1/3, and
.6 repeating is equal to 2/3, so
.9 repeating is equal to 3/3 which is equal to 1.
it's just one of those crazy things in math.

Erm, no, the problem is totally wrong, and even my first explaination was wrong, in a way.

The problem looks like this.

x = .9r
10(.9r)=9.9r
10(.9r)-.9r=9.9r-.9r
8.1r=9.0r
r=1.11r
1=1.11 (Which is the correct answer, although of course, the not true one... something like that.)

and .3333333 does not equal 1/3rd. .33 repeating is irrational and can't be expressed as a fraction, .3333 repeating is a close approximation of 1/3rd.

So no, it still doesn't work.

i hate to tell you, my friend, but you're wrong about the whole 1/3 thing. .3 repeating isn't irrational; it's rational because a rational number is defined as any number whose decimal repeats or terminates: 1/3 is exactly equal to .3 repeating, not a close approximation. if you take 1 and divide it by 3 (hence 1/3), you'll see that all you get is .33333...and so on. and as a side note, an irrational number is any number whose decimal component goes on, without terminating and without a distinguishable pattern, such as pi (3.1415...) or e (2.71...). so the whole .9 repeating = 1 is indeed true.

nah, sirsnekcip has it right. .3 repeating does exactly equal 1/3...

.99999999999 repeating into infinity is equivalent to 1.

This can be proven by using a geometric series. But i don't want to go through the whole proof. (though it follows the same idea that you have in your post)

1 / 3 = .3r is as much of an approximation as saying 1 = .9r

Gah! Ok, but I'm still right about the original problem, let's fight about 3rds some other time! :thumbsup: *Stomps off to Princeton to ask someone there.*

Here is another way
.9r = .999999999... = 9*(.1111111111...)
.9r = 9*sum(i=1 to inf, (.1)^i) <= read the sum of i=1 to infinity of (.1)^i

sum(i=1 to inf, (.1)^i) = an infinite geometric sum (.1^0 + .1^1 + .1^2 + .1^3... ) - 1
The sum (.1^0 + .1^1 + .1^2 + .1^3... ) is well known to converge to 1/(1-r) where r = .1, so,
.9r = 9*(1/(1-.1) - 1)
.9r = 9*(1/.9 - 1)
.9r = 9*(1/(9/10) - 1)
.9r = 9*(10/9 - 9/9)
.9r = 9*(1/9)
.9r = 1

Hate to be the one to break it to you, but when you multiply both sides by ten, .9 does not become 9.9.

.9 x 10 = 9 At least, by my calculations.

x = .99r
10x = 9.90r
10x - x = 9.9r - .9r
9x = 8.9r1 (because if we are dealing with say any number of 9s the multiplied number will have one less dp than the original number, so will produce a 1 at the end of that number...)
x = 0.9r

I do not see the point? x = 0.999999999999
10x = 9.99999999999
9x = 8.99999999991
x = 0.99999999 ~1

What is this meant to prove? That 0.999999999999999999999999999999999999 is ~1 for most things?


This is similar to what our maths lecturers told us... n/infinity = 0. X * n/infinty != 0... its an approximation.

nonono..

CoachAlan, in the first post, he used .9r to represent .9999999999... ,

so 10x .9999999999... = 9.99999999999...

I can understand the fact that .9r = 1 but take a look at your statement:

10x = 9.9r

Since you have already given the definition of x to be .9r, lets go ahead and plug it in here for the heck of it. This gives: 10x = 9x

This is the part that I cant understand :confused:

Thats because your algerbra is wrong there

You can't simply plug in .9r like that. You basically said if I have b=.123 and I have 4.123 then 4.123=4b which it isn't what you probably wanted to say is 4.123=4+b.

Thats the geometric series proof I mentioned ;)

Ouch, there are way too many people in here that are posting incorrect algebra and flawed logic it's making my head hurt. :)

Well, um, yeah. That's where you went wrong. How did you get 8.1?

What do you mean? When I multiply .9r by 10, I get .9almostr or .9r with one less decimal place? I think that inifinity minus any non-infinite number is still infinity. Especially one, since one is so... oneish. This is when you get into those freaky (*cough* obsessive *cough*) metaphysical debates over the nature of infinity.

Sorry about the misunderstanding of the "r" notation in my previous post.

I think what's basically going on here is that .9r is essentially equivalent to 1. It's as close to one as you can possible get, without actually being one. In calculus, if I'm not mistaken, it would be a number whose limit is one. That number, mathematically, is treated as a one because we don't have an adequate way of dealing with infinite numbers. At least, not in the way it's being used here.

By the way, I've only taken precalculus, so we haven't really gotten in to limits. If I'm off base about how this works, please correct me.

0.9r is exactly equal to 1. They both symbolically represent the same number.

Its not "..s as close to one as you can possible get, without actually being one.." , it IS 1.

0.9r is definied by a series sum, which can be shown to equal 1.

http://www.fortress-forever.com/uploads/converge.jpg

Look at it this way:

1/infinity = 0

.9r = 1 - 1/infinity

.9r = 1 - 0

.9r = 1

There is no need for any approximation anywhere, since 1/infinity exactly equals 0.

Also, .3r does exactly equal 1/3

Ok, turns out I made another mistake.

x=.9r
[Substitute Time!]
.9r=.9r
9r=9r (Multi both by 10)
9r-.9r=9r-.9r (Sub by .9r)
8.1r=8.1r (Doing da Math)
8.1=8.1 (Divide both by r)
1=1 (Divide both by 8.1)

Just weird algebra, don't know what they're trying to get at.

twistedmosaic, unfortunately you cannot do maths like that. "infinity" isnt a number such that you can use it in the arithmetic you did.

also, Paradise Lost's proof is not correct either, there is no logical implication from one statement to the next. Unless you inteded to Show that that 1=1. Also dividing by r makes no sense. "r" isnt a number.

One might as well do

x=100r
[Substitute Time!]
100r=100r
1000r=1000r (Multi both by 10)
1000r-100r=1000r-100r (sub by 100r)
900r=900r(doing da math)
900=900(divide both by r)
1=1 (divide both by 900)

Do you see it makes no sense. The above doesnt mean that 100=1!

x = .9r
10x = 9.9r
10x - x = 9.9r - .9r
9x = 9
x = 1

9.9r-.9r does not give you 9. It would give you 9r...
Then 9x=9r.
so 9(.9r)=9r
so .9r=r
Which doesn't work.

x = .9r
10x = 9.9r
10x - x = 9.9r - .9r
9x = 9
x = 1

How on earth do you get 10x=9.9r? If you multiply both sides you get 10x=9r...

And if you added 9x and 9r, well you can't do that. And even if you could, and then divide by 10, you get x=.99r

Munku you are confused on what r means.

r just means that digit repeats into infinity.

so if i have .99999999999999999999999999999999999...
and i multiply by 10 i get 9.99999999999999999999999999999...

so in this proof you say
x=.9r multiplying both sides by 19 gets us
10x=9.9r subtracting x from both sides gets us

9x=9.9r-x substituting .9r in for the second x gets us

9x=9.9r-.9r then 9.9r-.9r=9 (remember r is not a variable it is just a repeating of digits)

9x=9
x=1

If you were only multplying 10 and 0.9, then you would get 9.

However, you are multiplying 10 by 0.9r, which is 0.99999999999999... nine's out to infinity.

0.999999999999999999999 * 10 = 9.9999999999999999999999999.

Trust me, having been through too many years of Algebra, Calculus, and other maths, there are about four different ways to prove that 0.9 repeating is EXACTLY equal to 1.

The geometric series mentioned above is one of them... the algebra is another. The simple fractional one is yet another. 1/3 = 0.3repeating, 2/3 = 0.6repeating, and 3/3 = (1/3 * 3)... which equals 0.9repeating, but as we all know, 3/3 = 1. Or, you could break it down into ninths.

1/9 = 0.11111...
2/9 = 0.22222....
3/9 = 1/3 = 0.3333333333...
Keep on going... 8/9 = 0.888888...
9/9 = 0.999999999, but a number divided by itself is 1. So they are the same.

I dont see how 1/3 = .3r is any less of an estimate as 1 = .9r

Damn, fur's flying in this thread :)

I think that's the point. Neither are estimates, they are just different ways of expressing the value.

Hey, daking, I forgot sigma notation right after the test, but I'll assume that what you're doing is correct, since you took the time to actually scan it in. :D

1/3 is easily provable to be .333333r

just do long division and you will see that it is not an estimate and it is exact. You have a pattern that will repeat into infinity.

Based on the proofs I have seen above, I would have to agree that 9.99999999999999... is in fact exactly one.

Math is cool.

edit: oops. I sit corrected. I meant 10.

Actually that would be ten! :D

Haha. Yeah, math can be cool. But it gave me one bugger of a headache today...

.99999999... converges to 1, and for all real-world applications is equal to 1.

However, there is the question of whether 1/infinity (the number that .9999... would have to be added to to equal 1) is equal to zero or not is really the point we are debating.

I say that since the numerator is a non-zero term, that it is not actually equal to zero, no matter how large the denominator is.

Of course, since 1/infinity isn't a real number (since infinity is also not a real number), it just gets messy.

I think your a little confused.

".99999999... converges to 1, and for all real-world applications is equal to 1. "

No , this is wrong. You state converges to, but how can you talk about converges in respect of a single number. A finite summation may converge as the number of terms increases to infinity. But 0.9r is not a finite summation it is a symbolic represntation which exactly equals one. We dont say 0.3r converges to 1/3 , convergance doesnt come into it, it is exactly equal to 1/3. To talk about convergance you must provide an ordered measure.

In summary you can say f(n)=9*sum(0.1^i,i=1,n) converges to 0.9r as n(in N) tends to infinity. 0.9r exacty is 1, in real world or the mathematical world.

Here you make the mistake of accepting the notation of 1/infinity as valid. It is not valid or correct notation, it makes no sense.

In such cases it does pay to consider convergance. Consider

y=sin(x)/x

at x = 0, we have y=0/0 which is undefined. But ofcourse by the limit as x tends to zero is 1 by l'hopitals rule .

See if you can work out what the limit is as x tends to infinity :)

The limit is zero.

Notation is often defined by the person employing it, how can it be invalid in this instance? Certainly the concept of one divided by infinity comes up daily in the study of calculus. The limit of 1/x as x approaches infinity can be thought of as 1/infinity, which by the definition of limit is zero. 1/infinity = 0

The problem with accepting such notation as 1/ infinity, is that engenders the idea that infinity may be used within the structure of arithmetic. You then get such nonsense as infinity + 1.

Sure as a short hand it can be noted , however in rigourous calculus such notation is often more damaging than useful.

If we permit this kind of notation to enter mathematical proof sin(x)/x as x tends to infinity would be represneted as sin(infinity)/infinity. Then one might theorize that sin(infinity) is always less than 1 and so a finite number leading to the conclusion that the limit is 0. Where as the limit is obviously 1.

That same function sin(x)/x (xinR) is undefined and discontinuous at 0. Even tho the left and right limits as x tends to 0 are convergent and the same. It is a removeable singularity, The function needs to be extended to be continuous and defined on all real numbers.

The concept of Infinity needs to be carefully applied as more serious mistakes and errors in proof than that pointed out above have occured.

To wit(or not :))

1/infinity=0

so infinity/infinity=0*infinity. well 0 times anything = 0 and anything divided by anything =1. so 1 = 0.

Notation is only useful if people know what they are saying. That being said, anyone who isn't clear on the concept of infinity would be confused anyway. You could also claim that the notation for sin^-1(x) could easily be confused with 1/sin(x), but the fact is that mistakes like that are made by people who don't haven't been paying attention in math class.

Agreed.

In the context of this thread the notation of 1/infinity is definitely inappropriate as we get proofs such as this :