Foo - .9999999999999999999999... = 1?

Prove it.

c = .9999
10c = 9.9999...
10c-1 = 9.99999... - .99999...
9c = 9
c = 1

There is no number you can find between them, thus they are the same.

Proof complete.

c = .9999
10c = 9.9999...
10c-1 = 9.99999... - .99999...
9c = 9
c = 1

This one is circular. In your third statement you're using -1 = -.99999..., so you're using what you're trying to prove as an axiom.

pretty much what KingTermite said. explicitly we need a theorem that says between any two reals there is another real. if .99999bar and 1 are different numbers, what could this number between them be? well, in order to be less than 1 and greater than .99999bar, this number would need to be not a 9 in some decimal place, but that means that it's less than .99999bar, a contradiction since the number needs to be between the two, so .99999bar and 1 must not be different numbers.

Okay, this one is thanks to jfmckenna. Wouldn't have thought of this approach if he hadn't planted the seed.

c = 0.99999...
10c = 9.99999...
10c - c = 9.99999... - 0.99999....
9c = 9.0
c = 1.0

Since c = 0.99999... and c = 1.0 this proves that 0.9999... = 1.0

This one is circular. In your third statement you're using -1 = -.99999..., so you're using what you're trying to prove as an axiom.
Good Eye!

Good Eye!

I think he simply typo'ed. I think he meant to subtract 'c' rather than 0.99999.... You'll notice that when I restated his proof with the correction, it then makes sense.

//

Okay, we have proof by contradiction and what I believe to be proof by induction. Any others?

Good Eye!
Yeah..but his doesn't have lye in it. :)

Couldn't you just round up?

I always hated this proof.

I think he simply typo'ed. I think he meant to subtract 'c' rather than 0.99999.... You'll notice that when I restated his proof with the correction, it then makes sense.

//

Okay, we have proof by contradiction and what I believe to be proof by induction. Any others?
yea that was a typo.

really it depends on how you look at it. If you accept that .9999... is just an infinite length of 9's then it is different then it's limit which is 1. That is interesting though, what is 1 - .999999?

Couldn't you just round up?

Horray for common knowledge!

until I am navigating space travel, I'll just consider 1=.99999etc for the sake of convenience

Basically you just keep measuring until your instrument runs out of precision, then you're done.

Here's a simpler proof

1/3 = .333...

3 x 1/3 = 1
3 x .333... = .999...

Therefore
1 = .999...

Wgasah?

I ain't never been much for math, so how about a practical example. I'll hit you up side the head with full force, then I'll hit you up side the head with .99999999999999999999999999999999999999 force....you tell me if you feel a difference.

Horray for common knowledge!
hahahahahaha

I can't prove it, but I bet this guy can.

I never knew 25 divided by 5 is 14.

http://www.zippyvideos.com/2309578745009296/ma___pa_kettle_math/

I remember in math class once some of the students passing around a proof that proved 0 = 1. I can't remember how it was shown. It didn't using any calculus or converging series, it only used algebra.

Anyone have that proof?

Isn't the answer that it doesn't?
0.999... approaches 1 asymtotically, but never in fact reaches 1.
0.999... = 1 - (1/10)exp(n) [i.e., raised to nth power]

As n approaches infinity, (1/10)exp(n) is always a positive, non zero number. At best, you can say that the limit of .999999... is one.

Isn't the answer that it doesn't?
0.999... approaches 1 asymtotically, but never in fact reaches 1.
0.999... = 1 - (1/10)exp(n) [i.e., raised to nth power]

As n approaches infinity, (1/10)exp(n) is always a positive, non zero number. At best, you can say that the limit of .999999... is one.
What he said! http://www.madhatter.free.fr/smileys/albert.gifhttp://www.madhatter.free.fr/smileys/albert.gif

Isn't the answer that it doesn't?
0.999... approaches 1 asymtotically, but never in fact reaches 1.
0.999... = 1 - (1/10)exp(n) [i.e., raised to nth power]

As n approaches infinity, (1/10)exp(n) is always a positive, non zero number. At best, you can say that the limit of .999999... is one.
I think you are missing a vital point.

You are correct if you take any particular (finite) number, but I believe the OP meant .9999 repeating. You can not take a "snapshot" and compute to prove it because any "snapshot" is not .9999 repeating. Being finite, of course it is not equal to 1.

Exactly. SpongeDad's outline of a proof by induction fails because proving something for n = 1 and for n +1 for arbitrary whole number values of n does not prove something for n = infinity.

I'm not sure how the proof goes because it all depends on what you take as a "given" headed into the proof, and I'm not brushed up on my number theory. I suspect the proof depends on the proof that real numbers within any arbitrary range are not quantized (readily proven using basic set theory, where all of math starts) and that therefore, for any two numbers, if the numbers are not the same, then there must be a number of a value intermediate between the two. (If you could find two numbers "right next to each other", then those numbers are quantized.) Once that fact is accepted, you show that you cannot find a value intermediate between 0.999... and 1. Therefore, 0.999... = 1.

For a "proof" that would satisfy a high school student who has learned about geometric series, I would simply ask the student to calculate the value of 0.9 + 0.09 + 0.009 + ...

The answer using the accepted formula for the sum of an infinite geometric series yields (starting value) / (1 - multiplier) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1

Yes,why? Because you round up.

Exactly. SpongeDad's outline of a proof by induction fails because proving something for n = 1 and for n +1 for arbitrary whole number values of n does not prove something for n = infinity.

I'm not sure how the proof goes because it all depends on what you take as a "given" headed into the proof, and I'm not brushed up on my number theory. I suspect the proof depends on the proof that real numbers within any arbitrary range are not quantized (readily proven using basic set theory, where all of math starts) and that therefore, for any two numbers, if the numbers are not the same, then there must be a number of a value intermediate between the two. (If you could find two numbers "right next to each other", then those numbers are quantized.) Once that fact is accepted, you show that you cannot find a value intermediate between 0.999... and 1. Therefore, 0.999... = 1.

For a "proof" that would satisfy a high school student who has learned about geometric series, I would simply ask the student to calculate the value of 0.9 + 0.09 + 0.009 + ...

The answer using the accepted formula for the sum of an infinite geometric series yields (starting value) / (1 - multiplier) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1

So, for all those words, all you are really saying is that my first answer was correct. :p

Of course my bro is correct. :) But it sounds so much more authoritative when you throw in terms like "set theory" and "quantized". :p

I remember in math class once some of the students passing around a proof that proved 0 = 1. I can't remember how it was shown. It didn't using any calculus or converging series, it only used algebra.

Anyone have that proof?

Yes, if I had the time I could reconstruct it. The start is something like x=y. The key step is when you divide both sides of the equation by (x-y), which is dividing by 0 and thus not a valid step.

Of course my bro is correct. :) But it sounds so much more authoritative when you throw in terms like "set theory" and "quantized". :p
show off :D

Damn! If 0.99999... is not < 1, I wonder how many other answers I got right on math tests I took 40 years ago that the teacher got wrong.

Come on, technically .9 cannot be 1.

Come on, technically .9 cannot be 1.
Nobody yet said that .9 was equal to 1.

Yes, if I had the time I could reconstruct it. The start is something like x=y. The key step is when you divide both sides of the equation by (x-y), which is dividing by 0 and thus not a valid step.

That's it. Let's see:

x = y
x - y = 0
(x-y)/(x-y) = 0/(x-y)
1 = 0/(x-y)
1 = 0

Nobody yet said that .9 was equal to 1.
Oh ok:D

I like slvoids method the best :D

These type of proofs annoy me. I thought #'s were infinite, I see no reason they could not be. Couldn't you just keep adding 9's for lightyears in your size 12 font? Isn't the universe said to be expanding? Maybe your fingers can move fast enough to keep up with the rate of expansion.

This makes me think of that stupid x^0 = 1. NO IT DOES NOT. Arrgh! Power means multiply a number by itself the given amount of times the exponent indicates. so 3^2 = 3*3
3^3 = 3*3*3

Well if the exponent says zero, and zero is out way of quantifying nothing. Thats 3 * nothing. How does 3*nothing equate to 1? That means you are multiply by the absence of quantity.

This makes me think of that stupid x^0 = 1. NO IT DOES NOT. Arrgh! Power means multiply a number by itself the given amount of times the exponent indicates. so 3^2 = 3*3
3^3 = 3*3*3

Well if the exponent says zero, and zero is out way of quantifying nothing. Thats 3 * nothing. How does 3*nothing equate to 1? That means you are multiply by the absence of quantity.
Would you like your math to be internally consistent?

3^(-1) = 1/3
3^0 = ???
3^1 = 3

Keep in mind that 3^(x+1) = (3^x) * 3

Keep in mind also that exponents do not have to be integers. For that matter, exponents don't even have to be real numbers. How would you make sense of such beautiful equations as e^(i theta) = cos theta + i sin theta (brownie points if you know what equation this is) if you arbitrarily set x^0 = 0? You would end up with a discontinuity right at theta = 0.

I just say x^0 is 1 for my own calculations, so I don't lose pts on a test ;p

In that equation, what do e and I standfor? Current and electromotive force is the only thing I can think of that use those abbreviations

e is "euler's number", the base for natural logarithms.


edit: http://en.wikipedia.org/wiki/Natural_logarithm

I just say x^0 is 1 for my own calculations, so I don't lose pts on a test ;p

In that equation, what do e and I standfor? Current and electromotive force is the only thing I can think of that use those abbreviations

The first thing you have to remember is that some things are axioms. They just are what they are because everyone agreed to call them that. Out of convenience. From the axioms you start building.

ooooooooh! 2.78 or something of that like? I believe you can use that # to find continuous interest, I remember "pert". I was playing with my Ti the other day in history and the number .7071067812 kept showing up with functions I am not even familiar with.
I know this from Cosine and Sin of 45, and RMS conversions. I don't remember what functions I used. Seems like a special # in nature. I might make a T-shirt to show geek pride :D

If you like funky numbers, you should read up on the Fibonacci series.

Is it safe to say that between any whole number there are infinite amounts of that said whole number with a decimal and infinity? So really is it possible to even go from one whole number to the next without "cheating" in the count?

EX.

1
2
3

or is it really
1
1.999999999999infinity and 2 doesnt exist as it isnt possible number wise??

ooooooooh! 2.78 or something of that like? I believe you can use that # to find continuous interest, I remember "pert". I was playing with my Ti the other day in history and the number .7071067812 kept showing up with functions I am not even familiar with.
I know this from Cosine and Sin of 45, and RMS conversions. I don't remember what functions I used. Seems like a special # in nature. I might make a T-shirt to show geek pride :D

Yes, it's like a special number in nature that Euler discovered. It's kinda like Pi, but for other schtuff. ;)

Or could you really even make it to 1?

as in 0 would be followed by 0.0000001infinity and you could never make to 0.02 because you could go on forever with 0.0 with a 1 never being added to the end. I dunno...

In that equation, what do e and I standfor? Current and electromotive force is the only thing I can think of that use those abbreviations
e is the natural number 2.718...
i is the square root of -1

Yes, that equation is used in electricity/magnetism since certain phenomena have oscillations that allow easier math if you work in the complex plane (with an unimportant imaginary component coming along for the ride just to simplify math) and use that equation rather than only dealing with the real component. The equation allows you to switch back and forth between radial description (sin, cos are oscillating terms) and linear description (ie a point on the complex plane a + bi), allowing you to use whichever description is more convenient.

e^(a + bi) = e^a * (cos b + i sin b)

I vaguely remember doing some natural log stuff utilizing e in my algebra 2 days.
e^norris = ?

e is the natural number 2.718...
i is the square root of -1

Yes, that equation is used in electricity/magnetism since certain phenomena have oscillations that allow easier math if you work in the complex plane (with an unimportant imaginary component coming along for the ride just to simplify math) and use that equation rather than only dealing with the real component. The equation allows you to switch back and forth between radial description (sin, cos are oscillating terms) and linear description (ie a point on the complex plane a + bi), allowing you to use whichever description is more convenient.

e^(a + bi) = e^a * (cos b + i sin b)

oh, you are talking imaginary #'s. Complex plane is where imaginary #'s take the place of the Y axis, real take the place of the X, correct? NOw with imaginary #'s, isn't 4 some sort of a ratio? This is good to jog my memory from a last period class. I think you can use the complex # plane to determine impedance for an AC circuit. Something like capacitance and inductance are treated as real, while their respective reactances are treated as imaginary? I also see imaginary #'s written has "j" sometimes, not i.

Yeah, j is sometimes used, especially among physicists. As for questions about AC circuits, I refer you to your book. I don't remember that stuff. :p

Yeah, j is sometimes used, especially among physicists. As for questions about AC circuits, I refer you to your book. I don't remember that stuff. :p

I have never seen you stumped with any science/math related topic before. Don't let it end now, fill your mind with more knowledge!