Originally posted by Sam The Sham.99 = 1
Simple algebra will demonstrate it, no calculus needed. Here's the proof:
.9999.....= 1
10(.99999....)= 10 (1) Multiply each side by ten, you get:
9.9999...... = 10 Then subract 9 from each side
9.999999.... -9 = 10 - 9
.999999......= 1
Learned that in the 7th grade. Your problem Ivan is you're thinking in finite terms, the infini ...[text shortened]... 33333.....
3 (1/3) = 3 (.333333.....) (multiply each side by 3)
1 = .999999999........
10(.99)= 10 (1) Multiply each side by ten, you get:
9.9 = 10 Then subract 9 from each side
9.9 -9 = 10 - 9
.9 = 1
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Originally posted by ThinkOfOneGood point, Sam the Sham's solution assumes the result 1 = 0.9999... by using it as the starting point. I believe he meant something more along these lines:
.99 = 1
10(.99)= 10 (1) Multiply each side by ten, you get:
9.9 = 10 Then subract 9 from each side
9.9 -9 = 10 - 9
.9 = 1
(1) 0.9999... = c
Multiply both sides by 10:
(2) 9.9999... = 10c
Subtract (1) from (2):
9 = 9c
c = 1
QED
2 edits
Originally posted by PBE6Thank you, yes.
Good point, Sam the Sham's solution assumes the result 1 = 0.9999... by using it as the starting point. I believe he meant something more along these lines:
(1) 0.9999... = c
Multiply both sides by 10:
(2) 9.9999... = 10c
Subtract (1) from (2):
9 = 9c
c = 1
QED
2 edits
Originally posted by PBE61) a = c
Good point, Sam the Sham's solution assumes the result 1 = 0.9999... by using it as the starting point. I believe he meant something more along these lines:
(1) 0.9999... = c
Multiply both sides by 10:
(2) 9.9999... = 10c
Subtract (1) from (2):
9 = 9c
c = 1
QED
Multiply both sides by 10:
(2) 10a = 10c
Subtract (1) from (2):
9a = 9c
c = a
2 edits
Originally posted by Sam The ShamAnd not only this number, but all decimal numbers with a finite number of decimals.
.999999999999...(repeating to infinity) = 1
Like 1.2345 = 1.234499999999... (never ending 9's)
Not approximately but exactly!
It's quite funny (but natural) that there is always two ways to express such a decimal number.
Originally posted by ChessJesterWhat do you mean "the smallest number possible above zero"?
Ok. Is 0.0(repeating)1 a number?
If you try to think of the smallest number possible above zero, you come to this conclusion. Yet, if you try to add it together, it will never amount to anything greater than itself because it is an infinitely small decimal place. Basically, this number is zero?
so 0.0(repeating)1 = 0?
and does 0.0(repeating)2 = ...[text shortened]... it is nothing? it isn't a difference at all just a different representation of the same number?
I have no idea what you are talking about with that question.
Think Chemistry:
1 ppm = 1 g / 1000000 g = .00001%
That is easily a smaller amount than 0.0111111 and easily measured.
OK, maybe this doesn't help...just random knowledge.
😛
Ahh yes, it is because infinity has reared its ugly (or beautiful) head.
And beauty is in the eye of the beholder... or in this case, the mind of a mathematician.
From what I've gathered, we use limits to describe an infinite equation like this. For example 0.00000000...(to infinity)1 is actually describing the equation: [0.1 - 0.09 - 0.009 - 0.0009...etc] (to infinity) or the sequence, [0.1, 0.01, 0.001...] (to infinity) The limit (that point it cannot reach) of this equation is zero because the equation will approach zero but never reach it. Zero is its limit. To say that 0.000...(to infinity)1 is a real number is wrong, because it is technically not a number, it is the equation: [0.1 - 0.09 - 0.009 - 0.0009...etc] (to infinity) or the sequence [0.1, 0.01, 0.001...] (to infinity). The real number is the limit of the equation or sequence!
Meaning, that 0.00000000...(to infinity)1 = 0
and 0.999999999(to infinity) = 1
IF thats correct, I've come a long way in my understanding of this today.
http://en.wikipedia.org/wiki/Limit_%28mathematics%29#Limit_of_a_function_at_infinity
And, yes, I see that I should have posted this is posers and puzzles. Sorry.
Originally posted by ThinkOfOneIf you believe that's a counter-argument to the statements above, I think you're missing the basics.
1) a = c
Multiply both sides by 10:
(2) 10a = 10c
Subtract (1) from (2):
9a = 9c
c = a
In the previous proof, we assumed that 0.999... was some number "c". By following the rules of algebra, we showed that "c" was in fact 1. In your proof, you assume that two numbers are the same. After following the rules of algebra, you show that these two numbers are in fact the same. Since that works for any particular number you choose, all you have is a tautology.
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Originally posted by ChessJesterThe number you are talking about, the smallest number > 0 there is, let's call this epsilon for a while.
Ok. Is 0.0(repeating)1 a number?
If you try to think of the smallest number possible above zero, you come to this conclusion. Yet, if you try to add it together, it will never amount to anything greater than itself because it is an infinitely small decimal place. Basically, this number is zero?
so 0.0(repeating)1 = 0?
Another number, 1/epsilon, or "almost infinity", is the highest possible number < infinity.
Sorry to say, none of these numbers exist. You can define them, but you cannot do anything with them with normal arithmetics rules.
Even infinity itself, you cannot use within R. Let's call this number inf.
inf+1 = inf, inf-1 = inf
inf*2 = inf, inf/2 = inf
inf^2 = inf, sqrt(inf) = inf
and so on. Whatever you do with infinity it stays infinity.
But inf-inf can be whatever.
inf/inf can also be whatever.
So the sense moral of this is to never treat infinity, almost infinity, or infinitly near zero as a number, just don't.
Originally posted by FabianFnas(inf+1)-(inf) = 1 ??
The number you are talking about, the smallest number > 0 there is, let's call this epsilon for a while.
Another number, 1/epsilon, or "almost infinity", is the highest possible number < infinity.
Sorry to say, none of these numbers exist. You can define them, but you cannot do anything with them with normal arithmetics rules.
Even infinity itsel ...[text shortened]... to never treat infinity, almost infinity, or infinitly near zero as a number, just don't.
Originally posted by ChessJesteryou may not add or subtract Infinity (note of the director)
(inf+1)-(inf) = 1 ??
So what we really are talking about is the problem of systems.
1/3 is a number which cannot beexpressed in the decimal system without tricks.
since 1 divided by 3 is =.3 rest 0.1 we can perform the operation until the hell freezes over and still ahve the same (principal problem). so we invent a trick and state 1/3 = 0.(period 3).
This is a math trick and due to the nature of the trick 0.(period 9) is exactly equal to 1.
Then we have a completely different problem: wthis is is that every propertiy is not infinitly dividible. This is known since the very Ancient philosophers days. There is something which is not dividible the ATOM (greek for not dividable. Thus if we try to divide 1 frog under three persons either one gets the whole (living) frog or nobody gets any (just a heap of material which was a frog before dividing).
Or: There is absolutely no way to draw a perfect circle. If we look to the level of atoms (which we can treat as undividible when talking about drawing regardless of the fact that we can indeed smash them) the irregularities show up.
Then another system: pocket calculator. This machine is of course designed to calculate accroding to the rules and comes (if you read the manual) with alimit of accuracy. Thus you can easily get false results from it, just step beyond the boundaries...
Originally posted by ChessJesterNo. That's not true. inf is not a number.
(inf+1)-(inf) = 1 ??
Or is it? What is your proposal of that number, x1? Then I have another that is 10 times higher, x2=10*x. Therefore your expression is (x1+1)-(x2)=1 and we can all agree that this is wrong.
inf is not a number.