- #141
Tom McCurdy
- 1,020
- 1
It seems like the same principles used here are the ones used to solve zeno's paradox... I posted some of zeno's paradoxes in another part in the math forum... correct me if I am wrong.
However, if 0.9 repeating = 1 then they wouldn't have named them differently in the first place.
They are 2 numbers that could be represented as 2 lines, one line being just a 'point' longer than the other.
Integral said:Consider this inequality.
[tex] 1 - .1^n < x < 1+ .1^n[/tex]
It seems clear that there is only a single number for which this is true for ALL values of n>0, x= 1 . This is a simple statement that any number added to x results in something greater then 1, or any number subtracted from x results in something less then one. I doubt that you will find many people who will argue with the truth of the statement, x=1.
Now, in the link I posted above, I show that using simple arithmetic, involving only valid rational numbers, one can construct this inequality.
[tex] 1- .1^n < .999... < 1+.1^n [/tex]
notice that is the exact inequality as above, thus we have x=.999... If the original statement is correct that the inequality can only be satisfied by 1 you must be lead to the conclusion that
1=.999...
Hurkyl said:Ah, I see. So [itex]\frac{6}{4}[/itex], [itex]\frac{3}{2}[/itex], [itex]1\frac{1}{2}[/itex], and 1.5 are all different numbers! I can't believe I thought they were all the same.
How can a number be represented as a line? Oh, I see, you meant line segment. Well, as I'm sure you know, between any two distinct points there is another point. If both have their left endpoint at the origin, what lies between the right endpoint of the line segment representing 0.999... and the line segment representing 1?
Simply pick an n, any n in the integers and do the arithmetic. The whole point is that for ANY n you choose (you must pick an n) the relationship holds.musky_ox said:I don't get what you are saying here. It seems to me that while 1- .1^n = .999... maybe be true, this doesn't make sense... 1- .1^n < .999... < 1+.1^n. You are saying that 1cm - (and infinitely small amount of space aka a point) is greater than 0.999...cm, which is the same thing as far as i can tell.
Integral said:Simply pick an n, any n in the integers and do the arithmetic. The whole point is that for ANY n you choose (you must pick an n) the relationship holds.
Please show me an integer for which the bolded statement holds? Remember that infinity is not a valid integer, or is it a valid real number. I am not talking about anything physical, we are discussing math and not physics here.
Anyways, since mathematics is abstract, i don't see why there can't be an infinitely small amount.
So basically you are saying that mathematics (or whatever system of it we are talking about) is incapable of handling infinities
Well i can see that 0.99... is infinitely close to 1.
Its just a flaw that exists in any base system, because you can never acount perfectly for all the fractions.
We know that infinity has a place in abstract ideas and in the universe.
musky_ox said:It is just the way that our base system rounds up 0.33... to be 1/3. Obviously whatever mathematics we are talking about cannot handle infinity if it has to precisely say that infinity doesn't have a place in it.
musky_ox said:So basically you are saying that mathematics (or whatever system of it we are talking about) is incapable of handling infinities, and is thus not capable of representing our universe? Then what is the use of everyone studying it, if there are other systems that can better account for everything in life? In higher level mathematics/physics do they switch to different systems so we can handle the infinite?
you can never acount perfectly for all the fractions.
Hurkyl said:I challenge this. Would you care to demonstrate an actual error that arises from using decimals instead of fractions, even if you accept 0.999... = 1?
I can think of a number between them...
0.99...9 + 0.00...1 = 1
0.99...9 + nothing = 0.99...9
In base 12, 1/3 is a terminating number... so you don't get the small rounding error when you use it for calculations.
Hurkyl said:Warren didn't ask for a number between 0.{terminating sequence of 9's} and 1. He asked for a number between 0.{9 in every allowable position} and 1.
(And, incidentally, you didn't produce a number between 0.99...9 and 1)
In base 12, 1/3 is a terminating number... so you don't get the small rounding error when you use it for calculations.
This doesn't address my challenge.
But these numbers do not (and cannot) exist. There is no such thing as a number with an infinite number of nines, followed by an eight. There is no position to put an eight in "after" an infinite number of nines, because an infinite number of nines never ends.musky_ox said:Okay try this then. Using your logic:
0.99...9 = 0.99...8
0.99...8 = 0.99...7
Let me guess...somewhere, wayyyyy out there at the end of infinity there's room for just one more digit ?musky_ox said:I can think of a number between them...
0.99...9 + 0.00...1 = 1
0.99...9 + nothing = 0.99...9
Okay. So an infinitly large number exists, but an infinitly small number doesnt?
Hurkyl said:No, no infinitely large (real) number exists, and no infinitely small nonzero (real) number exists.
So how many 9s are after the decimal point in 0.99...?
Why can't we add an 8 after the 9s?