Poll: Does 0.99 Recurring = 1
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Soldato
Its defined as such. Can't argue with that can you?
This is the third "real" page in my n00by-degree calculus notes. I think it's designed so anyone can follow it.
So take f(x) = 1/x.
We start with something simple. Look at x=1. 1/1 is 1. Then look at x=2. 1/2 is 0.5. Continue. You'll never be able to get to "infinity" like this, because you will die, amongst other issues. So look at what is happening.
Lim n->inf [1/n] = 0
So take f(x) = 1/x.
We start with something simple. Look at x=1. 1/1 is 1. Then look at x=2. 1/2 is 0.5. Continue. You'll never be able to get to "infinity" like this, because you will die, amongst other issues. So look at what is happening.
Lim n->inf [1/n] = 0
Soldato
Terribly sorry we've offended you with our talk of numbers. Please except my most humble apologies.Originally posted by 50/50
Get over it you geeks.
Or rather, just ignore this thread?
Tell me your value for infinity in working that one out. Whatever number you give me, I can think of a bigger one.
Hence you can't "put a value" on infinity, only discuss its properties. One of its properties is that
Lim(1/n) as n->infinity = 0.
Soldato
BTW, you do realise = means "Defined to be", so you've just said "I define 0.9r = 1" therefore 0.9r = 1 axiomatically in your view.
Another vote for no, but then I haven't seen or understood the proof.
Is 0.0r1 0?
Is 1.0r1 1?
Thats the argument put forward. As long as there is a 1 at the end, 0.0r1 != 0
If 0.99r = 1.0, then 1.0 - 0.99r = 0.0r1 = 0
And it is my firm belief that 0.0r1 is NOT equal to 0. Only 0 is equal to zero, anything else tends to zero, but does not equal it. 1 / infinity != zero even though it should be, it will always come out slightly larger.
Is 0.0r1 0?
Is 1.0r1 1?
Thats the argument put forward. As long as there is a 1 at the end, 0.0r1 != 0
If 0.99r = 1.0, then 1.0 - 0.99r = 0.0r1 = 0
And it is my firm belief that 0.0r1 is NOT equal to 0. Only 0 is equal to zero, anything else tends to zero, but does not equal it. 1 / infinity != zero even though it should be, it will always come out slightly larger.
Are there three or four polls, one for each time I decided I have come to a different conclusion? Polls are useful for opinion only, and the first thing I do when I enter a poll is vote. Not get swayed by a hundred different conflicting "proofs" from various people of academia, of varying levels of technical competence. If you want to find out my opinion AFTER reading the thread, why not try getting everyone to post all of the comments etc, and once everyone has finished, THEN post a poll.
Thats why people are given the opportunity to vote after debates, not before them.
Intresting, yes i can grasp what bodak has said and that is a much better argument than yours.
So the bassis of your argument against my theory is bassed on the fact that as n aproches infinity the result of 1/n is defined/considered to be 0.
I will accept this, it was purly a theory and im happy to be wrong. In fact its nice for to be proved wrong instead of just ignoring me.
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alpha so your saying this site is wrong?
http://www.staff.vu.edu.au/mcaonline/tool/symbols/symbol.html
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This thread has great comedy value!
You kinda think some are finally getting it but then they just say something silly like "oh but the difference is so small you round it to one"!!!
I believe the most simple to understand proof is the following, its been stated before several times but i'll put it again, but even more simply:-
1/3 = 0.3r exactly i.e. 1/3 = 0.3333333333333...(insert infinite 3's here)
And you must know that 3 x 1/3 = 1. as in three times a third is exactly one, not a little bit less then one or approximatly one or roughly one, one.
Now seeing as 1/3 = 0.3r, and 3 x 1/3 = 1, you can substiture 0.3r for 1/3 in the equation, so 3 x 0.3r = 1. so if 3 x 1/3 is exactly 1, and 1/3 is exactly 0.3r, 3 times 0.3r must also be exactly one.
Another way of multiplying by three is to add the number to itself 3 times, i.e. 3x0.3r = 0.3r + 0.3r + 0.3r = 0.9r, this should be easy to see and obvious to all. So since 3x0.3r=0.9r, and 3x0.3r=3x1/3, and 3x1/3=1, 0.9r=1!!!!!
Now hopefully everyone should be able to follow this one easy enough and see that 0.9r is exactly equal to 1!
You kinda think some are finally getting it but then they just say something silly like "oh but the difference is so small you round it to one"!!!
I believe the most simple to understand proof is the following, its been stated before several times but i'll put it again, but even more simply:-
1/3 = 0.3r exactly i.e. 1/3 = 0.3333333333333...(insert infinite 3's here)
And you must know that 3 x 1/3 = 1. as in three times a third is exactly one, not a little bit less then one or approximatly one or roughly one, one.
Now seeing as 1/3 = 0.3r, and 3 x 1/3 = 1, you can substiture 0.3r for 1/3 in the equation, so 3 x 0.3r = 1. so if 3 x 1/3 is exactly 1, and 1/3 is exactly 0.3r, 3 times 0.3r must also be exactly one.
Another way of multiplying by three is to add the number to itself 3 times, i.e. 3x0.3r = 0.3r + 0.3r + 0.3r = 0.9r, this should be easy to see and obvious to all. So since 3x0.3r=0.9r, and 3x0.3r=3x1/3, and 3x1/3=1, 0.9r=1!!!!!
Now hopefully everyone should be able to follow this one easy enough and see that 0.9r is exactly equal to 1!
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Soldato
Belief and opinion don't come into "logic" though. Can you prove what you say?
I've proved 0.0r1 = 0, can see you see a flaw in it which you can counter in a similar manner (not just "IMO")?
Well, I kinda figured 0.0r0 = 0, 0.0rX would be greater than zero... but according to your arguments, 0.0rX is zero by definition, which actually makes sense to me now.
Which means that in turn 0.99r is 1.0, in the same way that 0.99rX is 1.0
Even if the X were 0, 0.99r0 would still equal 1.0 . Weird, but actually makes sense.
You can't have 0.0r1, 0.99r1, it just doesnt exist! If you were to have a load of 9's then stuck a 0 on the end it would not be recurring, there would have to be a finite number of 9's if you were to put another number on the end.
But you don't make sense, which makes me think the whole "infinite" thing isn't being stressed. The only thing other than 0.9r that I'd be "happy" to write (even, then at a push) is 0.9r9. Anything after the last 9 before the r, is what you will see until you're finished writing the number - which will never happen.
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