A question of numbers (one for the maths fans)

Mod should not have mentioned a zero way back there. You are correct when you say we never put it there because there is no end. There is also no 8 on the end because there isn't an end there.You might say that 2* 0.9 is 1.8 because 0.9 is 0.1 less than one and you end up doubling the missing bit less than one. 2 * .99 is 1.98 when you double the missing bit (now we re .02 less than 2).2 * 0.9... (endless 9's) is equal to 2 because there isn't any "missing bit" to double.

Message 161 Doesn't work according to what your saying.Don't we have to prove it, or is my math wrong?

From a quick glance, those equations appear to be slightly wrong. you divide the 1.5 by .5, not multiply by -1, to find out what x is.
x = 3, in that equation at the end. Here again, you do not multiply by -1. You would divide .499 . . . by .5 to get 1. And you should use .5, instead of .499 . . ..reason being, is that .499 . . . is not half of .999 . . ..
(number in line above have recurring decimals)
.499 is not even half of .999. .4995 is half of .999
(numbers in line above are finite) okay, this equation needs to be fixed.x = .333 . . .
.5x = .166 . . .
x = .333 . . .You have to divide the .166 . . . by .5 in order to find out what x is. This is going to sound wierd.
No. first, you would have to subtract by 1, but you still wouldn't get 8.999 . . ..That's because .999 . . . is 1. So you are in effect subtracting 1 from 10, without rounding the numbers.Now then, if the number was 9.999 and you subtract .999 you will get 9. If you subtract 1 from 9.999 you will get 8.999. That's because all the numbers in this segment have decimals that don't repeat.

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All a man's knowledge comes from his experiences

.999...=1 is not wrong. .999 does not equal 1, nor does .99999, but
.999... is another way to represent 1. So is 1/1, or 99999/99999 they are all different representations for ONE.Are you claiming to be a better mathematician than Newton, Leibniz, Cauchy, Cantor, Godel, Euclid, on and on and on? This is not Bible interpretation, it's mathematics. You can have you own private math if you want but who will use it? Well, it's a different way of representing things. Do you have a problem with Zeno's paradoxs or not?lfen

Why did you skip over Message 161 ?

They were just men like me, maybe with less knowledge of the world, I might add.However, I am not claiming I am correct. Don't know enough about it yet, I just glanced at it. I mentioned, that is how I always pictured .999... in my head.

quote:

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.9 is not equal to 1.
.99 is not equal to 1 but it is closer.
.999 is not equal to 1 but it is closer yet.
.9999 is not equal to 1 but it is still closer.
.999999999999999999 is not equal to 1 but it is very close indeed.

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.9... (the ... means an endless row of 9's. We need to be VERY careful with our notation) IS equal to 1 the difference is now zero.

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Heh. This is actually the correct way to prove this in mathematics, just without the technical language.In technical language, let a_n = 9*(1/10)^n. Then the correct way to write 0.99999... is sum_{i=0}^\infty a_nThe claim is that sum_{i=0}^\infty a_i = 1The proof is that sum_{i=0}^\infty a_n = \lim_{N\to \infty} \sum_{i=0}^N a_i (by definition of the infinite series)Now 1 - \lim_{N\to \infty} \sum_{i=0}^N a_i = \lim_{N\to \infty}( 1 - \sum_{i=0}^N a_i)= \lim{N\to \infty} (0.1)^N= 0So 1 - 0.99999... = 0 which means 1 = 0.9999....(Sorry for the LaTeX; I don't know any other way to write math symbols.)Of course, one should prove first that 0.99999... actually converges to a number, but this is obvious since the partial sums of the series are a Cauchy sequence.

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"These monkeys are at once the ugliest and the most beautiful creatures on the planet./ And the monkeys don't want to be monkeys; they want to be something else./ But they're not."
-- Ernie Cline

Yes, I know, I mentioned that it works. I had changed after you replied, go back and look at it.Question, if .4999... can be .5, why can't .333... be .4 ? Your missing the whole point then. The first line is subtracted from the second line. You wind up with mistakes, because you never attach a last number.

quote:

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x = 0.333333...
10x = 3.33333... (multiplying each side of the above line by 10)
9x = 3 (subtracting the 1st line from the 2nd)
x = 3/9 = 1/3 (simplifying)

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Repeating decimal - Wikipedia You are assuming it is 1 before you prove it. Isn't that wrong?
If the equation, and the idea of never adding the last number to an infinite set is correct, then it doesn't matter what number we put in front of x in the equation, it should always work, and it doesn't.
x should always = 1 if the formula is a valid way of proving .999... = 1

No it's not proof of anything. I don't know how you got
-0.5x=.166multiplying by -1 would give you -.5x=-.166...x= .166.../.5 =
1.666.../5 =
.333... = 1/3lfen

Using your logic, we could at any point in the equation replace .999... with 1, correct?x = 0.999...
10x = 9.999... (multiplying each side of the above line by 10)
9x = 8.999... (subtracting the 1st line from the 2nd) (using 1)I know what you'll say, and that is 8.999.. = 9

I wasn't saying that yuor answer's were wrong--well, not yet anyways. It was your setup that was wrong.There's no reason to multiply both sides of the equation by -1, and that's what you're doing. Also, without that last step of isolating the x by division, we've no clue what x is (unless we do it on our won, like I did). In order for .333 . . . to be .4, it would have to .3999 . . ., but .3999 . . . is not .333 . . .. and it is, if the equation you are referring to is the one we've been using the entire time. It doesn't matter what you multiply x by, you will always get 1 = .999 . . ..You say it doesn't--come up with a problem where 1 does not equal .999 . . . using that equation.

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All a man's knowledge comes from his experiences

It's not acceptable for a rigorous claim for the calculus and there are proofs. Engineers used to get it the theory wrong but the application right by saying "it goes to the limit". It's just that the proofs take time. You have to define what a limit is and a bunch of things. That is why I've tried to use Zeno's paradoxs to give you a more easily visualized solution to the problem. You keep confusing notation with concept.In the number 123, 1 is digit, 2 is a digit, 3 is a digit.
They are not the numbers 1, 2 , or 3 they are the digits of the number 123. How can you add "the last number to an infinite set"? if it is infinite then by definition there is no last number! Oh this is not about rounding decimals up! no, no.
.4999 rounded to the nearest tenth is approximately .5 but that is not at all what is happening here. No rounding is going on. The sum of an infinite series is not arrived at by rounding!!!! It may look that way but that is the diffence is as I stated .499 rounded up is APPROXIMATE but .4999... is an EXACT representation, that is to say is EQUAL to .5. Big difference.

That is correct. Just keep in mind, when replacing .999 . . . with 1, you have to add that 1 to the number that originally had .999 . . ..Otherwise, you would have:
8.999 . . . (replace .999 . . . with 1)
8.1,
which doesn't work out. SO just add the 1 to 8, and you'll get 9.

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All a man's knowledge comes from his experiences

You said you changed it, so I took another look--you now have the wrong value for x. Your equation is right up until you've added the negative (-) signs to both sides.First off, the (-) isn't necessary. Second, why is -.499 not -.4999 . . .? There's a big difference. Third, you divide -.499 by -.5, which will give you .998 (refer to second criticism).What you did was to divide -.5 by -.499, which would make your equation look like this:-.5x/-.499 = 0.It should look like this:
x = -.499/-.5

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All a man's knowledge comes from his experiences

"is not" is a bit unclear.0.49... = 0.9... / 2 does it not?