.9 repeating = 1? Is our numerical system flawed?, page 20
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reply posted on 29-10-2008 @ 10:56 AM by Sublime620
reply to post by stander





Math is conceptual. That's right, many of us are past the definitive days of 2 + 3. Without the idea of 1/infinity, we wouldn't have limits. Limits are the foundation behind calculus.

Perhaps you should try taking it, then you may understand the importance of them.

reply to post by Benarius



You might want to take a class on it. Calculus isn't the toughest match class to figure out, but the first few weeks of it require some pretty hardcore conceptual learning.

You start with reviewing limits. Then you have to master derivatives, a pretty large equation involving limits. Once you get that equation down forwards and backwards, the class gets a bit easy and you can start learning reverse derivatives.

[edit on 29-10-2008 by Sublime620]


reply posted on 12-12-2008 @ 02:22 AM by Iggus
reply to post by Johnmike



Just a pedantic point the equation for differentiation is given by lim x->0 [f(x+h)-f(x)]/h, it has the minus sign. It is after all just telling you the slope and is no different really from doing simple trig using the tan function. Here you just take a really small triangle.

As to the 0.99~ thing, this is really just writing the supremum (spelling might be off) of the numbers less than 1. Just think of it as taking the smallest number 'n-word' than 0 away from 1. They are not identical for if they were we would not have a continuous number line, but rather a dashed one with lots of wholes in it. I could simply argue that 0.99~8 is just as close to 0.99~9 as 0.99~ is to 1. For those who really want to understand go and look up supremum numbers and the axioms of the real number line.


reply posted on 13-12-2008 @ 06:40 PM by Johnmike
reply to post by Iggus



Typo, it wouldn't let me edit it either. This is the second time I had to correct it, though you did it for me this time. Last time I said it was h squared before I re-read it and realized.



reply posted on 14-9-2009 @ 01:33 AM by ARandomGuy
... ohhhhhh dear...

Something that needs to be made clear here is that just because something is comprised of infinitely many pieces does NOT mean that it is infinite. I'm calling on Zeno's Arrow Paradox here. Just because a number has the mystical "..." attached to the end of it does not render it unusable in standard mathematics - it simply means that it has an infinite decimal expansion.

Furthermore...

ALL NUMBERS ARE INFINITE DECIMAL EXPANSIONS.
This is true and has been since the day you were born, and most likely well before that. "1" is shortened notation for "1.00000...", 4/5 is shortened notation for "0.80000...", and "pi" is shortened notation for "3.14159...". These are three different numbers answering to three differing number sets - whole, rational, and real, and all THREE of them follow this rule. No matter what set your number belongs to, there are several ways to write it - in convenient shorthand (such as 1, 2/2, .5*2,) or with its infinite decimal notation (good luck writing out something comprised of infinitely many parts, like 1.00000000000... [and by 'write it out, I mean write out an infinite number of zeroes. Good luck.])

Now, to business...
.999... = 1.

It's true, I'm sorry if it disturbs or disappoints any of you. .999... is another way of writing the shorthand form of 1, and there are many mathematical proofs that prove this.

Take the Additive Property, for example. It states that any real number "a" plus the number zero will equal "a." Therefore, [a+0=a]. By rearranging this equation, we get what I like to call the Subtractive property - any number minus itself equals zero [a-a=0]. NOW, remember that "0" is actually the shorthand of 0.000... SO, when we plug the two different forms of writing the number "1" into the equation (1 and .999...), we get this:

1-.999... = 0
0.000... = 0
Therefore .999... = 1.

Imagine that.
But there should be a "...001" at the end of that "0.000..."!
The entire point of the "..." is that the expansion goes on forever, without end. There IS no end to tack the "...001" onto.

The Limit Argument
A sequence can have ONE and ONLY ONE limit.
Let us observe the sequence:
.9
.99
.999
.9999
...

therefore, the sequence gets closer and closer to .999... infinitely close.
It also gets infinitely close to 1. Therefore,
.999... = 1.

I cannot find for the life of me why THIS algebra is being disputed:
let x = .999...

Therefore,

10x = 9.999...
Subtracting x from both sides (legal) gives us

9x = 9.999...-x
Substituting ".999..." for 'x' on the right side (which is completely legal) gives us:

9x = 9.999... - .999...
9x = 9
x=1.
Above is pure, completely correct algebra.
But we already said that x=.999... therefore 0.999... equals one.

FOR THOSE OF YOU THAT DISPUTE THE ALGEBRA:
Assume that .999... is less than or equal to 1.
Let us next assume that .999... does NOT equal 1. (INCORRECT ASSUMPTION)
Therefore, .999... < 1.
Therefore, there should be some positive number N such that
.999... + N = 1.
HOWEVER, for ANY positive number N,
.999... + N > 1.
Therefore, one of our assumptions is wrong. I'll give you a hint: It's the second one.


".999... = 1" is INDISPUTABLE. There are NO PROOFS ANYWHERE THAT OPERATE UNDER STANDARD MATHEMATICS that can make this equation false. I've checked, but if you'd like to yourself, go ahead. Religious reverence of the concept of infinity or plain old disbelief does not make you correct - neither does the fact that 'so and so' agrees with you. Proof is everything in mathematics - and there are no proofs to the contrary. I rest my case.

Sam Hughes has an excellent article on this subject at his website "Things of Interest." [qntm.org]. I advise anyone who disputes this truth to look up the arguments posted there.

[edit on 14-9-2009 by ARandomGuy]

[edit on 14-9-2009 by ARandomGuy]

[edit on 14-9-2009 by ARandomGuy]


reply posted on 14-9-2009 @ 09:55 PM by BriggsBU
Originally posted by American Mad Man
Wow - this actually has my attention now - and I usually HATE math

OK so taking 10x - x = 9x, how did you get to 9x = 9?

As I recall, you can manipulate both sides, so long as you do the same to both (although I do recall there being exceptions to this rule).

If you were to devide by x, that would give you:

(10x-x)/x = 9x/x

this would then become 10x/x - x/x = 9x/x, each "x" cancels the other out (x/x = 1), thus:

10 - 1 = 9

so I don't think that does anything.


Your argument is flawed based on the mathematical Law of Equality.

10x - x = 9x.

This is true in the same way that 3x - x = 2x.

Example: Let x = 2

3(2) = 2 + 2 + 2 = 6
3x - x = 3(2) - 2 = 2 + 2 + 2 - 2 = 4
2x = 2 + 2 = 4

This relationship will be constant for any value of x.

Want another way to think of it? yx - x = (y-1)x.

To prove that, all you have to do is multiply through the right side.

Therefore, 10x - x = 9x.

QED ^.^

Edit to add some more information.

The reason we get 9x = 9 is simple as well. For reference, I'll write the entire proof here and put in some notes where applicable.

PROOF
---------
Let x = .999~

//We are declaring the variable x to be equal to .999~ where the nines repeat to infinity.

10x = 9.999~

//We multiply both sides by 10 which shifts the decimal one place to the right. It is important to note, however, that because the 9s repeat to infinity, they continue to do so after the multiplication. In essence, the 9s following the decimal are equivalent in both .999~ and 9.999~ due to the properties of infinitely repeating sets.

10x - x = 9x
//Now, as I described above, yx - x = (y-1)x through the Law of Equality.

9x = 9
//Now, we've shown that 10x = 9.999~ and x = .999~. If you subtract .999~ from 9.999~, you get 9. IE: 9.999~ - .999~ = 9

9x/9 = 9/9
//Divide both sides by 9.

x = 1
//Result shows that x = 1.

.999~ = 1
//Since we started out by setting x equal to .999~ repeating to infinity and have shown that x is equal to 1, by the Transitive Property, we have proven that .999~ is equal to 1.

QED ^.^

[edit on 14-9-2009 by BriggsBU]
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