Math: .99repeating = 1?

Factual Questions
41

The Ryan posted 04-14-2000 08:04PM the stuff between the quotes; my responses are in between the quotations.

Nobody can actually see an infinite number of things, nor can anyone see the square root of a negative number, let alone see a negative number; however, in mathematics there are the concepts of infinites (transfinites), imaginary numbers (square roots of negative numbers), and negative numbers (numbers less than zero). Since we’re discussing a purely mathematical issue and not a physical issue, your snide comment is irrelevant, not to mention that it shows you’ve no idea of what you speak. But, to answer your questions correctly: No (but I and others have postulated them, that’s part of mathematics), Yes (they exist in mathematics, the topic under consideration in this thread), and Yes (they have meaning in the realm of mathematics, the topic under consideration in this thread).

Well, apparently according to you, there’s no such thing with wich to be careful about throwing around. BTW, do you always contradict yourself in the very same posting? Anyway, discuss why there’s a whole notational system for infinities when “they don’t just go throwing them around,” please. I missed reading the funnies this morning and need a dose of humour which I’m sure your description of Aleph, Aleph-null, etc., will bring to the discussion. [My snide comment here, feel free to point it out.]

Nope, see ** below for an authorative answer on this.

**From Merriam-Webster’s online dictionary:

&

[quote]
terminating decimal (noun)

First appeared circa 1909

: a decimal which can be expressed in a finite number of figures or for which all figures to the right of some place are zero – compare REPEATING DECIMAL

From “Number,” Microsoft ® Encarta. {bolding and description of pictures in the article brought to you by Monty}

That rigorous enough for you?

Here’s a hint or two for you:

  1. The professional mathmetician (CKDext) has already shown that your take on this issue is incorrect
  2. If you really don’t know what you’re talking about, keep quiet and try to learn.
42

Actually, I don’t see much wrong with anything that The Ryan, Monty, or Dex said. Except when you say that the other is wrong.

I thought you guys were on the same side.

rocks

43

What kind of reply is this? You’re just reiterating my point! Do you understand the concept of a fraction?

Of course 1/4 = 0.25 why should it not?
And yes, 1/3 = 0.3333 repeating - get it?
As for 1 = 0.999 repeating…

Please go back to basic math class if:
a) You cannot comprehend the number 1
b) You fail to understand the meaning of the fraction 1/1 and how to set that up as a long division.

I’d really like to see a long division where 1 divided into itself can yield 0.9999…

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

44

I can see why Cecil feels like there is no hope for the teeming millions!

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

45

Oh, this is relevant. :rolleyes:

It is too clear, and so it is hard to see.

46

Proper treatment for trolls, IMHO: answer ONCE, to be sure that you are not dealing with honest ignorance, then drop it.

One more time, and then I’m done with this.

Several people have offered proof that .999…(repeating infinitely) = 1. The simple multiplication proof is usually sufficient:

(1) x = .99999…
(2) 10x = 9.999999…
Subtracting (2)-(1):
(3) 9x = 9, hence x = 1.

OK, that’s a proof. The testimony of every professor of mathematics in the country would be another sort of proof, I suspect.

You don’t buy those? OK, then, if you thinkf .999… is different from 1, then you should be able to pick a number BETWEEN them. Go ahead, tell us one.

For any two distinct real numbers, there are infinitely many real numbers between them. For instance, the midpoint. Go ahead, name one. We’re not talking about theoretical numbers, now, like “infinitesimals” or whatever, we’re talking real, honest-to-god, write-downable numbers. Name one, between .999… and 1.

If you can’t, then you have to concede that .99999… is just another way of writing 1, in the same way that .333333… is just another way of writing 1/3 or that .11111… is another way of writing 1/9.

Oh, and BTW, Monty, I’m no longer a professional mathematician. I did that back in the 70s, but then I became an actuary (pay was better) and now I’m a more generalized consultant. Contrary to the comment made in another thread, I did not predate Riemann, although I did have a class under Zygmund.

47

Your sarcasm shows you’ve missed the point.
Excuse me, but it is relevant! I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)?

Technically, these are two distinct points on the numberline. If you receive a paycheck, would you accept anything less than 1.0 times the amount you are owed?

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

48

Faulty logic, prof! Excuse me, but if you’re so sharp on mathematics, you should recognize the error in your logic! Here, you are attemtping to employ methods of solving simultaneous equations! The ground rule for aplying this method is that the two equations MUST BE independent! Since formula (b) is 10 times forumla (a), these are NOT independent equations! Thus, you have proved nothing!

Thus, it is improper to subtract (a) from (b)! Algebra cannot be applied at whim! :wink:

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

49

<< The ground rule for aplying this method is that the two equations MUST BE independent! >>

Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other.

A long division? Sure. (.99999…)/1 = .99999…

Happy?

The way that you toss out incorrect statements and ignore the comments of any others, leads me to suggest that this conversation is finished. Fighting ignorance is one thing; fighting stubborness is something else.

50

What’s interesting about this thread to me, is the logic. Many have issued mathmatical proofs that .999… = 1.

The detractors, on the other hand, have offered no mathmatical proof whatsoever. In fact, it would seem that the mathmaticians have won the point.

The only avenue left to those who do not beleive, is using “logic” to state why the proofs are wrong, not showing any mathmatical flaws in the proofs themselves.

Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999…?

51

CKDextHavn demonstrated 0.999 repeating = 1, so I won’t bother repeating him. I’d like to see you demonstrate that they are unequal.

Kbutcher writes

Nope. So far, all we’ve seen is handwaving.

It is too clear, and so it is hard to see.

52

Well, I tried this prove on the other thread on this subject, but I’ll try it again. I don’t know if it’s mathematically valid or not.

1 - 0.999… = D
D x 10 trillion = D
Therefore, D = 0

The idea is, multiply the difference between one and point nine repeating by the highest finite number you can think of, and the product will be no different. When multiplying yields no change, one of your factors is zero. QED

53

Jinx -

If 1/3= .3repeating (as you agree)

Is not .3repeating + .3 repeating + .3 repeating = .9 repeating true?

If so by substitution aren’t 1/3+1/3+1/3 = .9 repeating and since 1/3 * 3 = 1, 1=.9 repeating?
An

54

I know this is beating a dead horse, but like cold fusion fiasco, this is math chicanery!

a1) So, like what if I used 8x instead of 10x? I could fudge “x” to equal whatever I wanted! Ever studied matrices and determinants? What’s the value of a determinant when the equations are not unique?

a2) If your initial premise is let x=0.999…
then how can x <> x?

b) Your “long division” fails to prove that 1/1 = 0.999…

c) Equations must be consistent? No, the term is independent…otherwise, “x” can be shown to equal whatever value you wish by yielding x<>x.

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

55

Sorry if this might double-post because my first posting didn’t show for some reason. Anyway, I know this is beating a dead horse, but what we have here is math chicanery!

a1) The equations must be independent! What if a selected 8x instead of 10x? What would the solution be, then? I can “fudge” x to equal whatever I wish! You have simply proven x<>x…an old math mindbender.

a2) Even used matrices? What is the value (solution) would the determinant of your pair of equations have? And, then ask yourself, what is the significance of this solution? What is trying to tell me? It’s trying to tell you that the second equation is just a scalar multiple of the first (i.e.: not independent).

b) As for your example of long division, you failed to show that 1/1 = 0.999… All you have shown is that any number divided by unity is equal to itself! Hmmf!

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

56

It all depends on who taught you math. If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other.

If you understand the concept of why two equations are not unique if one is a mulitple of the other, then you would agree that x<>x is no proof.

Again, I’ll ask you, what if I used 8x instead of 10x in the example “proof” given by the Administrator on page 1 of this thread?

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

57

Sloth, the limit of 0.9999… is 1.0; I’d agree to that. In short, it will continually approach, but never reach, the value of 1.0

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

58

Untrue; I’m afraid you’re extrapolating the requirement that matrices be non-singular in order to solve the system of equations beyond its applicable range. This requirement doesn’t mean you can’t subtract two dependent equations, it just means you can’t solve a system of n equations in n variables if any of the equations are dependant. Not applicable to the current problem.

OK, easy enough,
(1) x = .99999…
(2) 8x = 7.999999…
Subtracting (2)-(1):
(3) 7x = 7, hence x = 1.

59

No, N linear equations in N unknowns must all be independent for a unique solution to exist. They need not be independent in order to validly combine the equations arithmetically.

If you have N independent linear equations in N unknowns then you may multiply any equation by a constant and/or add any equation to itself or any other equation without changing the solution (except you can’t multiply an equation by -1 and then add it to itself). This applies equally to the case N = 1, which is the case in the previous messages.

Demonstration for N = 1:

Any linear equation in one unknown “X” may be written as:

 aX = b

The matrix representation of this system is:

 {a}{x} = {b}

The determinant of the coefficient matrix is:

 a

and is non-zero if “a” is non-zero.

The determinant of the coefficient matrix with the coefficients of X replaced by the constant vector from the right side of the equation is:

 b

Therefore, by Cramer’s rule, if a <> 0 then:

 X = b/a

(gee, we really needed matrix algebra for that one, didn’t we?) {grin}

Now multiply the original equation by a constant “C” and add the result to the original equation:

    aX  =  b
 +  CaX = Cb
---------------
 aX + CaX = b + Cb

Simplifying:

(1+C)aX = (1+C)b

(can you see where this is going?)

By similar reasoning, the determinant of the coefficients is (1+C)a and the determinant with the coefficients of X replaced by the constant vector from the right side of the equals sign is (1+C)b, so the solution to the “system” of “equations” is, by Cramer’s rule:

 (1+C)b
 --------
 (1+C)a

or, provided C <> -1,

  b/a

which is the same as the solution of the original problem.

Therefore, a linear equation in one unknown may be multiplied by any constant (except -1) and added to itself without changing the solution.

The proof for larger values of N is left as an excercise for the student {grin}.

jrf

60

If A = B and C = D, then A-C = B-D. Always.

I’m still waiting for your proof that 1 and 0.999repeating are not equal.

It is too clear, and so it is hard to see.