The 1 = 0.999... question

[Just me]

Not wanting to get the war going again since many have proven the case, but just wondering about some of the logic.

#1. If 0.9999... is less than 1, find a number between 1 and 0.999.....
Since no number can be found, then 0.9999 must equal 1.
Question: What, then, is the number immediately preceeding 1?

#2. Primary solution is: x = 0.999999...
10x = 9.99999.....
10x - x = 9
9x = 9
x = 1
My question is with the math of 10x = 9.9999....
If my number 0.99999999 actually had a fixed number of digits, lets just say 10 for discussion sake. Then multiplying it by 10 would shift all digits over to the left by a decimal place, and leave a 0 in the tenth position. But since we're dealing with an infinite number of 9's, it almost seems that the logic cheats since the "infinity-eth" digit does not get dropped. Meaning that if we multiply a periodical by 10, we are magically adding a digit to the left of the number, without affecting the farthest digit to the right.


Just curious!?

[borschevsky]

[Tahnan]

1. There isn't one. Why should there be? The real number line is continuous, without breaks. If .999.... were the number immediately preceding 1--what would immediately precede .999...?

2. Multiplying by 10 has the effect of shifting the decimal point one place to the right, regardless of what follows the point. Consider, for instance multiplying .333... by 10. .333... is 1/3 (I hope that's uncontroversial); 1/3 * 10 = 10/3 = 3 1/3 = 3.333.... Nothing is "dropped," no zero is "inserted." (If you like, think of .99 as .99000000...; multiplying by 10 gives 9.9000000..., with nothing having been "added" after the .99)

I can't believe I'm having this discussion.

[Just me]

1. so argument that if 0.9999... is less than 1, then find a number between them is not a valid argument right?

2. I'm still confused...
If I shift the decimal one position to the right, then my last digit on the right becomes a 0
ie. if x = 0.99999999999999999999
0.99999999999999999999 * 10 = 9.99999999999999999990
then 10x - x = 8.99999999999999999991
so if instead x = 0.99999( insert infinite number of 9's here)99999
then 10x -x would be 8.99999( insert infinite number of 9's here)99991

====================

I'm obviously confused about the meaning of the "infinity" thing since...

1 > 0.9
1 = 0.9 + 0.1 and 0.1 > 0.09 therefore 1 > 0.99
1 = 0.99 + 0.01 and 0.01 > 0.009 therefore 1 > 0.999

The question is, "how long can I keep doing this until I get an equal sign?"
If I answer to myself, I'd think I could keep up this logic forever. Or infinitely. But it seems that actually I can't, because once I actually make it to infinity, then 1 = 0.99999...

Which is why I'm so confused!!???

Sorry, I'm just trying to understand so I can explain to some friends.

[Tahnan]

1. Quite the opposite. The number halfway between them, which by borchevsky's method necessarily exists, is 1.999.../2. What is that? Well, 2 into 1.9 is .9, carry the one, that's 19, and 2 into that is 9, so we have .99 so far, carry the one, that's 19, and, hey, wait. (Do this on paper if that wasn't clear.) (1 + .999...)/2 = .999...! But the difference between .999... and itself is 0. So since it's halfway between .999... and 1, the difference between .999... and 1 must also be 0. What does this tell you?

2. "If I shift the decimal one position to the right, then my last digit on the right becomes a 0." No, not exactly. You might think of it that way for convenience, but the last digit doesn't really "become" anything.

And even if it did, there is no "last digit" of .999.... By its nature, there is no place "after it" to put a 0.

[borschevsky]

I'm pretty sure this was all discussed in the old .99 threads, so you should check those out thoroughly before posting the same questions in this thread, or else people will start smackin' you around in here .

But quickly:

[ZutAlors!]

1. The argument is this (and it IS valid):
a. it has been established that the number line is dense; i.e., between any two distinct real numbers there are an infinite number of other numbers.
b. therefore, IF 0.999repeating does NOT equal one, then you should be able to find at least one number between them.
c. If you CAN'T find a number between 0.999repeating and one, then those two numbers are the SAME.

For the same reason, there are no numbers that are "next to" each other: you can't specify a number that is "just preceding" one. So the non-existance of a number "just preceding" one does not invalidate the above argument, rather, it stems from the above argument.

2. Yes, you are confused about the meaning of infinity. It's a mistake to try to treat "infinity" as "a really big number". If we were talking about a really big number of nines, your intuitive argument would make sense; however, this leads you astray here because the behavior of 0.reallybignumberofnines != the behavior of 0.infinitenumberofnines.

[Da5id]

Here's something that might help you wrap your brain around it. It helped me, anyway.....

I never had much of a problem with
S (1/2)^n where n 1®¥ = 1 (½ + ¼ + 1/8....)

Similarly

.999999..... is a representation of
S 9*10^-n where n 1®¥ = 1 (.9 + .09 + .009....)

Hope that helps.

[extropalopakettle]

quote:

---

I'm obviously confused about the meaning of the "infinity" thing since...

1 > 0.9
1 = 0.9 + 0.1 and 0.1 > 0.09 therefore 1 > 0.99
1 = 0.99 + 0.01 and 0.01 > 0.009 therefore 1 > 0.999

The question is, "how long can I keep doing this until I get an equal sign?"

---

Infinitely many times. How long will that take? Depends on how fast you do it.

[cha]

...although technically, it doesn't matter "how fast you do it", since there is no point on the number line that is identified as "infinity". No matter how fast or how long you go, you will never even be any closer to infinity, because there will always be an infinite distance ahead of you...

[dethwing]

how did you get the sigma marks da5id?

[cha]

Another question -
0.99 squared = .9801
0.999 squared = 0.998001
0.9999 squared = 0.99980001
0.99999 squared = 0.9999800001
0.9999999999 squared = 0.99999999980000000001
It would seem that the answer to 0.999... squared would consist of an infinite number of 9's followed by an 8, then an infinite number of 0's with a 1 at the end.
So, the first "half" of this number sequence (the 9's) would be infinitely long, and yet have more numbers after it, and the second "half" (the 0's) would be infinitely long and yet have a clearly identifiable "final number". Is this number also equal to one? Is this number greater, less than, or equal to 0.999...?


What is madness but nobiity of soul at odds with circumstance?

[Ghost Post]

without a final 9 in 0.999... to multiply there can be no final 1 or infinite number of 0s.

[borschevsky]

cha: ( .999999.... ) ^ 2 = .999999....

Proof:

( .999999.... ) ^ 2
= 1^2
= 1
= .999999....

[extropalopakettle]

[cha]

borschevsky - I used a bad example, but part of the question I have is can another infinitely repeating number like 0.222... be squared? How about pi? My thoughts on the first are that 0.222... = 2/9, which squared = 4/49, which can be written as a decimal (though not sure if it repeats). This still leaves the pi question...

[groza528]

(psst- 4/81, not 4/49)
Any fraction can be written as a repeating or terminating decimal. It will terminate if the ONLY prime factors of the DENOMINATOR are 2 and/or 5. Thus 4/81 will repeat.

[extropalopakettle]

[extropalopakettle]

An example of a defineable but non-computable real number.

[Da5id]