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>Of course then most students will insist that 0.999⋯≠1 which shows how ineffective this method is to understanding what the real numbers are. Okay, you got me

by mdm_ 13y ago

>Of course then most students will insist that 0.999⋯≠1 which shows how ineffective this method is to understanding what the real numbers are. Okay, you got me. Why is 0.999... != 1 incorrect?

lmartel 13y ago
Given that 1/9 = 0.111... You can see that 1 = 9/9 = 9 * 1/9 = 9 * 0.111... = 0.999...
justinpombrio 13y ago
If you want to be formal, 0.999... is by definition the limit as n goes to infinity of the sum from x=1..n of (9/10^x). This limit converges to 1. More intuitively, what is the difference (as in subtraction) between 0.999... and 1? I think you'll find that it doesn't make sense for it to be anything other than zero. And if the difference between two numbers is zero then surely they must be equal.
yumraj 13y ago
x = 0.9999... 10x = 9.99999.... 10x - x = 9.9999... - 0.99999.... 9x = 9 x = 1 therefore, 0.999... = 1
phdp 13y ago
Here is a demonstration of how they are equal. x = .99999... 10x = 9.99999... 10x - x = 9.999999... - .9999999... 9x = 9 x = 1
derleth 13y ago
1.000... - 0.999... = 0.000... That last value is, in fact, just zero, only written in a less-condensed form than usual. It is always the case in the real numbers that the only time a - b = 0 is when a = b, so that must mean 1.000... = 0.999... in the real numbers. More verbosely: At every point in the decimal expansion of the result of the subtraction, the result must be 0 because both 1.000... and 0.999... go on forever. This means that the result must be smaller than any positive real number, which means, because the real numbers contain no non-zero infinitesimal values, it must be zero, which means 1.000... must equal 0.999... because a - b = 0 if and only if a = b. There are other sets of numbers where this is not the case, such as the p-adic numbers for some p. Wikipedia has a whole article devoted to this: https://en.wikipedia.org/wiki/0.999... https://en.wikipedia.org/wiki/0.999....
cygx 13y ago
The value of a decimal number is by definition the sum of its digits weighted by the appropriate power of 10. If the fractional part is not finite, we need to compute an infinite series n 0.999… = lim Σ 9/10^k = lim ( 1 - 1/10^n ) = 1 - lim 1/10^n = 1 - 0 = 1 ^ n→∞ k=1 ^ n→∞ ^ n→∞ ^ ^ | ^ | ^ | | | definition | induction | algebraic basic now we're | | limit theorem proof getting silly 0.9 = 1 - 0.1 0.99 = 1 - 0.01 … = … which is the case of a geometric series ∞ Σ a⋅r^k = a / (1 - r) k=0 with a = 9/10, r = 1/10. Another way to look at this is via the constructive definition of the reals as equivalence classes of Cauchy sequences of rational numbers: The sequences (1, 1, 1, …) and (0.9, 0.99, 0.999, …) are just two of an infinite number of Cauchy sequences which make up the real number 1.
ColinWright 13y ago
You've had half a dozen answers already, and they are all correct, but you may find them unenlightening, because you may find they are not addressing what you feel the problem is. To you it may be "obvious" that 0.9999... can't equal 1, and if that's the case, no amount of explanation of why it is will satisfy you. If you want to talk about this further and to understand what's going on, I'm happy to try to work with you. My email is in my profile.
deleted 13y ago
[deleted]
saalweachter 13y ago
Starting at the very beginning: Stuff you write down aren't numbers, they're representations or names of numbers. There is a number named "1", there is a number named "2". "2" is not the number itself. You can't infer the two-ness of "2" by staring at the symbols. It turns out that our way of naming numbers isn't perfect. Each number has multiple names. "2" and "2.0" are both names for the same number, as is "2.00", "2.000", and "2.0000". It turns out that you can name "2" a different way, as "1.9999999999999999999999999999999999999999999999999999...", where there are an infinite number of nines. The names may be different, but the numbers they name are the same.

by mdm_ 13y ago

>Of course then most students will insist that 0.999⋯≠1 which shows how ineffective this method is to understanding what the real numbers are. Okay, you got me. Why is 0.999... != 1 incorrect?

lmartel 13y ago

Given that 1/9 = 0.111... You can see that 1 = 9/9 = 9 * 1/9 = 9 * 0.111... = 0.999...

justinpombrio 13y ago

If you want to be formal, 0.999... is by definition the limit as n goes to infinity of the sum from x=1..n of (9/10^x). This limit converges to 1. More intuitively, what is the difference (as in subtraction) between 0.999... and 1? I think you'll find that it doesn't make sense for it to be anything other than zero. And if the difference between two numbers is zero then surely they must be equal.

yumraj 13y ago

x = 0.9999... 10x = 9.99999.... 10x - x = 9.9999... - 0.99999.... 9x = 9 x = 1 therefore, 0.999... = 1

phdp 13y ago

Here is a demonstration of how they are equal. x = .99999... 10x = 9.99999... 10x - x = 9.999999... - .9999999... 9x = 9 x = 1

derleth 13y ago

1.000... - 0.999... = 0.000... That last value is, in fact, just zero, only written in a less-condensed form than usual. It is always the case in the real numbers that the only time a - b = 0 is when a = b, so that must mean 1.000... = 0.999... in the real numbers. More verbosely: At every point in the decimal expansion of the result of the subtraction, the result must be 0 because both 1.000... and 0.999... go on forever. This means that the result must be smaller than any positive real number, which means, because the real numbers contain no non-zero infinitesimal values, it must be zero, which means 1.000... must equal 0.999... because a - b = 0 if and only if a = b. There are other sets of numbers where this is not the case, such as the p-adic numbers for some p. Wikipedia has a whole article devoted to this: https://en.wikipedia.org/wiki/0.999... https://en.wikipedia.org/wiki/0.999....

cygx 13y ago

The value of a decimal number is by definition the sum of its digits weighted by the appropriate power of 10. If the fractional part is not finite, we need to compute an infinite series n 0.999… = lim Σ 9/10^k = lim ( 1 - 1/10^n ) = 1 - lim 1/10^n = 1 - 0 = 1 ^ n→∞ k=1 ^ n→∞ ^ n→∞ ^ ^ | ^ | ^ | | | definition | induction | algebraic basic now we're | | limit theorem proof getting silly 0.9 = 1 - 0.1 0.99 = 1 - 0.01 … = … which is the case of a geometric series ∞ Σ a⋅r^k = a / (1 - r) k=0 with a = 9/10, r = 1/10. Another way to look at this is via the constructive definition of the reals as equivalence classes of Cauchy sequences of rational numbers: The sequences (1, 1, 1, …) and (0.9, 0.99, 0.999, …) are just two of an infinite number of Cauchy sequences which make up the real number 1.

ColinWright 13y ago

You've had half a dozen answers already, and they are all correct, but you may find them unenlightening, because you may find they are not addressing what you feel the problem is. To you it may be "obvious" that 0.9999... can't equal 1, and if that's the case, no amount of explanation of why it is will satisfy you. If you want to talk about this further and to understand what's going on, I'm happy to try to work with you. My email is in my profile.

deleted 13y ago

[deleted]

saalweachter 13y ago

Starting at the very beginning: Stuff you write down aren't numbers, they're representations or names of numbers. There is a number named "1", there is a number named "2". "2" is not the number itself. You can't infer the two-ness of "2" by staring at the symbols. It turns out that our way of naming numbers isn't perfect. Each number has multiple names. "2" and "2.0" are both names for the same number, as is "2.00", "2.000", and "2.0000". It turns out that you can name "2" a different way, as "1.9999999999999999999999999999999999999999999999999999...", where there are an infinite number of nines. The names may be different, but the numbers they name are the same.