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Classic puzzle: .9999..... AND 1

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Here is a classic exercise: Which of the following are true and why?

0.99999..... < 1

0.99999..... = 1

0.99999..... > 1

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Posted · Report post

Sum_{i = 1 to infinity](0.1^i) = 1/9 as a geometric series (also easily proven algebraically). 0.999... = 9 * (0.111...) = 9 * Sum_{i = 1 to infinity](0.1^i) = 9 * 1/9 = 1. The equality is therefore true. By extension, both of the inequalities are not true.

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Posted · Report post

The easiest way to choose among the three options -- one that does not require the invocation of limits, Cauchy sequences, Dedekind cuts or delving into the theory of real numbers -- is to note the decimal representation of the rational number 1/3.

1/3 = 0.3333 ...

Multiply by 3.

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So are we saying only one case is true?

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So are we saying only one case is true?

Yes. Both sides of the relations specify real numbers, which are well-ordered. So, exactly one of the relations must be true.

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hmm. there's a mathematician on youtube, norm wildberger, whom i kinda like who would disagree.

the problem being that we cant actually "know" infinity. try writing infinite 9's. you can't.

so at best we can say 0.9999999 or wherever you care to stop approximately equals 1.

to prove this point, accurately calculate pi +e +sqrt(2). if you're being honest, you would calculate it as pi + e +sqrt(2).

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When I was in 5th grade, my math teacher gave the class the following proof:


(10x - x)/9 = x
10*0.999... = 9.999...
_ 9.999...
0.999...
9
9/9 = 1
QED
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hmm. there's a mathematician on youtube, norm wildberger, whom i kinda like who would disagree.

the problem being that we cant actually "know" infinity. try writing infinite 9's. you can't.

so at best we can say 0.9999999 or wherever you care to stop approximately equals 1.

to prove this point, accurately calculate pi +e +sqrt(2). if you're being honest, you would calculate it as pi + e +sqrt(2).

The notation convention 0.999 ... denotes an endless string of 9's.

You can't write the decimal representation of 1/3, but you can write 0.333 ...,

Thus, given the meaning of that notation, one can meaningfully write 1/3 = 0.333 ...

Similarly one can meaningfully write 1 = 0.999 ...

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well, not entirely sure i agree. for example, how would you denote 1/13?

you need a consistent way of denoting any rational number, not just ones that repeat the last digit.

personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part.

so

  _
0.9 = 1

would be more accurate (although more cumbersome.)

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Posted · Report post

here's my challenge for you.

accurately add 1/13 to 45/89 in decimal representation.

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well, not entirely sure i agree. for example, how would you denote 1/13?

you need a consistent way of denoting any rational number, not just ones that repeat the last digit.

personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part.

so

  _
0.9 = 1

would be more accurate (although more cumbersome.)

______

1/13 = 0.076923

The bar covers the part that repeats.

here's my challenge for you.

accurately add 1/13 to 45/89 in decimal representation.

This can be done, but would be rather tedious (particularly as the decimal expansion for 45/89 has a relatively long repeating digit string). The sum will also have a repeating pattern, of some size less than or equal to the LCM of the repeating digit string lengths of the two operands.

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Posted · Report post

well, not entirely sure i agree. for example, how would you denote 1/13?

you need a consistent way of denoting any rational number, not just ones that repeat the last digit.

personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part.

so

  _
0.9 = 1

would be more accurate (although more cumbersome.)

I might have missed the gist of this question the first time around...

Another way to express 1/13 is 0.076923076923076923...

It is clumsy compared with the bar method in this example (unlike .999...), but still understandable.

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