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A Maths Puzzle...

Tituscrow

Inactive
Feb 14, 2011
1,502
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NW England
..well, not so much a puzzle, more something that seems counter-intuitive yet logically bombproof. A prize for anyone who shoots a hole in it.

THREE WAYS TO PROVE 0.999(rec) = 1

1. Switching between fractions and decimals,
1/3 = 0.333(rec). Multiply both sides by 3 and you get 1 = 0.999(rec)

2. Take 1 - 0.999(rec) = 0.000(rec). Therefore the distance between 1 and 0.999(rec) is nothing, so they must be the same.

3. Let x = 0.999(rec). Multiply both sides by 10 to get 10x = 9.999(rec). Next, subtract the first equation from the second equation to get 9x = 9.
Therefore x = 1.

What am I missing (other than something better to do at this time of night) ?
 
IIRC, the proof involves a premise imagining that there exists some x that is smaller than 1 - .999...

And that such an premise leads to a conclusion that is know to be false. Therefore the premise that such an x exists must also be false. Or something like that.
 
Makes sense. An asymptotic philosophy as it were. It's not really a math problem, it's more of a philosophical debate in some respects (although mathematical proofs do venture more than a little bit into the realm of philosophy).

Basically, is an infinite number of 9's after the decimal is the same thing as 1? Literally, the answer is no, but practially, the answer is yes.
 
Makes sense. An asymptotic philosophy as it were. It's not really a math problem, it's more of a philosophical debate in some respects (although mathematical proofs do venture more than a little bit into the realm of philosophy).

Basically, is an infinite number of 9's after the decimal is the same thing as 1? Literally, the answer is no, but practially, the answer is yes.

If it's not exactly 1, then there would exist an X != 0 such that .999... + X = 1. Since such an X does not exist, .999... = 1 exactly.

edit: i'm dumb, here's a smarty pants with the answer...

Invalid Link Removed
 
0.999... is a repeating decimal number or an "irrational" number. To convert this to a "rational" number, you do the third problem.

3. Let x = 0.999(rec). Multiply both sides by 10 to get 10x = 9.999(rec). Next, subtract the first equation from the second equation to get 9x = 9.
Therefore x = 1.


A rational number can be put in the form a/b. So, any "irrational" number can be converted into a rational to be put in the a/b form.

eq1: x=0.666...
eq2: 10x=6.666...

subtract eq1 from eq2 gives 9x=6
solve for x x=6/9
reduce fraction x=2/3
 
0.999... is a repeating decimal number or an "irrational" number. To convert this to a "rational" number, you do the third problem....


a repeating decimal is a RATIONAL number.

it is rational because it can be written as the ratio of two integers...

in this case,eg, 999/1000= .999 or 99,999/100,000=.99999

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