 Extremeskins

# Does 0.9999 repeating 9 equal 1?

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Does 0.9999 repeating 9 equal 1?

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No but it's extremely close.

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No, but it's close enough as to make no difference.

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Does 0.9999 repeating 9 equal 1?

Is this a trick question?

1 = 1

0.999999999 rounds up to 1

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Is this a trick question?

1 = 1

0.999999999 rounds up to 1

Winner winner chicken dinner. Only if you round up. ##### Share on other sites

If you go to halfway closer to something and stop, then then go halfway closer again and stop, then go halfway again and stop, etc. etc., do you ever get there?

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Actually, I seem to recall mathematically proving, back in High School, that the answer is "yes".

Although I've always thought of myself as more of an engineer than a mathematician.

And to an engineer, "close enough" = "yes".

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no, but the limit as n approaches infinity where n is the number of digits, is 1. ##### Share on other sites

If you go to halfway closer to something and stop, then then go halfway closer again and stop, then go halfway again and stop, etc. etc., do you ever get there?

See above. the limit as n approaches infinity where n is the number of times you go half-way to y is y.

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Is this a trick question?

1 = 1

0.999999999 rounds up to 1

:geek:

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If you go to halfway closer to something and stop, then then go halfway closer again and stop, then go halfway again and stop, etc. etc., do you ever get there?

Had a Physics teacher that used to pose this question with respect to a girl (of course being a Physics class at a smaller school (so a small class) the class was all boys). His conclusion was 'Depends on what you want to do.'

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SKINSFAN89 fails at math.

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Suppose .9999999999 (repeating) is not equal to 1.

Then these are two distinct numbers, call them x and z. Then there exists a y such that x + y = z. i.e. (by the additive inverse property). So since x = .99999999 repeating, y = .00000000000 repeating and finally a 1. How many 0's are there?

If there are finitely many 0's (say n) then y = 10^-n. This contradicts that x has infinitely many nonzero digits since x + 10^-n = z

So y must have an infinite number of 0's before the 1. But if y has an infinite number of 0's then y = 0. Then it follows that x = y, i.e. .999999999 (repeating) equals 1.

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0.3 repeating 3 is one-third, right?

and 0.6 repeating 6 is two-thirds, right?

what is 0.3 repeating 3 plus 0.6 repeating 6? 0.9 repeating 9.

what is one third plus two thirds? 1.

no rounding. no calculus limits involved. 0.9 repeating 9 equals 1.

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Suppose .9999999999 (repeating) is not equal to 1.

Then these are two distinct numbers, call them x and z. Then there exists a y such that x + y = z. i.e. (by the additive inverse property). So since x = .99999999 repeating, y = .00000000000 repeating and finally a 1. How many 0's are there?

If there are finitely many 0's (say n) then y = 10^-n. This contradicts that x has infinitely many nonzero digits since x + 10^-n = z

So y must have an infinite number of 0's before the 1. But if y has an infinite number of 0's then y = 0. Then it follows that x = y, i.e. .999999999 (repeating) equals 1.

Any doubt that I had about your screen name was just shattered by this post right here.

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i was watching the show on the patriot missile where a 1/10th of a second couldnt be set in the systems.. 100hrs of use and it was off by i think a meter off = miss

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Are we talking 0.999 repeating cents, or 0.999 units of money?

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.3 repeating = 1/3

3 x .3 repeating = 3 x 1/3

3 x .3 repeating = 1

.9 repeating = 1

It's true.

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0.3 repeating 3 is one-third, right?

and 0.6 repeating 6 is two-thirds, right?

what is 0.3 repeating 3 plus 0.6 repeating 6? 0.9 repeating 9.

what is one third plus two thirds? 1.

no rounding. no calculus limits involved. 0.9 repeating 9 equals 1.

I started to say this too. But I was afraid that somebody would say that 2/3 is .666666666666667.

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My take: it's entirely based on your level of precision/granularity.

Check out the calculation of machine epsilon in the determination of Floating Point precision: http://en.wikibooks.org/wiki/Floating_Point/Epsilon

For IEEE 754 double precision you're looking at 1.11022302462516 x 10^-16 as the value for epsilon. So, for values smaller than epsilon, from a machine's point of view, the values are going to be indistinguishable.

As Peter said above, it ultimately depends on what you're doing with the value as to whether this makes a difference.

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0.3 repeating 3 is one-third, right?

and 0.6 repeating 6 is two-thirds, right?

what is 0.3 repeating 3 plus 0.6 repeating 6? 0.9 repeating 9.

what is one third plus two thirds? 1.

no rounding. no calculus limits involved. 0.9 repeating 9 equals 1.

.3 repeating is accepted to be 1/3, but in truth, it never does reach it. the limit as n approaches infinity where n is the number of digits is 1/3, but .3 itself is not itself truly 1/3.

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0.999999 = 1 in 99.99999% of everything involved in the universe. The unsolved fraction is the divine and not to be understood.

Mathematical proof of Supreme Creator existence.

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SKINSFAN89 fails at math.

Its true...its all true ##### Share on other sites

.3 repeating is accepted to be 1/3, but in truth, it never does reach it. the limit as n approaches infinity where n is the number of digits is 1/3, but .3 itself is not itself truly 1/3.

Thank you.

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.3 repeating is accepted to be 1/3, but in truth, it never does reach it. the limit as n approaches infinity where n is the number of digits is 1/3, but .3 itself is not itself truly 1/3.

yes it is, if there are truly an INFINITE number of 3's and not just A VERY LARGE NUMBER of 3's. 0.9 with a MILLION BILLION 9s on the end does not equal 1. 0.9 with an infinite number of 9s DOES equal 1.