O.9999999... = 1?

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Mercy
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Re: O.9999999... = 1?

#21 Post by Mercy » Mon Dec 18, 2017 6:45 am

A real proof would start with "Let ε > 0".

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Re: O.9999999... = 1?

#22 Post by ghug » Wed Dec 20, 2017 9:50 pm

I explicitly said NO MATH EQUATIONS

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Re: O.9999999... = 1?

#23 Post by Jeff Kuta » Sat Dec 30, 2017 11:46 pm

nyet

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Re: O.9999999... = 1?

#24 Post by ItsHosuke » Sun Dec 31, 2017 2:26 am

Testing signature

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Re: O.9999999... = 1?

#25 Post by Jeff Kuta » Sun Dec 31, 2017 6:41 am

I was told there would be no math.

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Re: O.9999999... = 1?

#26 Post by ItsHosuke » Sun Dec 31, 2017 8:22 am

O.9999999... doesn't equal 1, simply because it's an O in front, not 0 (Zero)

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Re: O.9999999... = 1?

#27 Post by yavuzovic » Sun Dec 31, 2017 9:01 am

The problem is not 0.99999999999...
10-9.9999999999999999.... seems more confusing. Can someone explain that smallest positive number?

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Re: O.9999999... = 1?

#28 Post by Rjmcf » Sun Dec 31, 2017 11:58 am

Can someone explain to me why “a” is essentially the same as “A”? Only don’t use the whole alphabet, only use the letters a, h, x, and z. Any answers that contain letters other than those I won’t accept for some arbitrary reason. Thanks.

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Re: O.9999999... = 1?

#29 Post by reedeer1 » Sun Dec 31, 2017 5:25 pm

Is the name color based on donation level?

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Re: O.9999999... = 1?

#30 Post by Wattsthematter » Thu Mar 28, 2024 5:19 pm

This is Wattsthematter testing the new forum.

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Re: O.9999999... = 1?

#31 Post by CaptainFritz28 » Thu Mar 28, 2024 5:25 pm

yavuzovic wrote:
Sun Dec 31, 2017 9:01 am
The problem is not 0.99999999999...
10-9.9999999999999999.... seems more confusing. Can someone explain that smallest positive number?
It's 1*10^-infinity.
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Re: O.9999999... = 1?

#32 Post by CaptainFritz28 » Thu Mar 28, 2024 5:28 pm

I disagree with the notion that 0.9999999... = 1. Usually people like to "prove" this by talking about 0.333333... = 1/3, and that 1/3*3 = 1, so 0.33333...*3 = 1, and thus 0.9999... = 1. But this fails to take into account the fact that 0.33333... is merely a decimal approximation of 1/3, albeit infinitely close, just as 0.9999... is infinitely close to 1.

The two numbers are differentiated by an amount equal to 1*10^-infinity.
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Re: O.9999999... = 1?

#33 Post by cdngooner » Thu Mar 28, 2024 5:59 pm

But this fails to take into account the fact that 0.33333... is merely a decimal approximation of 1/3
I'm not sure that is correct. If you ever stop the repeating sequence, THEN 0.3333333333333333333333333333 is an approximation of 1/3. But if you accept that 0.3333333333333333333333.... is a repeating rational number, then it IS EQUAL to 1/3. Its the infinite repetition that makes it equal. Any contrary argument implicitly assumes that the function stops repeating at some point.

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Re: O.9999999... = 1?

#34 Post by kingofthepirates » Thu Mar 28, 2024 6:02 pm

what's up with these random forums popping back up after so long? 7 years or so before the posts today... crazy! Anyway, I saw math and felt the urge to post/correct:

0.99999... infinitely repeating is equal to 1.

10*0.999999...=9.999999...
10*0.999999...-0.999999...=9*0.999999...=9.999999...-0.999999=9
therefore: 9*0.999999...=9, divide both sides by 9, and we get 0.999999...=1

furthermore, the limit as x->-infinity of 1*10^x is 0, so 1-0.99999... infinitely repeating is 0, so 1=0.99999... infinitely repeating.

the two numbers are mathematically equal to each other. Another way to think of is as the limit of the series 9*10^-n, as n goes to infinity.

for a more worded explanation: consider 0.999... repeating. to make exactly 1, a person would need to add 0.000...1, where there are infinity 0's before the 1. the 1 will never arrive, because there are infinite 0's, which is the same as 0. so 0.9... repeating is the same 1, since their difference is 0. This explanation was how I first comprehended the equality between the 0.9... repeating and 1.
CaptainFritz28 wrote:
Thu Mar 28, 2024 5:25 pm
yavuzovic wrote:
Sun Dec 31, 2017 9:01 am
The problem is not 0.99999999999...
10-9.9999999999999999.... seems more confusing. Can someone explain that smallest positive number?
It's 1*10^-infinity.
the smallest positive number also doesn't really exist, since you could get infinitely close to 0, but there will always be number between any number you give and 0. as a direct counter example, if 1*10^-infinity is a positive number (which implies non-zero), a smaller number would be 1/2*10^-infinity. and you could keep cutting that smaller and smaller, ad infinitum.
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Re: O.9999999... = 1?

#35 Post by Esquire Bertissimmo » Thu Mar 28, 2024 6:10 pm

Very glad the right answer is on the board :)

0.999999.... = 1 is a consensus view in mathematics.

KoP's proof is exactly right.

Another way to think about this is conceptually. What number separates 0.999... from 1? There is no such number if you mean the "..." to represent the concept of infinitely repeating 9s after the decimal. It would be a misapplication of the word "infinity" to say there is. Real numbers are separated by some value and that's impossible in this case, so 0.999... really is 1.

That's also why 1/3 really is 0.333... and why it's still true that 3*(1/3) = 1 and not some number less than 1. This makes intuitive sense - otherwise, repeating (1/3)*3 would give a different answer depending how many time you did the calculation.

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Re: O.9999999... = 1?

#36 Post by sweetandcool » Thu Mar 28, 2024 7:29 pm

Just a "proof" that 1/3= 0.33.... (infinitely many '3's). Since that seems to be tripping some people up.

We use some algebra here: 1/3=x --> 100/3=100x --> 33 + 1/3=100x --> (33/100)+ ([1/3]/100) --> .33 + ([1/3]/100)=x.

We now know the first two decimal digits of 1/3 to be .33. We can now use that to fact to calculate the remainder ([1/3]/100)= (.33+[1/3]/100)/100.

If we were to stop calculating 1/3 here it would be 1/3=.3333 + z, where z is some Real Number greater than 0.

We can keep substituting in the value for 1/3= .33 +([1/3]/100) infinitely many times, and it becomes clear that 1/3= 0.3333..... (infinitely many '3's).
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Re: O.9999999... = 1?

#37 Post by sweetandcool » Thu Mar 28, 2024 8:30 pm

sweetandcool wrote:
Thu Mar 28, 2024 7:29 pm
Just a "proof" that 1/3= 0.33.... (infinitely many '3's). Since that seems to be tripping some people up.

We use some algebra here: 1/3=x --> 100/3=100x --> 33 + 1/3=100x --> (33/100)+ ([1/3]/100) --> .33 + ([1/3]/100)=x.

We now know the first two decimal digits of 1/3 to be .33. We can now use that to fact to calculate the remainder ([1/3]/100)= (.33+[1/3]/100)/100.

If we were to stop calculating 1/3 here it would be 1/3=.3333 + z, where z is some Real Number greater than 0.

We can keep substituting in the value for 1/3= .33 +([1/3]/100) infinitely many times, and it becomes clear that 1/3= 0.3333..... (infinitely many '3's).
I should also note that you can use this method to calculate the value of any fraction where the denominator is <=100. For numbers greater than 100, I will leave that as an exercise for the reader (because it is bothersome to type out).

I love fractions and when I am bored and waiting in line or something I like to mentally calculate the values of fractions. 1/7 is my favorite, it is a pretty fun fraction.

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