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0.999...= 1

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Re: 0.999...= 1

#31

Post by tomtomtom1 »

saagarjha wrote: post #13
rs23296008n1 wrote: post #6

I'm actually curious what impact it would have on various proofs if 0.999... wasn't accepted as 1. What gets broken? What consequences do we hit?

Arithmetic breaks, as multiplication is no longer the inverse of division. (For example, 1/3 * 3 = 0.999… would no longer work.)

why?

1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense.

I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?

Re: 0.999...= 1

#32

Post by peanutz454 »

edflsafoiewq wrote: post #17
ltbarcly3 wrote: post #7

This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is half way between them. Conversely, if the average of two numbers is equal to either of the numbers, then the two numbers are equal. (this isn't a proof, just a way to convince yourself of this) What's the average of .9999... and 1?

Or: if they're different, what is their difference?

0.000...1 obviously

Re: 0.999...= 1

#33

Post by ginko »

larschdk wrote: post #25
ginko wrote: post #9

I remember being doubtful when being presented with this in middle school, but after being shown this as fractions makes it obvious: 1/3 = 0.333.. 3 * 1/3 = 3 * 0.333.. 3/3 = 0.999.. 1 = 0.999..

I don't mean to troll you, but if you were doubtful that 0.999... = 1, then you should also be doubtful that 0.333.. = 1/3. Any argument that 0.999... is not quite 1 can also be used to argue that 0.333... is not quite 1/3. I think it's mostly a matter of definition, since mathematicians consider sums of infinite series equal to their limit (if it's finite), i guess for many practical reasons. If you accept this, the…

See my other reply along the same lines. I was thinking about adding this caveat to my original message as well, but I think understanding that decimal numbers with infinite digits exist and that 0.999.. = 1 are separate things. The second being less intuitive.

Re: 0.999...= 1

#34

Post by [deleted] »

ginko wrote: post #28
ant6n wrote: post #23

Earlier quoted context omitted.

Well, here you reduced 1=.9999... to 1/3=0.333... What if I don’t believe that second equation.

As in the 0.333... will stop at some point? That would still mean that 3 time 0.333... with a LOT of 3s end up being being 1. I also figure it's a bit more intuitive for pupils to just try out calculating the decimal representation of 1/3 and seeing that it'll just keep going forever.

[deleted]

Re: 0.999...= 1

#35

Post by [deleted] »

ltbarcly3 wrote: post #7

This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is half way between them. Conversely, if the average of two numbers is equal to either of the numbers, then the two numbers are equal. (this isn't a proof, just a way to convince yourself of this) What's the average of .9999... and 1?

[deleted]

Re: 0.999...= 1

#36

Post by steerablesafe »

username90 wrote: post #11
bytedude wrote: post #2

Flame wars over this used to be common on the internet. People intuitively have the notion that the left side approaches 1, but never actually equals it. They see it as a process instead of a fixed value. Maybe the notation is to blame.

The intuition that there is something in between isn't really wrong, it make sense and they work, otherwise physicists wouldn't be able to work with them. So that intuition is correct, it is mathematicians who just don't understand it fully yet. Maybe fully formalizing this is what unlocks the final piece keeping us from creating a unified theory in physics?

Nonstandard analysis is a rigorous framework for working with infinitesimals (and infinitely large numbers).

https://en.wikipedia.org/wiki/Nonstandard_analysis

Re: 0.999...= 1

#39

Post by [deleted] »

rini17 wrote: post #8
klodolph wrote: post #4

An interesting consequence of this in proofs. You’ll see various proofs involving real numbers that must account for the fact that 0.999…=1.0. There are, of course, many different ways to construct real numbers, and often it’s very convenient to construct them as infinite sequences of digits after the decimal. For example, this construction makes the diagonalization argument easier. However, you must take care in you…

I never understood the fixation on diagonalization. Why can't ever exist another way for mapping any set to countables?

[deleted]

Re: 0.999...= 1

#40

Post by hinkley »

notRobot wrote: post #24
iso1631 wrote: post #12

Earlier quoted context omitted.

Another secondary school 'proof' x = 0.9999..... 10x = 9.9999..... (10x -x) = 9x = (9.9999.... - 0.9999....) = 9 x = 9/9 = 1

or, x = 0.9999... 10x = 9.999... 10x = 9 + 0.999... 10x = 9 + x 9x = 9 x = 1 Presented slightly more clearly https://en.wikipedia.org/wiki/0.999...#Algebraic_arguments

Yeah that’s way more complicated than it needs to be and I’m tempted to replace that whole section with:

x = 0.9999...

2x = 1.9999...

2x - x = 1

x = 1

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