A big controversy in math (0.999... = 1)

Discussion
1

Vote: I am curious!!!

0.999… = 1?

  • Yes
  • No
0 voters
1 Like
2

Here’s my take on it- 0.999… = 1 - 10^-n, as n->∞. What’s 10^-n approach as n->∞? It approaches 0. Thus, 1 - 10^-n approaches 1 as n->∞.

2 Likes
3

Uh… well I haven’t learned limits so I just said, if two numbers aren’t equal, you should be able to find a real number between them. You can’t find one between 0.999… and 1, so they have to be equal.

2 Likes
4

That’s some pretty sound reasoning for most ordinary systems. That’s a lot easier than the method I used…

5

there are dozens of proofs (from very intuitive to very formal) … and you can find lots of videos …
here’s a simple one:
1/3 = 0.3333… (just do the division)
now multipliy both sides times 3:
3 * 1/3 = 3 * 0.3333…
and you get
1 = 0.9999…

6

ntm dev ,there is one,if .9… is 1-1/infinity,then 1-1/2(infinity)
also if 1/infinity =0 so 1-1/infinity=1,then there’s infinite prime numbers and only 2 is even ,so your odds off randomly picking a prime number =1/infinity=0?

7

What’s wrong with:

Let x = 0.999..., so then 10x = 9.999...

Now subtract:

 10x = 9.999...
-  x = 0.999...
----------------
  9x = 9
   x = 1
1 Like
8

well what about 2(0.9…)
2
x .99999999999…9
1.99999999999…8
so there is a gap

1 Like
9

well, actually
2 * 0.999999…
is
1.999999 …
because 9*2 = 18 so each 8 gets an extra 1 added …

1.8
  18
   18
    18 
     ...
10

but,there will be after an infinite 9s ,on that’s an 8

1 Like
11

well that’s sort of related to p-adic numbers … but it does not work like that in real number

12

Nothing wrong with that

No offense but 0 idea what you’re trying to say @bhssingul

13

If it has infinite 9s, how will I put an 8 at the end? It’s like going to the end of an infinite line…

14

yes ,but there still has to be an eight somewhere,and how many even primes are there?
what the odds of when picking any of the infinite primes,the odds you pick 2 is 1/infinity,and you are saying 1- 1/infinity=1{which is just 9 repeating=1}
then from that 1/infinity =0 and circling around,the odds of picking 2=0?

15

You are assuming there are infinite primes, and how does “1/infinity” relate here? 0.999… is not prime.

Another thing, you are using a probability directly as a number. You say that because 1/infinity = 0, 2=0. But 2 doesn’t equal 1/infinity. You just basically changed a probability into an actual number.

And like Guzman was saying, it’s a p-adic number, so the 8 can’t exist.

16

Okay, being honest, there are like 100x more proofs for 0.999… =1 than there are for 0.999… != 1

0.999... - Wikipedia…

17

https://medium.com/@kenahlstrom/proof-that-99999-is-not-equal-to-1-5672e7dd58ce
also , if you give 1/3=.333
3/3=.9999
do long division…its not .333333 its bigger
and what is the 1- 1/(10^n) as n gets bigger:.999999999
and there is infinite primes

18

Oh i thought you meant something else about primes I was confused when you said about 1/infinity, never mind.
Also I don’t like the proof with algebra that’s why I don’t use it

19

and if 1/infinity = 0,and there is ifinite primes
if n->infinty,
the odds of picking 2 out of primes =0

20

1/infinty doesn’t equal 0