0.999... = 1 !?!?!?!?!

A

abroms

New member
May 17, 2009
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This can't be true! They're clearly different numbers. I mean, the first one has a lot of 9's and the second only has one 1! Can somebody please prove to me that this is true!?

/sarcasm
 


Yes. It's true.
It's about binary representation of floating point numbers.
 
it's true

338_Screen_shot_2010_03_30_at_4.23.png
 
no it isn't, it's about convergent series. .999 = 1 has been around longer than that.
Click to expand...

"0.9(9)" is not convergent series.

But yes, some things have been around longer than other things. I agree with you there :)
 
1 == 1/1

1/1 == 3/3

1/3 + 1/3 + 1/3 == 3/3

1/3 = .33~

.33~ + .33~ + .33~ == .99~

so .99 == 1
 
bb_wolfe said:
1 == 1/1

1/1 == 3/3

1/3 + 1/3 + 1/3 == 3/3

1/3 = .33~

.33~ + .33~ + .33~ == .99~

so .99 == 1
Click to expand...

Yeah, the other algebraic proof is usually:

0.999... = x
9.999... == 10x
9.999... - 0.999... == 10x - x
9 == 9x
1 == x
 
int a = 1;
double b = 0.9999;

if (a == (int)b) // yes it's true but with approximation
 
bb_wolfe said:
1 == 1/1

1/1 == 3/3

1/3 + 1/3 + 1/3 == 3/3

1/3 = .33~

.33~ + .33~ + .33~ == .99~

so .99 == 1
Click to expand...


When I was in 7th grade me and my friends were in advanced math (whatever that means for 7th graders) and my best friend had long and heated arguments with the teachers using this same explanation. PLUS OMG

1/1 == 3/3
.99 == 1
bb wolfe == Mrs. Flannigan
 
That solution has a flaw. (Aside from the fact that it was an April fool's joke.)

The only way a periodic fraction can be equal to an integer is if you represent both using integers. And that would only make them equal because of a finite resolution of the system.
 
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