Is 1 = 0.9999......

[dighn]

Originally posted by: Jeff7181
I'm no rocket scientist, but how to do you get 9x=9 from 10x - x = 9.9999... - 0.9999... ??????

Also... you're assuming right off the bat that x=.999... if that's true, then your conclusion is false, because that formula proves that x-.999... not that 1 = .999...

it goes like this
x = 0.9r
10x = 9.9r (shift by 1 place)
10x - x = 9.9r-0.9r = 9
10x-x = 9x
9x = 9
x = 1 = 0.9r

although, that's probably not a good proof because of the 9.9r - 0.9r = 9 step

 

[Jeff7181]

oh... nevermind... caught my error and now I see the error in my math =)

 

[Jeff7181]

 

[dighn]

nm

 

[Muzzan]

dighn: your proof is a bit off... 10x - x = 9x essentially states that x = x (woah, 9 = 9, I never knew that before ). 10x - x = 9 (which is how it's supposed to be) on the other hand...

 

[ElFenix]

Originally posted by: dighn

Originally posted by: Jeff7181
I'm no rocket scientist, but how to do you get 9x=9 from 10x - x = 9.9999... - 0.9999... ??????

Also... you're assuming right off the bat that x=.999... if that's true, then your conclusion is false, because that formula proves that x-.999... not that 1 = .999...

it goes like this
x = 0.9r
10x = 9.9r (shift by 1 place)
10x - x = 9.9r-0.9r = 9
10x-x = 9x
9x = 9
x = 1 = 0.9r

although, that's probably not a good proof because of the 9.9r - 0.9r = 9 step

you inserted an x where not needed in the 4th line

 

[bleeb]

ITS ALIVE! ALIVE!! ALIVE!!!

 

[Jeff7181]

x=0.9r
10x=9.9r
10x-x=9.9r-x
(subtract x from both sides)
10x=9.9r
(10x)/10=(9.9r)/10
x=0.9r

 

[Jeff7181]

Originally posted by: Jeff7181
x=0.9r
10x=9.9r
10x-x=9.9r-x
(subtract x from both sides)
10x=9.9r
(10x)/10=(9.9r)/10
x=0.9r

The origional formula substitues 0.9r for x... assuming x=0.9r, which is what you're trying to prove, so you can't assume that. Err wait... nah, that doesn't work... lemme rephrase...

 

[Muzzan]

Originally posted by: Jeff7181
x=0.9r
10x=9.9r
10x-x=9.9r-x
(subtract x from both sides)
10x=9.9r
(10x)/10=(9.9r)/10
x=0.9r

10x - x != 10x.

 

[dighn]

Originally posted by: ElFenix

Originally posted by: dighn

Originally posted by: Jeff7181
I'm no rocket scientist, but how to do you get 9x=9 from 10x - x = 9.9999... - 0.9999... ??????

Also... you're assuming right off the bat that x=.999... if that's true, then your conclusion is false, because that formula proves that x-.999... not that 1 = .999...

it goes like this
x = 0.9r
10x = 9.9r (shift by 1 place)
10x - x = 9.9r-0.9r = 9
10x-x = 9x
9x = 9
x = 1 = 0.9r

although, that's probably not a good proof because of the 9.9r - 0.9r = 9 step

you inserted an x where not needed in the 4th line

which line? i don't see it


anyway

u nkow now that i think about
9.9r = 9 + 0.9r so 9.9r - 0.9r should be ok unless 0.9r != 0.9r

0.9r * 10 = 9.9r should be fine also

 

[ElFenix]

Originally posted by: dighn

Originally posted by: ElFenix

Originally posted by: dighn
it goes like this
x = 0.9r
10x = 9.9r (shift by 1 place)
10x - x = 9.9r-0.9r = 9
10x-x = 9x
9x = 9
x = 1 = 0.9r

although, that's probably not a good proof because of the 9.9r - 0.9r = 9 step

you inserted an x where not needed in the 4th line

which line? i don't see it

bolded

 

[dighn]

Originally posted by: ElFenix

Originally posted by: dighn

Originally posted by: ElFenix

Originally posted by: dighn
it goes like this
x = 0.9r
10x = 9.9r (shift by 1 place)
10x - x = 9.9r-0.9r = 9
10x-x = 9x
9x = 9
x = 1 = 0.9r

although, that's probably not a good proof because of the 9.9r - 0.9r = 9 step

you inserted an x where not needed in the 4th line

which line? i don't see it

bolded

oh i did that for emphasis because it's unarguable that 10x-x = 9x

 

[Muzzan]

Originally posted by: Jeff7181

Originally posted by: Jeff7181
x=0.9r
10x=9.9r
10x-x=9.9r-x
(subtract x from both sides)
10x=9.9r
(10x)/10=(9.9r)/10
x=0.9r

The origional formula substitues 0.9r for x... assuming x=0.9r, which is what you're trying to prove, so you can't assume that.

Perhaps we should try another number in order to see that the method does in fact work.

x = 0,333...
10x = 3,333...

10x - x = 3,333... - 0,333...
9x = 3
x = 3/9
x = 1/3

x = x <=> 0,333... = 1/3, which anyone with a grade-school eduction knows is true So the formula works.

But, if we use this formula with x = 0,999... instead, we get that x = 1, which means that 0,999... = 1. Maybe it's clearer now?

 

[Jeff7181]

The origional formula substitues 0.9r for x on the right side, but not on the left side, that's against the rules of math if I'm correct. If you follow the rules of math...

x=0.9r
10x=9.9r
10x-0.9r=9.9r-0.9r
10x=9.9r
(10x)/10=(9.9r)/10
x=0.9r

You say it doesn't matter that you didn't substitute it on both sides? Ok, well in my formula I choose to substitute it, and it doesn't show that 0.9r = 1

 

[dighn]

Originally posted by: Jeff7181
The origional formula substitues 0.9r for x on the right side, but not on the left side, that's against the rules of math if I'm correct. If you follow the rules of math...

how is it against the rules if x is eqaul to 0.9r

1 = 1
x = 4
1 + x = 1 + 4

we're not trying to solve for x here, we're just using x as a symbol for the number

 

[Syran]

Don't know if anyone linked this yet: http://www.math.fau.edu/Richman/html/999.htm

 

[Muzzan]

Nope, that's not "illegal".

 

[Jeff7181]

Originally posted by: dighn

Originally posted by: Jeff7181
The origional formula substitues 0.9r for x on the right side, but not on the left side, that's against the rules of math if I'm correct. If you follow the rules of math...

how is it against the rules if x is eqaul to 0.9r

1 = 1
x = 4
1 + x = 1 + 4

we're not trying to solve for x here, we're just using x as a symbol for the number

ok... so...

x=0.9r
10x=9.9r
10x-x=9.9r-0.9r=9
10x-x=9
(10x-x)/10=9/10
x-x=9/10
0=0.9r =) how's that?

yeah, I guess that doesn't work either does it, lol ... nevermind

 

[Muzzan]

Originally posted by: Jeff7181

Originally posted by: dighn

Originally posted by: Jeff7181
The origional formula substitues 0.9r for x on the right side, but not on the left side, that's against the rules of math if I'm correct. If you follow the rules of math...

how is it against the rules if x is eqaul to 0.9r

1 = 1
x = 4
1 + x = 1 + 4

we're not trying to solve for x here, we're just using x as a symbol for the number

ok... so...

x=0.9r
10x=9.9r
10x-x=9.9r-0.9r=9
10x-x=9
(10x-x)/10=9/10
x-x=9/10
0=0.9r =) how's that?

Distributive property, damnit!

(10x-x)/10=9/10
x - x/10=9/10

 

[dighn]

nm already answered

 

[spidey07]

(10x-x)/10=9/10

you need to learn how to divide first.



-edit- seems others are quick as well.

 

[Jeff7181]

Lemme try again... lol...

x=0.9r
10x=10(0.9r)
10(0.9r)=10(0.9r)

see what I'm saying?

 

[Syran]

Originally posted by: Jeff7181

Originally posted by: dighn

Originally posted by: Jeff7181
The origional formula substitues 0.9r for x on the right side, but not on the left side, that's against the rules of math if I'm correct. If you follow the rules of math...

how is it against the rules if x is eqaul to 0.9r

1 = 1
x = 4
1 + x = 1 + 4

we're not trying to solve for x here, we're just using x as a symbol for the number

ok... so...

x=0.9r
10x=9.9r
10x-x=9.9r-0.9r=9
10x-x=9
(10x-x)/10=9/10
x-x=9/10
0=0.9r =) how's that?

yeah, I guess that doesn't work either does it, lol

(10x-x)/10 != x-x

You have to simplify it somehow.

(10x-x)/10 = 10x/10 - x/10 = x - x/10 = 9x / 10

Back to the same spot.

The problem seems to be that x is really already solved. You cannot subsititue it in the final answer, and have it look proper... it's not like getting a 2nd possible answer due to a abs or some form of exponential raise. seems the real issue is you already have a solved equation, either x = 1 or x = 0.9r, not both.

 

[Muzzan]

Why not both? A single number can be represented many, many ways (2 = 8/4 = 16/8 = 1000/500). If you look a few posts up, I posted an example which used 0,333... instead of 0,999... Would you be willing to argue that 1/3 != 0,333...?