Mathematical error 1=0.9 repeating

let x=0.9(repeating)
10x=9.9(repeating)
10x-x=9.9(repeating)-0.9(repeating)
9x=9
x=1
therefore 1 is actually 0.9(repeating)

repeating is the line above meaning goes on forever

right
it's not actually an error, rather, .9(repeating) is so close to 1 that there are no numbers in between .9(repeating) and one, but this is impossible, and therefore .9(repeating) is equal to 1
as you proved.

The' error' rises in the (in)ability of maths to understand the concept of infinity and this is up to the user.

If x = 0.999(recurinng); then when one really multiplies this by 10 you translate the whole thing across. So for example if x = 0.9999 then 10x = 9.999 and thus one 'loses' 0.0009. Now if you take recurring to mean infinity, this of course does not happen when you multiply by 10 so 0.999999recurring becomes effectively 1 by the user, in the first instance. Really, then, the outcome is up to the belief of the user. This assumption appears in the second line of these equations.

math is just a language used for mathematicians to describe size, quantity, and order... there is no "one" answer to any math problem. math is always evolving as we need to describe more things, think of calculus for that matter.... newton needed to make a language so he could describe his laws to the world....

on the math thought, do you agree or disagree with this formula proving 1=2...

a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1

run that by ur math teachers and see what they think.

i know the answer.. im seeing if anyone else can get it.

10x-x=9.9(repeating)-0.9(repeating)
9x=9

10(.9)=9
9 - .9= 8.1
9.9 - .9= 9
so 8.1=9... right?

10 times 90% is 9.

.

no it works out. Anything devided by itself is equal to 1. for instence if (a-b) is equal to three well then 3/3 is equal to 1. So it cancels out on both sides.

a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1

In line 4 both sides of the equation have the factor (a-b). Since a=b a-b=0. The fifth liine may be rewritten as (a+b)(a-b)/(a-b) = b(a-b)/(a-b). Since a-b=o what we have is a classic case of division by zero, or the question "How many times does zero go into..." since no matter how many times you add a factor of zero you remain at zero zero goes into any number an infinite amount of times (it never reaches its goal). In other words the final line should read 2/0=1/0, infinity equals infinity, a true statement.

let x=0.9(repeating)
10x=9.9(repeating)
10x-x=9.9(repeating)-0.9(repeating)
9x=9
x=1
therefore 1 is actually 0.9(repeating)

Now let X equal a finite number: .99
10x=9.9
10x-x=9.9-.99
9x=8.91
x=.99

Let x=.99999
10x-9.9999
10x-x=9.9999-.99999
9x=8.99991

This patern (an 8 followed by (n-1) 9s where n is the amount of digits in x followed by a 1) continues no matter how many digits X goes out to. You should think of .9(repeating) times 10 not as 9.9(repeating) but as 9.9(repating)0, an infinite amount of nines followed by a zero. This way the zero in that number is always by the final nine in .9(repeating) and when you subtract them you will gett 8.9(infinity-1 times)1 which divided by 9 yields .9(repeating)

heyo! we have a winner!... sort of. when you divide a number by 0, lets say 3/0... you're asking how many groups of 0 can go into 3 and that answer does not exist, it is not infinity... this is because if it was infinity and you put a number as an answer then check it... lets put 4... you get 4*0=3 which is by all means false, this happens with all numbers you put in, therefore you have no answer.

HOWEVER

what if you do 0/0, THEN you get infinity. because any number times 0 = 0. so, what if

a=b=0
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1


??

Drummajor, infinity is not a number: the concept of infinity-1 times is not valid in the context in which is is used.
Jacker_Jones, have you worked with limits and open intervals in calculus yet?

lim ƒ(x) = 1
x -> infinity

No i haven't had the pleasure perhaps in the future

You still can't devide by zero as you assume to in this problem.