mathematical problem

I've got a mathematical problem.
I deeply hope I have to re-think or I have just miscalculated.

Beware it caused some serious headache when I thought about it, the first time.


(I underlined to express that the digit(s) repeat infinite)

We have this formula:
( 10 * x - 1 * x ) / 9 = x

=> ( 9 * x ) / 9 = x
=> x = x

So, it's true for all numbers.




But if x = 0.9

it says

( 10 * 0.9 - 1 * 0.9 ) / 9 = 0.9

=> ( 9.9 - 0.9 ) / 9 = 0.9
=> 9 / 9 = 0.9
=> 1 = 0.9 ?? ?? ??


Anyone who can help?

 

9/9 is 1

 

Line 1 after 'it says' should be:

(10x- 1x)/ 9 = x
[x in this case is 0.9]

line 2:
(9- 0.9)/ 9 = 0.9

line 3:
(8.1)/ 9 = 0.9

Ans:
0.9 = 0.9

 

( 10 * 0.9 - 1 * 0.9 ) / 9 = 0.9---do you not see what you did wrong?!!

=> ( 9.9 - 0.9 ) / 9 = 0.9---do you not see what you did wrong?!
=> 9 / 9 = 0.9
=> 1 = 0.9 ?? ?? ??

[10*.09]=9, not 9.9

how did you get 9.9?

 

If this only was my error, I'd be happy *sigh* :-\




Thanks for replying, anyways.

But...



The underlined part is very important

I don't know a (quick/good) way to say it in english.


0.9 means 0.9999999999999999999999999999999999999... ( the number of digits is endless )


0.9 != 0.9



so 10 * 0.9 = 10 * 0.9999999999999999999999999999999999999... = 9.9999999999999999999999999999999999999... = 9.9

that's why 10 * 0.9 is 9.9



And further

9.9 - 0.9 = 9.9999999999999999999999999999999999999... - 0.9999999999999999999999999999999999999... = 9




Has anyone a solution for this?

 

you can't use an infinity number. but lets assume x=.(9) as having to the hundred thousandth, that is .99999

[10(.99999)-1(.99999)]/9=.99999, you get,

[9.9999-.99999]/9=.99999, then

8.99991/9=.99999,and your answer is,

.99999=.99999

there is a reason why you get a 1=.(9) and not a .(9)=.(9)
it has something to do with 10 that you probably don't see and notice with the other number. you can get a hint and understanding from the answer i gave above.

paper is truly the ultimate tool to your understanding!

also, it was a human error!

 

Math > 8th grade = brain fart.

 

math is really fun when you know what you're doing.

i hated math in college because professors do things according to the book, and literally by the book. they spent most or rather all of the class time on proving and explaining the definition of the equation. and of course they do give a few examples which were primarily the simple ones. once they were done with one they would go to the next. homework was assigned, but knowing that it is college you sometimes put things a side. basically, professors should provide classwork time for the students to get hands on when they learn firstly learn the concept. in that one instance, they would at least get a hands on instead of a lecture.

i had this one professor where she showed up for class and said she would not explain the definition of the equation. she said there isn't any reason for it as it has been proven over the centuries. but what is important is knowing on how to properly use them, that is applied. we spent in groups and individually figuring out problems that she gave. she even made us put our problems on board and explained them to the class to show our understanding. it was like high school but it worked [80 to 90% of the class passed].

also, if you want an explanation for the problem above, i can provide one for you. i never gave on because some people may want to solve it or figure it out. btw, it felt really good when i figured out the reason and why.

 

... Yes, very funny, guys ...


I found a solution on my own.

I'll explain it, but first ...

--------------------------------------------------------------------

Calculators got limited memory, so they cannot hold an infinite number of digits
rounding numbers is like cutting throats

I can, I did!

100000 != ∞

reason:
100000 / 2 = 50000
∞ / 2 = ∞
let's take all integral numbers starting with 1 - that's an infinite amount
ok, now let's take only every second (all even numbers) - still an infinite amount

paper - finite space
0.(9) - infinite number of digits
=> digits won't fit!


... ok, now it's better




-----------------------------------------------------------------------------------------------------------------------------------------------------------------------




So, let's start with the solution and explanation thingy.

uhu, that's right it was a human error - My human incapability of understanding infinity

and the reason why I got 1=.(9) is because


it's right!


Yes, 1 is equal 0.(9)! - It's the same number

--------------------------------------------------------------------

explanation:

Let's think of a function 'f(x)' - x is the number of digits of 0.99999999...

i.e.
f(0) = 0.0
f(1) = 0.9
f(2) = 0.99
f(3) = 0.999
f(4) = 0.9999
f(5) = 0.99999
...

It's easy to see if x gets high, f(x) will get closer to 1

--- only closer, so far ---

at this point we need some analysis


Let's throw the x against infinity.

x->∞ (that means an infinite number of digits... one moment, that's what we want)

now f(x) is super mega extremely near 1 - that near, that it's equal

f(∞)->1


Maybe you want a proof - look at the first post there is one


--------------------------------------------------------------------

another explaination:

Let's take one third = 1/3 = 0.(3)

now we multiply the whole thing with three

and we got 3/3 = 0.(9)

or 1 = 0.(9) !!!!!!!!!

--------------------------------------------------------------------


Nonetheless, thanks to everyone replied.

Some Few of you really increased my knowledge.


If someone want some explanation on it (again), I'm always here to help.

 

you could always copy and paste a wiki page to give an example of how 1=.(9).

 

And 2+2=5. Just a matter of how you define it.

Furthermore, 1 != 0.9. I don't know about the 1 = 0.(9), or why you would want to use it.

First post: you have an error in your calculation. You assume that 9.(9) - 0.(9) gives 9. The error generated by that calculation, is the difference in your 1 = 0.(9).

Every computer rounds numbers. Ever wondered how your calculator can calculate the SIN/COS/TAN functions so fast? Look up CORDIC algorithm. If you want an explaination I can give one.

 

Marcy: why did'nt you tell us its a limit problem? or what kind of math it was (algebra, calculus etc.)
Oh and congrats on getting the problem right

 

I would have, if I had found it on the english one.

But, in case someone is interested. Here's the german wiki ->http://de.wikipedia.org/wiki/Periodizit%C3%A4t_(Mathematik)
examples are far down.

OK
in my first post:
+ is addition with numbers
- is subtraction with numbers
* is multiplication with numbers
/ is division with numbers
= is equality with numbers

??? Sorry, I don't see my error, could you explain it a bit detailed.

No, I wondered why my own cosine is that excessively lame compared to the pre-defined one
;D CORDIC looks nice

I didn't know it is a limit problem when I wrote the first post. --- Yeah, it's calculus, I think.

 

http://en.wikipedia.org/wiki/0.999...

 

Now this i understood, and i also agree with it

 

Somebody who likes knowing what CORDIC is! Wow! My study doesn't seem so pointless after all.

Your definition of substraction does not match your calculation.

You say that 9.(9) - 0.(9) = 9, assuming that 0.(9) - 0.(9) = 0. Like (9 - 0) + 0.(9) - 0.(9) = 9 if you break it into terms.

Example:

Say the first 0.(9) is 0.99 in your case, or 0.9 + a where a = 0.09
Say the second 0.(9) is 0.9 in your case, or 0.9 + b where b = 0

Substituting this in your equation:
(0.9 + a) - (0.9 + b) = a - b = 0.09 - 0 = 0.09

a and b is the difference between the assumption that the first 0.(9) is 0.99 and the second 0.(9) is 0.9.

The error you made in your assumptions is the error that you will find in the result.