math guys.. here's a question

[habib89]

i got this off another site.. the math, at least to me, sounds alright... what do you guys say? i remember hearing something about this in math, but i guess i wasn't paying much attention.. thoughts???

x = .99999999 (repeating forever)
10x = 9.9999999 (again repeating forever)

Now, subtract these equations.
10x-x = 9x.
9.999 - 9.999 = 9

Now you have
9x = 9 <-- Basic Algebra now. Divide both sides by 9.

9 divided by 9 is 1. Therefore you get x = 1

In the equation it stated x = .9999999999 so, .9999999 = 1

 

[diskop]

Originally posted by: habib89
i got this off another site.. the math, at least to me, sounds alright... what do you guys say? i remember hearing something about this in math, but i guess i wasn't paying much attention.. thoughts???

x = .99999999 (repeating forever)
10x = 9.9999999 (again repeating forever)

Now, subtract these equations.
10x-x = 9x.
9.999 - 9.999 = 9

Now you have
9x = 9 <-- Basic Algebra now. Divide both sides by 9.

9 divided by 9 is 1. Therefore you get x = 1

In the equation it stated x = .9999999999 so, .9999999 = 1

9.999 - 9.999 = 9 ?
Where did that come from

 

[dlaw]

Firstly he means 9.999-.999 =9

While I don't think it is ture if they are repeating forever(9.999-.999!=9 here). Just like infinity-infinity is not zero?

 

[Haircut]

0.999... where the 9s repeat infinitely is indeed equal to 1.

It can be shown that there is no number between 0.999... and 1 and by the continuum property of the real line they must be the same number.

There was a huge thread about this in Highly Technical forum last year sometime, but it appears to have been deleted now.

 

[max105]

there's a few websites that point out alot of weird and sometimes contradictory stuff in math....if I find the website again, I'll post a link.

 

[kt]

9.9999... - .9999... != 9

The proof assumed that it is, that's why you get different x that originally stated.

It's like a guy stealing a penny out of a money vault containing a trillion trillion trillion pennies.. sure, know one will noticed the penny is missing, but it is still missing.

 

[murphy55d]

math is hard.

 

[dlaw]

Originally posted by: murphy55d
math is hard.

Agree

 

[BurnItDwn]

This proof is true, an analogy of sorts ....

think of one third as being .3333333 repeating. Multiply it by 3 and you get .999999 repeating. Three Thirds is exactly 1, just as .9999999 repeating is exactly 1. Remember that there are an infinite number of digits after the decimal and they are all the higest amount possible.

To simplify it

Let N be the number of 9s after the Decimal point. As N increases to infinity, this increases to 1.

 

[eakers]

THIS MIRACLE IS CALLED SUBTRACTIVE CANCELATION.
IT HAPPENS MORE OFTEN THAN YOU WOULD THINK.

 

[Alphathree33]

Originally posted by: eakers
THIS MIRACLE IS CALLED SUBTRACTIVE CANCELATION.
IT HAPPENS MORE OFTEN THAN YOU WOULD THINK.

Eakers, I'm impressed. But... "subtractive cancellation"? That sure sounds impressive. Did you make it up?

 

[kt]

Originally posted by: BurnItDwn
This proof is true, an analogy of sorts ....

think of one third as being .3333333 repeating. Multiply it by 3 and you get .999999 repeating. Three Thirds is exactly 1, just as .9999999 repeating is exactly 1. Remember that there are an infinite number of digits after the decimal and they are all the higest amount possible.

To simplify it

Let N be the number of 9s after the Decimal point. As N increases to infinity, this increases to 1.

You're walking in circle with this. You're trying to prove that x = .9999... being substituted into an equation will turn x = 1 by assuming .9999... is equal to 1? It DOESN'T MAKE SENSE! If it were this simple, I would've A'sed all my math classes by assuming all my answers are correct.

 

[eakers]

Originally posted by: Alphathree33

Originally posted by: eakers
THIS MIRACLE IS CALLED SUBTRACTIVE CANCELATION.
IT HAPPENS MORE OFTEN THAN YOU WOULD THINK.

Eakers, I'm impressed. But... "subtractive cancellation"? That sure sounds impressive. Did you make it up?

I MAY OR MAY NOT BE A MATH STUDENT

 

[jeremy806]

Not quite correct.

For the mathematicians out there, it is improper to use regular old addition and subtraction on these types of numbers (i.e. 0.999...)

0.999... cannot equal x, 0.999... is shorthand for limit as n--> infinity of summation...

So, the fractional part of 0.999... and 9.999... are NOT the same and do not cancel out during subtraction, they only become the same as n goes to infinity thus the expression .999... only equals 1 as n goes to infinity.

Do I make sense?

Kind of like 2*infinity does not exactly "equal" infinity except in the limit.

Jeremy

 

[Alphathree33]

Originally posted by: kt

Originally posted by: BurnItDwn
This proof is true, an analogy of sorts ....

think of one third as being .3333333 repeating. Multiply it by 3 and you get .999999 repeating. Three Thirds is exactly 1, just as .9999999 repeating is exactly 1. Remember that there are an infinite number of digits after the decimal and they are all the higest amount possible.

To simplify it

Let N be the number of 9s after the Decimal point. As N increases to infinity, this increases to 1.

You're walking in circle with this. You're trying to prove that x = .9999... being substituted into an equation will turn x = 1 by assuming .9999... is equal to 1? It DOESN'T MAKE SENSE! If it were this simple, I would've A'sed all my math classes by assuming all my answers are correct.

Actually it does make sense.

Look, start with something you know to be true:

1/3 = 0.3 repeating

Now multiply both sides by 3

3/3 = 0.9 repeating

Now simplfy the left hand side

1 = 0.9 repeating

 

[kt]

Originally posted by: Alphathree33

Originally posted by: kt

Originally posted by: BurnItDwn
This proof is true, an analogy of sorts ....

think of one third as being .3333333 repeating. Multiply it by 3 and you get .999999 repeating. Three Thirds is exactly 1, just as .9999999 repeating is exactly 1. Remember that there are an infinite number of digits after the decimal and they are all the higest amount possible.

To simplify it

Let N be the number of 9s after the Decimal point. As N increases to infinity, this increases to 1.

You're walking in circle with this. You're trying to prove that x = .9999... being substituted into an equation will turn x = 1 by assuming .9999... is equal to 1? It DOESN'T MAKE SENSE! If it were this simple, I would've A'sed all my math classes by assuming all my answers are correct.

Actually it does make sense.

Look, start with something you know to be true:

1/3 = 0.3 repeating

Now multiply both sides by 3

3/3 = 0.9 repeating

Now simplfy the left hand side

1 = 0.9 repeating

You just restated what the above poster just said which is ONLY true when the limit is bounded by infinity. If you look at the original proof, it started out saying x = .99999 repeating. At a later stage of the proof, it introduces a new "variable" into the proof by saying that 9.999 repeating - .999 repeating = 9. This can only happen if the limit is infinite. When that "variable" is introduced into the proof, then everything changes. It's like if you all your life you have been living in a 2 dimensional world, and when a third dimension is introduced all your surroundings change.

 

[Alphathree33]

Originally posted by: kt

Originally posted by: Alphathree33

Originally posted by: kt

Originally posted by: BurnItDwn
This proof is true, an analogy of sorts ....

think of one third as being .3333333 repeating. Multiply it by 3 and you get .999999 repeating. Three Thirds is exactly 1, just as .9999999 repeating is exactly 1. Remember that there are an infinite number of digits after the decimal and they are all the higest amount possible.

To simplify it

Let N be the number of 9s after the Decimal point. As N increases to infinity, this increases to 1.

You're walking in circle with this. You're trying to prove that x = .9999... being substituted into an equation will turn x = 1 by assuming .9999... is equal to 1? It DOESN'T MAKE SENSE! If it were this simple, I would've A'sed all my math classes by assuming all my answers are correct.

Actually it does make sense.

Look, start with something you know to be true:

1/3 = 0.3 repeating

Now multiply both sides by 3

3/3 = 0.9 repeating

Now simplfy the left hand side

1 = 0.9 repeating

You just restated what the above poster just said which is ONLY true when the limit is bounded by infinity. If you look at the original proof, it started out saying x = .99999 repeating. At a later stage of the proof, it introduces a new "variable" into the proof by saying that 9.999 repeating - .999 repeating = 9. This can only happen if the limit is infinite. When that "variable" is introduced into the proof, then everything changes. It's like if you all your life you have been living in a 2 dimensional world, and when a third dimension is introduced all your surroundings change.

I didn't introduce any mysterious variable and by saying 0.9 repeating, it is implied that it repeats to infinite. You wouldn't normally 'specify' how long it repeats.

 

[Chaotic42]

There's a better way to prove that .99999999... = 1

1/3 = .33333333333333...
+2/3 = .66666666666666...
---------------------------------
3/3 = .99999999999999...

Since any number divided by itself is 1, .99999... = 1

 

[Dhruv]

speaking of math, did you guys read this: link

pretty interesting...

 

[LordSnailz]

it's simpler to understand if you covert x to fraction form -
x= 1/9
10x = 10/9
10x - x = 9x .. plug in the fraction form of x you get 9*(1/9) which is equal to 1.

 

[silverpig]

Here we go again. There WAS a huge thread on this in HT last year of which I was a part.

There are numerous proofs that show that 0.9... does indeed equal 1.

The simplest one (and least rigorous) is the 1/3 = 0.3.... and then muliply by 3 one.

There is also long division of 1/1 (which I think I may have invented hehe).

Also, in order to show that 0.9r (abbreviation for 0.9 repeating forever) is indeed LESS than 1, you must show explicitly that there exists a number in between 0.9r and 1. Example: 4 is less than 5 because there exists a number 4.5 that is larger than 4 but less than 5. Try doing that with 0.9r and 1. It can't be done. I can even show you how to go about trying.

Let 0.9r = w + x for some numbers w and x.
Let 1 = y + z for some numbers y and z.
Let a = b + c for some numbers a, b, and c.

Now write 0.9r < a < 1 -----> w + x < b + c < y + z

If you can come up with numbers for w, x, b, c, y, and z that satisfy all of the conditions above, then you may say that 0.9r is indeed less than 1.





Also, if you change into a different number base and re-try the 1/3 = 0.333... proof and choose such a base that the fraction doesn't repeat when converted into a decimal, it works out very nicely. (I can't remember the proof, or which number base to switch to (4 maybe?) but I'm sure google will help you.

 

Originally posted by: eakers

Originally posted by: Alphathree33

Originally posted by: eakers
THIS MIRACLE IS CALLED SUBTRACTIVE CANCELATION.
IT HAPPENS MORE OFTEN THAN YOU WOULD THINK.

Eakers, I'm impressed. But... "subtractive cancellation"? That sure sounds impressive. Did you make it up?

I MAY OR MAY NOT BE A MATH STUDENT

I CAN POST IN BOLD CAPITAL LETTERS TOO!

 

[silverpig]

Oh yeah, and you can also do it by expressing 0.9r as a sum and then subtracting off the terms and stuff...

 

[notfred]

Originally posted by: habib89
i got this off another site.. the math, at least to me, sounds alright... what do you guys say? i remember hearing something about this in math, but i guess i wasn't paying much attention.. thoughts???

x = .99999999 (repeating forever)
10x = 9.9999999 (again repeating forever)

Now, subtract these equations.
10x-x = 9x.
9.999 - 9.999 = 9

Now you have
9x = 9 <-- Basic Algebra now. Divide both sides by 9.

9 divided by 9 is 1. Therefore you get x = 1

In the equation it stated x = .9999999999 so, .9999999 = 1

so if 0.9999... = 1, then 10x-x = 8.9999..., right?

 

[silverpig]

Originally posted by: notfred

Originally posted by: habib89
i got this off another site.. the math, at least to me, sounds alright... what do you guys say? i remember hearing something about this in math, but i guess i wasn't paying much attention.. thoughts???

x = .99999999 (repeating forever)
10x = 9.9999999 (again repeating forever)

Now, subtract these equations.
10x-x = 9x.
9.999 - 9.999 = 9

Now you have
9x = 9 <-- Basic Algebra now. Divide both sides by 9.

9 divided by 9 is 1. Therefore you get x = 1

In the equation it stated x = .9999999999 so, .9999999 = 1

so if 0.9999... = 1, then 10x-x = 8.9999..., right?

Yep. If you multiply 9(0.9r) out, you will get 8.9r. This can be also broken down into a sum of 8 + 0.9r = 8 + 1 = 9.