1 = .99999999...?

IB Maniac · #1 03-06-2002, 10:54 PM

Does 1 really equal .9999....???

Let's say:

x = .99999...

Multiply by ten and you get:

10x = 9.99999...

Subtract one 'x' and you get:

9x = 9.0

Divide by 9 and you get:

x = 1

:)

Cobra · #2 03-06-2002, 11:02 PM

What are you talking about? You can't just add and multiply and subtract random numbers to another number to say that two numbers are equal. Whatever you do to one side of the equation you have to do to the other. I'll use the "=" symbol even though it's not correct.

Therefore
.99999 = 1

Multiply by 10 and you get:
9.99999 = 10

Subtract one 'x' and you get:
9 = 9.00001

Divide by 9 and you get:
1=1.00001

So no, .99999.... does not equal 1. It equals .99999

What you had was (10x) - (1x) / 9, or 9x/9, which equals x, not 1.

IB Maniac · #3 03-06-2002, 11:06 PM

Uh...I'm sorry, that is not correct. It's not 9x/9....to get x by itself in the equation 9x = 9, you divide both sides by the nine on the left side, so it would be:

x = 9/9

Therefore, x = 1

IB Maniac · #4 03-06-2002, 11:09 PM

"Multiply by 10 and you get:
9.99999 = 10

Subtract one 'x' and you get:
9 = 9.00001"

How in the world did you get this? If x = .99999...subtract that out of 9.99999...., you do not get 9.000000......1, you simply get 9.0

Aaron · #5 03-06-2002, 11:18 PM

My geometry teacher actually showed this to us. He spent about 10 minutes explaining it. According to him, it is true. I personally don't know whether to believe it or not.

CrystalKat · #6 03-07-2002, 07:41 AM

I remember a similar exercise years and years ago -- go over whatever arithmetic they showed you VERY carefully! As I recall, our teacher used it to see if we could catch the fact that at one point (out of a couple dozen steps) he'd divided by 0 -- of course, mathematically impossible, making the whole thing a nice gag.

1of4 · #7 03-07-2002, 08:13 AM

Oh brother. :rolleyes:

Only on a rollercoaster website will I find a discussion about algebra. :p

PhantomRevnge67 · #8 03-07-2002, 10:56 AM

you guys got me lost after the second step! but how could 1 equal that? i dont see any ones in .999999!

Skye · #9 03-07-2002, 01:40 PM

i've seen similar tricks - all include dividing by 0, which as melissa said is impossible.

RickyD · #10 03-07-2002, 02:02 PM

[QUOTE][i]Originally posted by Skye [/i]
[B]i've seen similar tricks - all include dividing by 0, which as melissa said is impossible. [/B][/QUOTE]

The only trick I've seen with dividing by zero is where you can "prove" that 1=2 (I can post that one, if you like. It's really kind-of funny that you can twist math that way.)

However, as far as I know, there are no "tricks" in the proof that 0.9999999 = 1, just some basic algebra. In fact, you can use the EXACT SAME STEPS to figure out things like .333333333 = 1/3, or .142857142857142857 = 1/7, or even .8181818181 = 9/11 (as long as it's a repeating decimal)

eg.

x = .81818181818181...

Multiply times 100 in this case (since it's 2 digits that are repeating)
100x = 81.818181818181....

If x really does equal .818181818181..., you can subtract x from one side and .8181818181..... from the other (since it's the same thing) and you get

99x = 81

Then you divide both sides by 99 and get x = 81/99 or x=9/11

(Of course, you could save all that trouble and just punch in 9/11 on the calculator for that example ...)

Make any sense?

Cobra · #11 03-07-2002, 03:03 PM

[QUOTE][i]Originally posted by IB Maniac [/i]
[B]Uh...I'm sorry, that is not correct. It's not 9x/9....to get x by itself in the equation 9x = 9, you divide both sides by the nine on the left side, so it would be:

x = 9/9

Therefore, x = 1 [/B][/QUOTE]

How did you get 9x = 9? Nevermind, that doesn't even matter, because you already defined x as .99999. You aren't solving for x, you already have it and are trying to prove that it equals 1. Therefore 9(.99999) = 9, and you have 8.99991 = 9. It never works, you can't prove that .99999 = 1 because it doesn't. What you had was 9x/9, which cancels out to 1x/1, which is x.

The reason you have 9x/9 is:
First you multiplied x by 10 (you get 10x). Then you subtracted 1x (9x). Then you divided that by 9. 9x/9 = 1.

[quote]How in the world did you get this? If x = .99999...subtract that out of 9.99999...., you do not get 9.000000......1, you simply get 9.0[/quote]

Because whatever you do to one side of the quation you have to do to the other. If you were trying to prove that .99999 = 1, you would need to do to the right side of the equation whatever you are doing to the left side, and therefore the two can never equal each other.

Again, you are saying that x= .99999, and we discovered that 9x/9 = 1. If we were solving for x, then x would indeed be 1, but you already defined x as .99999. Therefore 9(.99999)/9 = 1, turns out to be .99999 = 1, which is not true.

You cannot just add numbers to a number to prove that it equals another number. If that were the case, you could do this:

3 x 10 = 30
30-5= 25
25/5 = 5, and therefore 3 = 5. That is exactly what you did, except you used a variable (x).

[quote]However, as far as I know, there are no "tricks" in the proof that 0.9999999 = 1, just some basic algebra. In fact, you can use the EXACT SAME STEPS to figure out things like .333333333 = 1/3, or .142857142857142857 = 1/7, or even .8181818181 = 9/11 (as long as it's a repeating decimal[/quote]

The difference between .999999 = 1 and .33333333 = 1/3 is that .3333333 really does equal 1/3 while .99999 does not equal 1. Your proof of .81818181 = 9/11 is correct because what you did to one side of the equation you did to the other side. IB Maniac did not do that in his "proof."

IB Maniac · #12 03-07-2002, 03:56 PM

"How did you get 9x = 9?"

If x = .999... then you should be able to subtract one of the x's from the left side, and .999... from the right:

10x = 9.999.....

Subract x from the left and .999... from the right (because x IS EQUAL TO .999...) so you get 9x = 9

Now you have the following:

9x = 9

ONCE AGAIN I WILL SAY, YOU DO NOT DIVIDE THE 9X BY 9, YOU DIVIDE THE 9 ON THE RIGHT SIDE BY THE 9 ON THE LEFT SIDE TO GET THE X BY ITSELF:

9x = 9
x = 9/9

If you do what you said, and divide the other way to get 9x/9, one side of the equation will equal 0, which you do not want to do.

IB Maniac · #13 03-07-2002, 04:02 PM

"3 x 10 = 30
30-5= 25
25/5 = 5, and therefore 3 = 5. That is exactly what you did, except you used a variable (x)."

How in the world did you get this? 25/5 IS NOT 3, it is five, therefore both sides of the equation are equal. If you do know that both sides are equal, you can do whatever you want (with certain limits) to both sides of the equation Ex:

6 = 2 x 3

You can add a four, if you do it to both sides, and if you know that both sides of the equation are equal:

6 + 4 = (2 x 3) + 4
10 = 10

You can also subtract from each side, divide, multiply, square, cube, etc....as long as you do it to BOTH sides, like I did.

Cobra · #14 03-07-2002, 04:09 PM

Now I see what you are saying, but you still can't do that.

What you are doing is:
x=.99999
10x = 9.9999
9x=9

Unfortunately, since you know the value of x, it doesn't come out to x=9/9 and then x=1.

x=.99999, therefore
10(.99999) = 9.9999 (this is true)
9(.99999) = 8.99991 (NOT 1!)

Face the fact that .99999 = .99999 and 1 = 1 but .99999 does NOT = 1.

[quote]How in the world did you get this? 25/5 IS NOT 3, it is five, therefore both sides of the equation are equal. If you do know that both sides are equal, you can do whatever you want (with certain limits) to both sides of the equation Ex: [/quote]

Obviously that example went way over your head, so I'm not even going to bother explaining it.

phantompenguin · #15 03-07-2002, 05:49 PM

*stares blankly at screen*

:confused: