Yesterday at work we had a pretty heated debate over whether .999~ (repeating) is equal to 1. I can’t believe that in all of my years of math, I had never heard of this argument before. Basically, there are a couple “proofs” for this, which seem to make sense to me and verify the legitimacy of this statement. Somehow though, there is just something about it that doesn’t seem right. Here’s one proof.
.999~ = 1
So let’s set x = .999~:
x = .999~
Multiply both sides by 10:
10x = 9.999~
Subtract x from both sides:
10x – x = 9.999~ – x
9x = 9.999~ – x
Substitute .999~ in for x on the right side, which was our original declaration:
9x = 9.999~ – .999~
9x = 9
x = 1
But we said initially that x was equal to .999~!
I know that I have a lot of good math people that read my blog, so I’d be interested in hearing your opinion on this. This whole discussion started after someone came across this blog entry on the topic. There are tons of comments there arguing both sides. It just seems to me that .999~ will never quite get to 1, but these arguments for it really do make sense to me. I just still can’t believe that yesterday was the first time I ever heard this.
19 comments:
I learned something new today. Being an engineer, I of course decided to find proof and instead, I am now a convert in agreement that .99999... = 1! Here's something interesting I found that helps readjust my thinking:
When you think of 0.999... as being 'a little below 1', it's because
in your mind, you've stopped expanding it; that is, instead of
0.999999...
you're _really_ thinking of
0.999...999
which is not the same thing. You're absolutely right that 0.999...999
is a little below 1, but 0.999999... doesn't fall short of 1 _until_
you stop expanding it. But you never stop expanding it, so it never
falls short of 1.
Kahn, the proof of this was actually part of your math curriculum, so unless you were sick that day, you have heard it before.
There's lots of ways to look at it.
"one ninth" = .111~
"two ninths" = .222~
...
"nine ninths" = .999~
But... nine ninths = 1.
1 - .999~ = .000~1 ???? No, there is no such thing as .000~1 (you can't put a 1 after an infinite number of 0's!)... it's just .000~, more commonly known as 0. Thus .999~ = 1.
I should note that whenever I get in this argument, it does become heated because I'm a complete prick about it. People are basically convinced by my argument, but want to settle on some sort of, "we're both sort of right" kind of thing. This is, of course, wrong, and if you're still under this impression then the argument isn't over.
It's kinda like the probability that a realization of a continuous, normal distribution is 0 is exactly 0. Not really, really, small, but 0.
Dan, what year did we cover this? Was this in Algebra II with Norm? That year was the fuzziest for me. I never really nailed algebra until I used it all the time with calculus.
I can't quite pull from memory who the teacher was. I originally wanted to gravitate to Mrs. Doughty, but I'm not sure that's right.
It may be mathmatically correct, but in my opinion, never socially accepted. I'm no math wiz, but isn't this similar to an asymptote? Doesn't it infinatly not reach 1? In my opinion .999~ = 1 only at infinity, which is never reached.
However, the proof is in the pudding...
if x = 0.9999~
10x = 9.9999~
10x - x = 9.9999~ - 0.9999~
9x = 9
x = 1.
I think this proof is easier to follow...
Hmmm, I can't edit my post. Please ignore the proof that was stuck at the end. I ment to end with the punny joke...
I had this argument a few months ago here at work. Actually, it wasn't an argument so much as my coworker's refusal to accept my proof of it, which was really irritating since he was the one who initially asked me to prove it.
I can't remember which year we learned it either, but am also leaning towards it having come from Doughty, so probably 11th grade then? I remember it being explained as Dan said, with the 1/9 = .11111~ x 9 = 9/9 = 1 logic.
Way to go Mathletes!
But I go with the logic that .999~ never quite reaches 1 infinitely, thus is not equal to 1.
If someone offered me $1 or $.999~, I would take the $1 in one hand, and punch the math dork with the other. O'Doyle Rules!!
So Dan... "we're both sort of right." Heyyyy-ooohhh.
I may be a math dork, but I get paid to be a math dork.
It is an asymptote, if you were thinking of a function where the x axis is "number of nines after the decimal point." But, the repeating symbol actually stands for infinity -- you've reached the value the asymptote never reaches. 1.
Now, let's talk about something less dorky. Hey, I got the Firefly DVD's for Father's Day...
You aren't helping yourself Dan. But I guess a hardcore fan would have gotten the DVDs when they first came out. Next time you are in town, I have a couple of guys to introduce you to. They will love talking to you about Firefly and why FOX was dumb to cancel the show.
Your proof is not complete. You substituted .999~ for x on only one side of the equation and that is not legal in math. You have to substitue it for both sides since thats how you started the proof. So..
9*.999~=9.999~ - .999~
8.999~1 = 9
divide both sides by nine and you get.
.999~=1
Which is your orginal equation, which we know is wrong. You can't start with something that is known to be false and try to use math to prove otherwise.
Drew, I thought of that yesterday too, but I believe that 9* 0.999~ is actually 8.999~. If you carry that out to infinity, it never has that "1" tacked on the end. Thus 8.999~ = 9, which is the same as saying 0.999~ = 1.
You have to have a ...9991 as 9 times 9 is 81. The 8 and the 1 on the ends will always be there no matter how far you carry it out. That is the only way it works. try it out on your calculator. :)
It seems like we need to have a little talk about infinity. Infinity isn't just a really big number -- it's little girls being sucked into TV's and babies being born in outer space. It's a stairway to heaven that's not for sale and a gnu doing newscasts. It's a journey to the center of the mind that can't be had even on Paula Abdul's best junk. Making a bunch of 9's into a 1 is a trifle for a sideways 8. Did I just blow your mind? That's infinity.
I think I caught references to Poltergeist and The Great Space Coaster in the last post. Get on board!
Well here's another proof that might be a little easier to swallow.
1/3 = .333~
1/3 + 1/3 + 1/3 = .333~ + .333~ + .333~
3/3 = .999~
1 = .999~
To me, this one makes the most sense.
Now I'm confused - where did the 81 come from?
9x = 9
divide both sides by 9 to get x. The whole point is that x = 1 or x = 0.999~ is the same exact thing for the equation above.. read my original (and first) comment to see why you will never have the 81 because that assumes a finite end to the number.
I won't be able to appease the math dorks, but this equation just proves that we have a fundamentally flawed number system.
This relies on mathematical operations against an infinitely repeating number, when mathematical operations against infinity concepts don't hold up.
For instance, say ∞ remains ∞ whether you add or subtract a number from it (∞ + 1 isn't somehow more than ∞). So take:
∞ = x + 1
Now, let's try to prove x is infinity:
x = x + 1 would mean x = 0, and since 0 = 1 is not true, x cannot be infinity.
But, let's try to prove x is not infinity; then:
∞ = x + 1
∞ - 1 = x + 1 - 1
which would mean that ∞ = x, which again is a contradiction.
Some might also argue .999~ = 1 because there is no number in between them (because any different real numbers will have another number between them), but I think this is again an artifical limitation of an infinite (unlimited) number line.
I'm not going to argue strongly either way though, unless 1 = .9999~ can somehow save me money at the gas pump.
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