Does 0.9999 repeating 9 equal 1?

[757SeanTaylor21]

Startin from page one i stopped at psge three; brain overload, brain mAlfunction, brain shutdown

[Mark The Homer]

There was already a thread on this a couple years ago.

The correct answer is yes.

---------- Post added April-9th-2011 at 02:13 PM ----------

There was already a thread on this a couple years ago.

The correct answer is yes.

Edit: Wait - this is the other thread. Nevermind.

[Pwyl]

That was a .99999999999999% weird bump.

so it was just under 1% weird? That's not very weird.

[Woofer]

Can 15 minutes really save me 15% or more on my car insurance?

Albert Einstein once made the comment:

"As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality."

and

"Everything that can be counted does not necessarily count; everything that counts cannot necessarily be counted."

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I'm sensing infinity in there somewhere. If only there were some way to quantify it.

[GhostofSparta]

[SonOfWashington]

^^^ Maybe if that proved that they weren't equal, I'd question where my god is.

[GhostofSparta]

^^^ Maybe if that proved that they weren't equal, I'd question where my god is.

Well then clearly this was not directed at you. Perhaps more at those like this...

Oh my gawd 1 is 1 not .999999999999999999999999

[polywog999]

A "repeating" 9.would be indicative of infinity, as any fraction of infinity is infinity.

[Hubbs]

Ghost, you just made my head explode.

[Mickalino]

And to an engineer, "close enough" = "yes".

Then you must define "close enough" in mathematical terms

[Hitman21ST]

[PokerPacker]

I've decided that .999999999999999... = 1- (the minus should be a superscript). Meaning that it is equal to the left-boundary of 1.

[DjTj]

2=1

---------- Post added April-11th-2011 at 02:04 PM ----------

I've decided that .999999999999999... = 1- (the minus should be a superscript). Meaning that it is equal to the left-boundary of 1.

Which means that it approaches 1 as the number of 9s reaches infinity. Therefore, 0.9999... = 1.

---------- Post added April-11th-2011 at 02:13 PM ----------

ooh, ooh, I have a question!

lets say, hypothetically, that a hockey goalie has 9 save on 10 shots. his save percentage is .900. Now, lets say that his number of shots increases to infinity and he saves every single one of those shots. what is his save percentage?

His save percentage is (9 + x) / (10 + x), where x is the number of shots after the first ten. The limit as x approaches infinity is 1, so his save percentage approaches 1 as the number of shots goes to infinity.

[RiggosMohawk]

If 0.9 repeating equals 1, then it satisfies multiplicative identity.

∞ x 1 = ∞

∞ x (0.9 repeating) != ∞

Therefore 0.9 repeating does not equal 1.

[PokerPacker]

Which means that it approaches 1 as the number of 9s reaches infinity. Therefore, 0.9999... = 1.

Except that while for most practical purposes 1- equals 1, it is not for all scenarios. If dealing with periodic impulse functions, you'll have a period of, say, T. If you integrate from 0 to T where there is a unit impulse on both 0 and T, only one of them will be in the integral (well, depending on how you deal with impulse functions. you COULD make it exist half before and half after as opposed to just before or just after, but then you'd count half of each of those impulse functions so it makes no difference)

The unit impulse itself has infinite height, and infinitely little width, but with an area of 1. convergence would dictate that it has a width of 0, and 0 times anything is 0 therefore it has 0 area, but that is not the case. it has an infinitely small width (1- .99999999999999999999...) and an infinite height such that area of height times width = 1.

[Thinking Skins]

If 0.9 repeating equals 1, then it satisfies multiplicative identity.

∞ x 1 = ∞

∞ x (0.9 repeating) != ∞

Therefore 0.9 repeating does not equal 1.

really? I thought infinity times any number > 0 was infinity, and infinity times any number less than 0 was negative infinity. So infinity times 1 = infinity, and infinity times .9repeating is infinity as well. But that doesn't prove anything because infinity times 2 is also infinity.

---------- Post added July-8th-2012 at 10:02 PM ----------

http://mathtricks.org/wp-content/uploads/2010/01/1-2-proof.jpg

if A = B, then you can't go from line 4 to line 5 because A - B = 0, and since going from line 4 to line 5 implies dividing by A - B, it implies dividing by 0, which is not allowed.

[Hitman21ST]

if A = B, then you can't go from line 4 to line 5 because A - B = 0, and since going from line 4 to line 5 implies dividing by A - B, it implies dividing by 0, which is not allowed.

I was wondering when someone would catch that

[greenspandan]

I was wondering when someone would catch that

only took a year and a half!

ah, back when i could make new threads. the good ol' days.

[SonOfWashington]

If 0.9 repeating equals 1, then it satisfies multiplicative identity.

∞ x 1 = ∞

∞ x (0.9 repeating) != ∞

Therefore 0.9 repeating does not equal 1.

What the actual ****.

Infinity isn't a number, it's a concept. You can't just multiply it like that.

[greenspandan]

Infinity isn't a number, it's a concept. You can't just multiply it like that.

That used to be true, but we've got some pretty powerful computers now. with Signetics' advances in write-only memory ( http://en.wikipedia.org/wiki/Write-only_memory ) and Chrysler corp's pioneering research on the turbencabulator (

), multiplying by infinity is no longer the stuff of science fiction.

[balki1867]

If you subtract 1 - 0.9999999, the answer is 0.0000000000000000000 with the zeros repeating forever. Therefore IMO 0.9999999999 = 1

[mistertim]

What the actual ****.

Infinity isn't a number, it's a concept. You can't just multiply it like that.

Depends on what you're doing. If it is just pure math then sure you can, why not; there are plenty of "concepts" that don't really make sense in the real world that are incorporated in advanced mathematics.

If its for physics then when you end up with infinities as you solve an equation that usually means something is borken. That's one of the first ways they found out that relativity and quantum mechanics just wouldn't work together without some other overarching theory: when you try to use the equations together you end up with a bunch of nonsensical infinities, even though when done individually the equations were know to be highly accurate and consistent with test results.

[SonOfWashington]

Depends on what you're doing. If it is just pure math then sure you can, why not; there are plenty of "concepts" that don't really make sense in the real world that are incorporated in advanced mathematics.

If its for physics then when you end up with infinities as you solve an equation that usually means something is borken. That's one of the first ways they found out that relativity and quantum mechanics just wouldn't work together without some other overarching theory: when you try to use the equations together you end up with a bunch of nonsensical infinities, even though when done individually the equations were know to be highly accurate and consistent with test results.

I agree, but in the post I quoted it seems way out of context. Besides, it's already been proven in this thread that .999... = 1.

[nonniey]

If you go to halfway closer to something and stop, then then go halfway closer again and stop, then go halfway again and stop, etc. etc., do you ever get there?

No but when it comes to certian things, e.g. sexual intercourse, you get close enough.

[Thinking Skins]

I just wanted to do this proof because I people were saying that .9 repeating was the closest number to 1 that is less than 1.

Suppose (as some say) that (.9 repeating) is the closest number to 1 that is less than 1. Lets say x = .9 repeating.

Then x is clearly a real number, and thus it has a point on the number line. And obviously 1 is a real number and has a point on the number line. And because (we're assuming) these two numbers are distinct/different, there is a gap between these two numbers, we have that x < 1. Then we know that (x + 1) / 2 is a number that is different than both x and 1.

Suppose (x + 1) / 2 <= x

Then (x + 1) <= 2x.

Then 0 <= 2x - x - 1

Then 0 <= x - 1, which implies that x >= 1, which says that (.9 repeating is greater than or equal to 1), which we know is false.

So (x + 1) / 2 > x.

Now suppose (x + 1) / 2 >= 1

Then x + 1 >= 2

Then x >= 1, which we again know to be false.

So (x + 1) / 2 < 1.

So we have that x < (x + 1) / 2 < 1. But this contradicts our assumption that x, or .9 repeating, is the closest real number to 1 that is less than one, because we just found another number that is also less than one, but is closer to 1. This means that our assumption of such a number is wrong (since our steps since then are all logical step). So we must reject this hypothesis and conclude that the two numbers are the same.

So ((.9 repeating) + 1) / 2 is between (.9 repeating) and 1.