Why is 0.99999… = 1?
A cool thing I learned a long time ago is that 0.99999… = 1. And I mean exactly equal. Not just “very close to”. EXACTLY equal.
Behold the proof!
Let a = 0.111111…
Then 10a = 1.11111…
10a – a = 1.11111… – 0.11111…
9a = 1.But 9 times 0.11111… is 0.99999….
So 0.999999…. is exactly equal to 1!
This has some fun consequences. Like… decimal expansions are not unique! In other words, some numbers have more than one possible way of being written as a decimal.
Of course, whenever you hear a fancy proof like this, you have to be very careful. Like I showed in Why convergence matters, you can sometimes use seemingly impeccable logic to show complete nonsense.
For more great examples of simple proofs of nonsense with VERY subtle errors, check out the website Find the error.
March 17th, 2009 at 3:22 pm
Let’s assume a has k ones after the decimal, 10a will then have k-1 ones. There’ll be an error in the k:th decimal, you’re back to 0.9999… To save the proof you has to accept that infinity minus one is exactly equal to infinity.
Infinity is such a weird thing ~~
Btw Vlad, remember your old “Why I love mathematics[...]” post? Well, recently we were discussing riddles on a Starcraft forums I visit often, so I threw your ball weighting riddle at them. Quite surprisingly for me, one guy came up with an answer that was WAY different from my own. So I got a little curious, did you go by a process of elimination for your solution? Or some kind of system to give the balls identity?
March 17th, 2009 at 5:06 pm
@Shadowart:
You’re right, infinity IS such a weird thing
Mathematicians have been afraid of tackling infinity for centuries, and then one guy came along and started playing with it, and showed that:
- there are exactly as many even numbers as there are whole numbers (even though logically there should only be half as many)
- there are exactly as many rational numbers (fractions) as there are whole numbers!
- there are MORE real numbers than rational numbers
No wonder everybody gets freaked out by infinity :p
And as for the ball weighing riddle – I’m not sure what you mean by your question. If by elimination you mean trying every single possibility until I came across the right one then no, I didn’t use elimination. If you mean saying “ok, let’s say I put this many balls on each scale. What do I do if one scale is heavier than the other? Mmm ok. And what do I do if they’re the same?”… then yes, I did use elimination.
Did that answer your question at all?
March 17th, 2009 at 6:12 pm
That’s what I call elimination, I did the same.
There’s another way to do it where you assign a combination of placements on the weighting scale to each ball and determinate the odd one and it’s nature from these. That method can also be generalised into answering the question “Given n balls and k weightings, when is the answer to the riddle ‘yes’?”