Comments on: Darn Tricky Math http://smidg.in/2007/11/21/darn-tricky-math/ The Smidgin: A Sporadic Blogger Sun, 05 Feb 2012 06:39:23 +0000 hourly 1 http://wordpress.com/ By: Steve http://smidg.in/2007/11/21/darn-tricky-math/#comment-3733 Mon, 26 Nov 2007 17:45:23 +0000 http://smidg.in/2007/11/21/85/#comment-3733 On second thought, I believe our Prof’s story was that while there is a complicated Calculus-level proof for 0.(9) = 1, there is also the simple algebraic proof. Thus, the point was: don’t make something harder than you have to.

]]> By: Steve http://smidg.in/2007/11/21/darn-tricky-math/#comment-3732 Mon, 26 Nov 2007 17:43:34 +0000 http://smidg.in/2007/11/21/85/#comment-3732 I believe I learned about this mathematical conundrum in Pre-Calc class in high school. We were talking about limits and series in the class, and I think the example was

Lim (terms -> inifinity) of 0.9 + 0.09 + 0.009 … = 1

Or something like that. (I don’t know how to do the fancy overbars and junk, so bear with me.)

Anyway, that’s the only place limits came into the story at all. Our prof. went on to show us that using simple algebra, you can prove that 0.(9) = 1

x = 0.(9) (parentheses representing the overbar)
10x = 9.(9) (multiplying by 10 just moves the decimal point)
Now, subtract x from both sides.
-x … -0.(9)
9x = 9
Now, divide by 9
x = 1

And yes, effectively 0.(9) is just an imperfect decimal representation of 9/9, which is of course 1.

]]> By: edgarsr http://smidg.in/2007/11/21/darn-tricky-math/#comment-3734 Thu, 22 Nov 2007 16:13:03 +0000 http://smidg.in/2007/11/21/85/#comment-3734 Actually, you’re right about this last equation as well:
1 = 0.(9) + 0.(0)1
The thing is:
0.(0)1=0
And so everything goes into the right places again..

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