Yes, 0.9 recurring (0.9...) is equivalent to 1. It's related to the idea that every number has a non-truncating decimal equivalent (i.e. a decimal which doesn't end in heaps of zeros). Here's what I remember from an informal 0.9... = 1 proof I saw a few years back:
Consider an infinite set S of numbers {0.9, 0.99, 0.999, ...}. Each element of S has a finite number of 9s and is marginally smaller than 1. 0.9... doesn't belong to S as it has an infinite number of 9s (and hence is bigger than every element of S).
Now imagine a number which is just smaller than 1 - let's call it 1-ϵ, where ϵ is an infinitesimally tiny number. Since S is an infinite set, there inevitably exists a number in S which is bigger than 1-ϵ. Hence 0.9... is also bigger than 1-ϵ. This leads us to the corollary that 0.9... is larger than every number smaller than 1. Now obviously every number bigger than 1 is also bigger than 0.9...
So if every number smaller than 1 is smaller than 0.9... and every number bigger than 1 is bigger than 0.9..., then 0.9... must equal 1.
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