We all know that $\frac{1}{3} = 0.\overline{3}$, and that when we multiply it by $3$, we get $1$. But, $3\times \frac{1}{3} = 0.\overline{9}$.   Are there any ideas on how to prove that $0.\overline{9}= 1$?  I have no ideas, I only wrote down that $0.\overline{9}+x= 1$ for $x\neq 0$, but I don't know how to proceed.
I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?
Max dimension of a subspace of singular $n\times n$ matrices I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.  The following question was given in a contest I took part. I had an approach but it didn't solve the problem.     Consider $V$ a linear subspace of the real vector space $\mathcal{M}_n(\Bbb{R})$ ($n\times n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?   A quick guess would be $n^2-n$ since if we consider $W$ the set of $n\times n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.  Now the only thing there is to prove is that if $V$ is a subspace of $\mathcal{M}_n(\Bbb{R})$ of dimension $k &gt; n^2-n$ then $V$ contains a non-singular matrix. 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