“Proof” that zero is equal to one by infinitely subtracting numbers Recently, I came across a “proof” that $0=1$. Here is how it goes:

Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. Then, we get $1-x=0-x$. So, $1-x+x=0-x+x$. Hence, $1+0=0+0$ and so $1=0$.

I could not figure out what went wrong in this proof. The result is clearly not true but the proof seems to be true. I then asked a few people and they all could not figure out what went wrong. Can someone come please help me to identify what went wrong? Thank you.
 A: So called infinite sums in mathematics have formal definition as series and is based on concept of partial sum:
$$a_1+a_2+ \cdots =\sum\limits_{n=1}^{\infty}a_n=\lim\limits_{n \to \infty}\sum\limits_{i=1}^{n}a_i$$
where $S_n=\sum\limits_{i=1}^{n}a_i$ is partial sum.
Now let's come to your example: if you consider  $1-1-1-1-1-1-1-...$, then we should construct partial sum for it
$$\begin{array}{}
S_1=1 \\
S_2=1-1=0 \\
S_3=1-1-1=-1 \\
S_4=1-1-1-1=-2 \\
S_5 =1-1-1-1-1=-3 \\
\cdots \\
S_n=2-n \\
\cdots
\end{array}$$
As you see partial sum have no finite limit, which means, that expression $1-1-1-1-1-1-1-...$ is not finite number and cannot be used as such.
Funny example of such "proof" can be obtained if you consider expression $1-1+1-1+1-1+1-...$ and do not investigate convergence:
$$0=(1-1)+(1-1)+\cdots= 1+(-1+1)+(-1+1)+\cdots=1$$
A: When you write out an infinite series, you should first check if it converges. If not, then normal procedures like bracketing doesn't work anymore.
For example, here's a similar (false) proof that all integers are $0$:
Let $x = 1 + 1 + 1 +  \cdots $. For any integer $n > 0$, bracket the first $n$ terms so that $x = (1+1+\cdots+1) + 1 + 1 + 1+ \cdots = n + x$. Hence $n=0$.
A: 
Let $x=1−1−1−1−1−1−1-\cdots.$
Since $1−1=0$
$x=0-1-1-1-1-1-1-\cdots$.
Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get
--> $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. <--

Here is the error. Having a minus before the brackets negates everything inside.
So it actually becomes:
$$x = 0 - (1 + 1 + \cdots)$$
And I don't think it's a valid math operation to cross out $\infty$ on both sides, as $\infty$ is just a placeholder for very large number (not a concrete large number, so $\infty_{left} \ne \infty_{right}$).
A: Leaving questions of convergence aside, note that subtraction is not associative.
With just a three term expression of the same type:
$(1-1)-1=-1$, and
$1-(1-1)=1$
Did I just prove $1 = -1$???
