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.

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    $\begingroup$ $1-1-1-1-1-1-1-...$ is not a real number; and in extended real numbers $a+\infty=b+\infty=\infty$ does not imply $a=b$ $\endgroup$ – J. W. Tanner Nov 24 '20 at 1:47
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    $\begingroup$ With infinite sums, you can’t use the associative property like that, nor can you switch terms around. Just for the record, I believe there is a theorem stating that an infinite series can converge to pretty much any value if you switch the order of addition. $\endgroup$ – Gauss Nov 24 '20 at 1:48
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    $\begingroup$ @Gauss: cf. Riemann series theorem $\endgroup$ – J. W. Tanner Nov 24 '20 at 1:51
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    $\begingroup$ I'm not following the step where you bracket things. If you expand the brackets aren't you getting -1+1+1+1+1 which is not what you had before? I think possibly you meant 1- (1+1+1+1...)? $\endgroup$ – Chris Nov 24 '20 at 10:30
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    $\begingroup$ It is impressive that there are in fact three logical errors in such a short proof, when just one would suffice to reach some absurd conclusion like $1=0$. $\endgroup$ – Servaes Nov 24 '20 at 10:41

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: $$0=(1-1)+(1-1)+\cdots= 1+(-1+1)+(-1+1)+\cdots=1$$


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 $n=0$.


Let $x=1−1−1−1−1−1−1-\cdots.$

Since $1−1=0$


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 $\infty_{left} \ne \infty_{right}$).


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


Did I just prove $1 = -1$???

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    $\begingroup$ It is a well-known fact that $0=2$, so yes, ${-}1=1$. $\endgroup$ – Asaf Karagila Nov 24 '20 at 10:57

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