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Yesterday, while scribbling on the back page of my maths copy, I accidentally came across this

If $ x\in R\ $ and $ x+x+x+.....\infty = m $ where $ m $ is any real number.

Then we can write

$ x+x+x+.....\infty = m $

$\Rightarrow $ $ x+m=m $

$ \Rightarrow $ $ x=0 $

In the same way we can prove that $ 6=0 $ or $ 7=0 $

But how is this really possible? Please explain. I am in 10th standard.

Jyrki Lahtonen
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Aniruddha
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    Apparently, $x+x+\cdots$ is not a real number unless $x=0$. – David Mitra Sep 16 '17 at 06:50
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    Yes, anyone can scribble on the back of their maths book! But this shows the danger of treating divergent series as if they were convergent. – Angina Seng Sep 16 '17 at 06:50
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    The short answer is the sequence does not converge so it is impossible to have infinite number of $x$s summing up to m. For infinitely small $x$ you need to use integral. – cr001 Sep 16 '17 at 06:52
  • But you won't get any conclusion for $m$, you just get $0+m=m$, how can you get $6=0$? – john doe Sep 16 '17 at 06:54
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    There are many issues in the argument. 1) is $x + x + \cdots$ well defined? 2) if it is well defined, is it a real number? 3) in a infinite sum like this, is $x + (x + x + \cdots ) = x + x + \cdots$ a legal operation? 4) When all this make sense, you find the only possible choice is $x = 0$. So the conclusion $x = 0$ is correct. However, it doesn't mean when you substitute $x$ by a random real number, say $6$, you still get $6 = 0$. – achille hui Sep 16 '17 at 06:59
  • As per as my limited understanding of infinity, $x+x+..\infty=m\Rightarrow x+m=m$ because infinity + any number = infinity; but that does not prove $x=0$ – Vikram Sep 16 '17 at 07:10
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    @Vikram Infinity is not a number, so you cannot do arithmetic with it. Some people pretend that they do, but in that case it's usually something else they're really doing, like looking at the sizes of sets and their (disjoint) unions and (direct) products. – Arthur Sep 16 '17 at 09:20

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What you have written down is a valid argument in the sense that all the conclusions are inescapable consequences of assuming that the premise is true. However, that is not to say that your premise is true, since in this case it is actually false.

To be concrete, your premise $P$ is the following:

$P(x,m)$: If $x$ and $m$ are any two real numbers, the equality $$ x+x+x+\dots = \sum_{i=1}^\infty x = m \tag{$\ast$} $$ holds.

You now want to prove $Q$:

$Q$: If $x$ is any real number, then $x = 0$.

Well if we assume that $P$ is true, then we can prove $Q$ as follows:

Proof of $Q$: Let $x$ be an arbitrary real number. By $P(x,0)$, we have $$ \lim_{N\to\infty} Nx = \sum_{i=1}^\infty x = 0. $$ However, this implies that $Nx = 0$ for every $N$, and since $\Bbb R$ has no zero-divisors and $N\ne 0$, it follows that $x = 0$, as desired.

Therefore, if $P$ holds then all sorts of identities hold, such as $6 = 0$ and $7 = 0$. It's a straightforward exercise in analysis to show that if $m$ and $x$ are real numbers such that $(\ast)$ holds, then $x = m = 0$, so that $P$ doesn't actually hold.

Alex Ortiz
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