Can this really happen?

Question

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.

Answer 1

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.