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A friend of mine showed me this proof to demonstrate that any algebra variable does not equal any number.

his proof relies on this idea $$0\neq1$$ $$0x\neq1x$$ $$0\neq1x$$ $$0\neq x$$ Full Proof: $$\forall n (n \in \Re)$$ $$0\neq1\Rightarrow0(x-n)\neq1(x-n)$$ $$0(x-n)\neq1(x-n)\Rightarrow0\neq x-n$$ $$0\neq x-n\Rightarrow n\neq x$$

I feel like it is a divide by zero fallacy but I'm yet to pinpoint the exact error

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    $\begingroup$ $0\neq 1$ does not imply that $0x\neq 1x$. Just let $x=0$ to see a counter example, $\endgroup$ – lulu Mar 30 '18 at 22:06
  • $\begingroup$ MultiplyIng both sides with 0 you turn the inequality into equality. $\endgroup$ – user Mar 30 '18 at 22:07
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The false implication $x \neq y \to ax \neq ay$ is used. This is true iff $a$ is non-zero (in a field), because then the equivalent (by contraposition) $ax = ay \to x=y$ can be shown by division by $a$. So indeed it's a division by $0$ error implicitly.

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The first line in your extended (non)proof is wrong when $x=n$. So it's not true for all $x$ and $n$.

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You can't multiply inequations like this: $a\neq b$ does not imply $ac\neq bc$. Indeed, if $c=0$, then $ac=0=bc$ is always true, even if $a\neq b$.

(If you know that $c\neq 0$, then this step would be valid, since if $ac$ were equal to $bc$ you could divide both by $c$ to get $a=b$ which contradicts the fact that $a\neq b$. So, in a sense, this is indeed a division by $0$ error.)

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When we multiply for $(x-n)$ we need to set $x\neq n$ that is precisely the solution we obtain.

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But with x=0 ,0x=1x (=0) so line 2 is a fallacy (false)

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