Please can someone tell accurately and precisely the actual fallacies of this fake proof that 0.999... ≠ 1.000...?

Question

Please can someone tell accurately and precisely the actual fallacies of this fake proof that 0.999... ≠ 1.000... ?

Define $F: \{\text{decimals in } [0,1]\} \to \{\text{decimals in }[0,1]\}$ by
$a.bcdefg... \to a.0b0c0d0e0f0g..$.

Define $G: \{\text{decimals in } [0,1]\} \to \{\text{decimals in }[0,1]\}$ by $h.ijklmn... \to h.i0j0k0l0m0n0...$.

then

$$F(0.99999...) + G(1.00000...) = 1.0909090909... $$ $$F(0.99999...) + G(0.99999...) = 0.9999999999... $$ $$F(1.00000...) + G(1.00000...) = 2.0000000000... $$

If it were true that $ 0.999... = 1.000...$, then $ F(0.999...) + G(1.000...) = F(0.999...) + G(0.999...) = F(1.000...) + G(1.000...) = 1.090909... = 0.999999... = 2.000000...$, which is absurd.

Hence $0.999... ≠ 1.000...$ QED

Answer 1

Let $S$ denote the set of all sequences $(a_n)_{n \in \mathbb Z}$ with $a_n \in \{0,1,2,3,4,5,6,7,8,9\}$ for all $n$ and $a_n = 0$ for all but finitely many positive indices $n$. You may of course formally write $$(a_n) = \ldots a_{4} a_{3} a_{2} a_{1}a_0.a_{-1} a_{-2} a_{-3} a_{-4} \ldots $$ with "decimal point" after $a_0$, i.e. in the form of the decimal represention of a real number. However, you should be aware that this notational convention does not mean that $(a_n)$ is a real number. It is a sequence of digits which has to be distinguished from the real number $$\sigma((a_n)) = \sum_{n\in \mathbb Z} a_n 10^n .$$ The assocation $(a_n) \mapsto \sigma((a_n))$ gives us a function $$\sigma : S \to \mathbb R $$ which is clearly surjective. It is not injective, but let us assume for the moment that we do not know this fact. The fake proof in your question claims that $$\sigma(\ldots 0000.9999 \ldots) \ne \sigma(\ldots 0001.0000 \ldots) .$$ Let us analyze the "proof" and try to understand the meaning of $$F(0.99999...) + G(1.00000...) = 1.0909090909... \tag{1}$$ $$F(0.99999...) + G(0.99999...) = 0.9999999999... \tag{2}$$ $$F(1.00000...) + G(1.00000...) = 2.0000000000... \tag{3}$$ These equations involve the two functions $F, G : S \to S$ defined in your question.

The fake argument is a sleight of hand confusing the reader by sloppy notation which does not distiguish between the sequence $(a_n)$ and the real number $\sigma((a_n))$. Technically $(1) - (3)$ are equations in $S$. The problem here is that it is not obvious how to form the sum of arbitrary sequences in $S$. However, in some cases we can do it componentwise, namely if the sum of any two digits having the same index is $\le 9$. This interpretation is possible in $(1) - (3)$. But sloppily $(1) - (3)$ can also be regarded as equations in $\mathbb R$ which should more precisely be written as $$\sigma(F(\ldots 0000.9999 \ldots)) + \sigma(G(\ldots 0001.0000 \ldots)) = \sigma(\ldots 0001.0909 \ldots) \tag{1'}$$ $$\sigma(F(\ldots 0000.9999 \ldots)) + \sigma(G(\ldots 0000.9999 \ldots)) = \sigma(\ldots 0000.9999 \ldots) \tag{2'}$$ $$\sigma(F(\ldots 0001.0000 \ldots)) + \sigma(G(\ldots 0001.0000 \ldots)) = \sigma(\ldots 0002.0000 \ldots) \tag{3'}.$$ So far everything is completely okay. But now also the arguments of $F$ and $G$ are regarded as real numbers and that is the fundamental mistake. The fake proof is based on the "conclusion"

If it were true that $ 0.9999... = 1.0000...$ [an equation of real numbers meaning that $\sigma(0.9999 \ldots) = \sigma(1.0000 \ldots)$], then $F( 0.999... ) = F(1.000...)$ and $G( 0.999... ) = G(1.000...)$. Now insert in $(1) - (3)$ or in $(1') - (3')$.

This argumentation is pure nonsense. We definitely have $F( 0.999... ) \ne F(1.000...)$ and $G( 0.999... ) \ne G(1.000...)$ as well as $\sigma(F( 0.999... )) \ne \sigma(F(1.000...))$ and $\sigma(G( 0.999... )) \ne \sigma(G(1.000...))$.

The fake argumentation essentially pretends that $F$ (and similarly $G$) induces a function $\bar F : \mathbb R \to \mathbb R$ such that $$\sigma \circ F = \bar F \circ \sigma .$$
This is of course false, but can easily be overlooked.

Answer 2

The "proof" assumes that $0.999\cdots\ne1$, since they are mapped to different numbers by $F$ and $G$.

In other words, $F$ and $G$ are not functions on $[0,1]$.

Answer 3

Define $F(a+b)=a+a+b.$ Then $F(2+2)=2+2+2\neq 4+4+0=F(4+0),$ so $2+2\neq 4+0?$

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A more normal error is defining $\sqrt{x^2}=x.$ Then we’d get $-1=\sqrt{(-1)^2}=\sqrt{1^2}=1.$

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At heart, your definition assumes you can’t have two different digital representations of the same number.

In my example, $F(4)$ takes a lot of values because there are a lot of ways to write $4=a+b.$ This just means that I haven’t defined a function on the real numbers.

Your definition of $F$ is only a function if $0.999\dots\neq 1.000\dots.$ So you are assuming what you are trying to prove.