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I don't understand how the proof above shows that two distinct real numbers correspond to different infinite decimal.

All I got out of the explanation is given any two distinct real numbers $a$ and $a'$, we can always pick an integer $n$, such that $\frac{1}{10^n}<|a-a'|$, which to me means that there is always a number that is smaller than the absolute difference between the two given real numbers $a$ and $a'$. What is the significance of this and how does it relate to the objective of showing that two distinct real numbers correspond to different infinite decimal? I am also puzzled at the conclusion "...will certainly differ by the nth decimal place"

I tried to follow workings with some example numbers say $a=6$ and $a'=6.1$ but still did not make any progress.


To go through your example:

We note that $|a - a'| > 1/10^{2}$. So, according to the proof, the decimals should differ by the $2$nd decimal place.

Indeed, this is the case. We note that the $0,1,$ and $2$ digits of the representation $a$ must be could be $6,0,0$ since $$ 6 \leq a \leq 6+1 = 7\\ 6.0 \leq a \leq 6.0 + 0.1 = 6.1\\ 6.00 \leq a \leq 6.00 + 0.01 = 6.01 $$ So, is the same true for $a'$? $$ 6 \leq a' \leq 6+1 = 7 \quad \text{. . . good so far}\\ 6.0 \leq a' \leq 6.0 + 0.1 = 6.1 \quad \text{. . . nothing wrong yet}\\ 6.00 \leq a' \leq 6.00 + 0.01 = 6.01 \quad \text{. . . this isn't true!} $$ So, we see that the two numbers can't have the same decimal representation.

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  • $\begingroup$ In my example, $a=6$ and $a'=6.1$. So, shouldnt the decimals differ by the 1st decimal place? Is your illustration actually for $a=6$ and $a=6.01$? $\endgroup$ – mauna Aug 26 '14 at 4:57
  • $\begingroup$ @mauna note that $6.1$ can also be written as $6.0999...$, in which case the two decimals would agree for the 0th and 1st decimal places, but not the second. In general, real numbers need not have unique decimal representations, but every decimal representation describes exactly one real number. $\endgroup$ – Omnomnomnom Aug 26 '14 at 15:25

It does not just mean that there is always a number that is smaller than the absolute difference between the two given real numbers. Rather it means that there is such a number of the form $1/10^n$, so that digits in the $n$th decimal places in the two numbers differ from each other.

PS: For example, two numbers differing in the $5$th place and not before are two numbers that differ by $1/10^5$. Two numbers differing by at least $1/10^5$ are numbers that differ at or before the $5$th place after the decimal point. \begin{array}{cccccccccc} & 3 & . & 1 & 4 & 1 & 5 & 9 \\ -& 3 & . & 1 & 4 & 1 & 5 & 8 \\ \hline & 0 & . & 0 & 0 & 0 & 0 & 1 & = & 1/10^5 \end{array}

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  • $\begingroup$ I still don't get how that is supposed to work...can you please illustrate with a small example say $a=6$ and $a=6.1$ or something similar? $\endgroup$ – mauna Aug 25 '14 at 19:18
  • $\begingroup$ Absolutely - it is using the general Archimedean axiom to deliver a specific instance of the form you require - many mathematical proofs are of this kind: I can find a number less than any $\epsilon$ - how do I choose my $\epsilon$ to solve this problem.(Something equivalent to the Archimedean axiom needs to be stated as part of the definition of the real numbers or - more usually - proved as a consequence of a different definition - a huge edifice is built on a seemingly trivial observation) $\endgroup$ – Mark Bennet Aug 25 '14 at 19:22
  • $\begingroup$ @mauna : I added an example. Does that help? $\endgroup$ – Michael Hardy Aug 27 '14 at 14:24
  • $\begingroup$ yes, it does clarify what you mean and I thank you for that. Your answer combined with @Omnomnomnom answer made me understand the proof. But I think Omnomnomnom is a little bit better as it walk's through the proof with the example numbers I asked for. $\endgroup$ – mauna Aug 27 '14 at 15:03

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