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A way to write a number in various bases is: 1012, 3Fa16, 51910.

The thing here, is that we apparently specify the base in decimal by default. This makes sense in everyday life, since we're not really doing base conversion when grocery shopping. But 10 is ambiguous when working with different bases. So why is 10110 not binary or ternary or octal when working specifically with base conversion?

How about 101102? Is 1111113 = 18310 = 183?


Related:

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    $\begingroup$ You just answered your question and then asked it in the next sentence. Why is $101_{10}$ not binary or ternary or octal? The thing here, is that we specify the base in decimal, by default. $\endgroup$ – Kenny Lau Oct 3 '17 at 7:51
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    $\begingroup$ And as for why base ten is default, it's because base ten is our default numbering system, which presumably is because we have ten fingers. $\endgroup$ – Kenny Lau Oct 3 '17 at 7:52
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    $\begingroup$ If you really are worried about ambiguity you can always end the chain like such: $101_{1111111111_1}$ as base $1$, no matter what base your reader assume you wrote it in is always unary. (Please don't do this.) $\endgroup$ – orlp Oct 3 '17 at 8:00
  • $\begingroup$ This is an interesting observation! It reminds me somewhat of metalanguage in logic. Specifically, say we're in the middle of defining the natural numbers $0,$ $1,$ $2,$ $\ldots$ in basic set theory, and we call the reader's attention to something $2$ sentences back. Here, the first usage of $2$ is $2$ in the object language and the second usage of $2$ is $2$ in the metalanguage. $\endgroup$ – Dave L. Renfro Oct 3 '17 at 8:02
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    $\begingroup$ I have wondered this as well. An alternative would be to specify the base in itself but then the subscript would always be $10$ and hence not very useful. Maybe the base should use a different system such as Roman numerals. This is not a serious suggestion; the choice of $10$ as the default is arbitrary but very well established. $\endgroup$ – badjohn Oct 3 '17 at 8:08
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Yes, you could indicate the base of the base too, but eventually you need a base-indicator that is specified in some "default" base. We normally use base ten so if nothing else is stated that is what the base that is assumed.

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  • $\begingroup$ Do you really assume it's binary if nothing else is stated? (Joke, but partially serious). $\endgroup$ – Stewie Griffin Oct 3 '17 at 7:57
  • $\begingroup$ @StewieGriffin Yes, of course. There are 10 kinds of people - those who know binary and those who don't:) $\endgroup$ – skyking Oct 3 '17 at 8:01
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    $\begingroup$ Actually, there are 10 kinds of people! Those who know binary, those who don't, and those who expected that joke to be in ternary. $\endgroup$ – Stewie Griffin Oct 3 '17 at 8:14
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As far as I know, the base is always specified in decimal, but it is up to you to change that convention, provided you state it clearly.

The following shorthands are universally adopted: $b$ for two, $o$ for eight, $d$ for ten, $h$ for sixteen.

If you want a system that can denote any base without extra conventions, you can resort to unary:

$$10_{||||||||||||||||}=16_{||||||||||}=51_{|||}.$$

This is why th Babylonians dropped their base $60_{||||||||||}$ numeration ($_{||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||}$).

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  • $\begingroup$ Another "convention free" option would be $10_{1+1+1+1+1+1+1+1+1+1}$, which is just another way of writing unary, of course. $\endgroup$ – goblin Oct 3 '17 at 9:08
  • $\begingroup$ @goblin What about succ(succ(succ(succ(succ(succ(succ(succ(succ(succ(0)))))))))), more Peanoish ? $\endgroup$ – Yves Daoust Oct 3 '17 at 9:14
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    $\begingroup$ Love it :) $\;\!$ $\endgroup$ – goblin Oct 3 '17 at 9:31

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