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I don't know if there exists a special, separate term for this operation in mathematics (maybe "to ascribe"?), but here is an example. Let's say I have a variable $m=2$ and I want to write it at the end of number $1$ in order to get another variable $n$. How to indicate this intention? I should write something like $n=1\{<m>\}=12$. That is, if $m=20$, then $n=120$, and so on... Is there a professional notation for this in math?

lyrically wicked
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    When we talk about "words" or "strings" we refer to this operation as concatenation. I have heard people refer to "concatenating numbers" as well. The wiki-math page suggests the notation $a| b$. E.g. $57| 123 = 57123$. – JMoravitz Oct 02 '15 at 05:37
  • Some people use notation $\overline{xy}$ to denote number $10x+y$. – cr001 Oct 02 '15 at 05:40
  • @JMoravitz: thank you. But how to indicate that a number should be inserted in the middle of another number? $n = 1||m||9$ (to get 129, from the above example)? – lyrically wicked Oct 02 '15 at 05:45
  • I don't think there's a convention. I think if use simply include a sentence "I will use the convention of a string of variables to represent the number with those variable as digits. I will intersperse actual digits amid variable to represent known digits." Then just write n = 1m9. – fleablood Oct 02 '15 at 05:50
  • @cr001: yes, yes, I also remember that I've seen the overbar/overline in some book for a similar operation. But when I checked wikipedia.org/wiki/Vinculum symbol page, I did not see an example of such usage... – lyrically wicked Oct 02 '15 at 05:50
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    It might be best to simply state what you are attempting to do by using words rather than symbols, especially since such operations are not commonly used. If you were to define $|$ as a binary operation as described in my link, then it follows that it is associative (but not commutative), at which point $a|b|c$ would be perfectly well defined, so if you wanted to take $a=31$, $b=33$ and $c=7$, then $a|b|c = 31337$. Explicitly, in base 10 you would have $a|b = a\cdot 10^{1+\lfloor\log_{10}b\rfloor} + b$ – JMoravitz Oct 02 '15 at 05:51
  • Just be sure to explain to the reader that you're working in base 10. – goblin GONE Oct 02 '15 at 06:19

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