Purpose
The Conway base 13 function was created as part of a "produce" activity: in this case, the challenge was to produce a simple-to-understand function which takes on every real value in every interval, that is, it is an everywhereSketch of definition
* Every real number ''x'' can be represented inB34C128
.
* If instead of , we judiciously choose the symbols , something interesting happens: some numbers in base 13 will have representations that ''look'' like well-formed decimals in base 10: for example, the number 54349589 has a base-13 representation of −34.128
. Of course, most numbers will not be intelligible in this way; for example, the number 3629265 has the base-13 representation 9+0−−7
.
* Conway's base-13 function takes in a real number ''x'' and considers its base-13 representation as a sequence of symbols . If from some position onward, the representation looks like a well-formed decimal number ''r'', then ''f''(''x'') = ''r''. Otherwise, ''f''(''x'') = 0. (Well-formed means that it starts with a + or − symbol, contains exactly one decimal-point symbol, and otherwise contains only the digits 0–9). For example, if a number ''x'' has the representation 8++2.19+0−−7+3.141592653...
, then ''f''(''x'') = +3.141592653....
Definition
The Conway base-13 function is a function defined as follows. Write the argument value as a tridecimal (a "decimal" inProperties
* According to the intermediate-value theorem, every continuous real function has the intermediate-value property: on every interval (''a'', ''b''), the function passes through every point between and The Conway base-13 function shows that the converse is false: it satisfies the intermediate-value property, but is not continuous. * In fact, the Conway base-13 function satisfies a much stronger intermediate-value property—on every interval (''a'', ''b''), the function passes through ''every real number''. As a result, it satisfies a much stronger discontinuity property— it is discontinuous everywhere. * From the above follows even more regarding the discontinuity of the function - its graph is dense in . * To prove that the Conway base-13 function satisfies this stronger intermediate property, let (''a'', ''b'') be an interval, let ''c'' be a point in that interval, and let ''r'' be any real number. Create a base-13 encoding of ''r'' as follows: starting with the base-10 representation of ''r'', replace the decimal point with C and indicate the sign of ''r'' by prepending either an A (if ''r'' is positive) or a B (if ''r'' is negative) to the beginning. By definition of the Conway base-13 function, the resulting string has the property that Moreover, ''any'' base-13 string that ends in will have this property. Thus, if we replace the tail end of ''c'' with the resulting number will have ''f''(''c'') = ''r''. By introducing this modification sufficiently far along the tridecimal representation of you can ensure that the new number will still lie in the interval This proves that for any number ''r'', in every interval we can find a point such that * The Conway base-13 function is therefore discontinuous everywhere: a real function that is continuous at ''x'' must be locally bounded at ''x'', i.e. it must be bounded on some interval around ''x''. But as shown above, the Conway base-13 function is unbounded on every interval around every point; therefore it is not continuous anywhere.See also
*References
* {{cite journal, last = Oman, first = Greg, title = The Converse of the Intermediate Value Theorem: From Conway to Cantor to Cosets and Beyond, journal = Missouri J. Math. Sci., volume = 26, issue = 2, date = 2014, p = 134–150, url = http://www.uccs.edu/Documents/goman/Converse%20of%20IVT.pdf, archive-url=https://web.archive.org/web/20160820112316/https://www.uccs.edu/Documents/goman/Converse%20of%20IVT.pdf , archive-date=2016-08-20, url-status=live Functions and mappings John Horton Conway Special functions