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 Weird function challenge (Posted on 2006-08-15)
Find a function f:R->R (R the set of real numbers), such that

1. f has a discontinuity in every rational number, but is continous everywhere else, and
2. f is monotonic: x<y → f(x)<f(y)

Note: Textbooks frequently present examples of functions that meet only the first condition; requiring monotonicity makes for a slightly more challenging problem.

 See The Solution Submitted by JLo Rating: 4.3000 (10 votes)

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 Needless to say this is very confusing | Comment 11 of 33 |

To be continuous, the limit of the function as it approaches a given value must be the same as the evaluation of the function itself at that value.

But then, despite the fact that there are a larger order of infinity of irrational numbers than rationals, it's still true that between any two irrationals there are infinitely many rationals (* see below); so how don't the discontinuities at all the rationals break that of the irrationals?

* Two irrationals differ in their decimal representation starting at some point. The representation up to that point represents some rational number. Make the next digit the lower of the two digits that differ in the next position between the irrationals or some digit that falls between the differing digits, then continue in some pattern that eventually or immediately continues in any repeating pattern.

 Posted by Charlie on 2006-08-19 16:44:09

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