Let g be the function defined on the set of all real numbers by
g(x)= 1 if x is rational
e^x if x is irrational
then the set of numbers at which g is continous is???
2007-03-22 16:11:40 · 4 answers · asked by Akash S 1 in Science & Mathematics ➔ Mathematics
hmmm... it's really tricky, but I would risk saying it's not continuous anywhere, since it's very similar to the Dirichlet function (1 if x is rational, 0 if x is irrational).
But I'm not sure what happens around 0, since e^0 = 1
Edit 2: I'm so dumb, my previous explanation was completely and very obviously wrong.
A friend, who's studying for a Mathematics degree, told me it's continuous in 0 and tried to explain it to me by email, but I didn't get it either (I can't copy/paste the email, because it's in spanish, our native tongue)
2007-03-22 16:19:13 · answer #1 · answered by javier S 3 · 0⤊ 0⤋
Hmm, interesting question. It clearly cannot be continuous at any x not equal 0, but is it continous at x = 0?
2007-03-22 23:17:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
OMG just looking at it hurts my brian
2007-03-22 23:14:30 · answer #3 · answered by narsi_hunter 1 · 0⤊ 1⤋
48573
2007-03-22 23:16:11 · answer #4 · answered by Anonymous · 0⤊ 1⤋