→ Madroach :其實也可以用反證法來證明 我另外打一篇好了 01/16 02:47
※ 編輯: Madroach 來自: 111.248.14.130 (01/16 03:19)
※ 引述《yueayase (scrya)》之銘言：
: 看到這個敘述,我想到我看到得一題極限證明題:
: Let f be continous on [a,b] and let f(x) = 0 when x is rational.
: Prove that f(x) = 0 for every x in [a,b].
: 嘗試去做證明:
: Proof:
: Assume there is a irrational number x' in [a,b] such that f(x') = a ≠ 0
: Then, we want to show that lim f(x) = 0 ?
: x->x'
: or want to show that f is not continuous?
: 到底該怎麼取?
Since f is continuous on [a,b]
then for all x in (R\Q)∩[a,b] implies
lim f(r) = f(x)
r→x
make a sequence{r_n} which r_n→x as n→∞
each r_n is in Q for n in N
then lim f(r_n) = lim f(r_n) = f(x)
n→∞ r_n→x
but each f(r_n)=0, hence f(x)=0 for all x in (R\Q)∩[a,b]
so f(x)=0 for all x in [a,b]
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.248.14.130