Recent content by Changbai LI

It is a complex problem. If you are interising,there is some paper.but it was write by Chinise. www.lcbmaths.com

0.9... is not =1. You can see 0.9.../=1(English) My emai is Changbai LI

0.99…≠ 1 1. 10÷3 = 3+ 1 ÷3，so 10÷3 = 3 …1 is not correct. The values on each side of an equal sign means both values are strictly equal. 9÷3 = 3. it is right. It can be checked by direct computations (by times 3). 10÷3 =（9+1）÷3 =3 + 1÷3 is right now. It can be checked by direct computations...

1\=0.9999... I know 1/3=0.3…, then multiply by three to get 1=0.9…. Anyone knows it . But it is easy to see that 1/3 was finished. In other words, it got the end. And 0.9… is going on on the contrary. In other words, it is not finished forever. So 1/3 is not equal to 0.9…, I think.

Prove:0.9…\=1 1. 1^n=1^10^n=1(even n->infinite) (1) 2. lim(1+1/n)^n=e(when n->infinite) (2) 3. lim(1-1/n)^n=1/e(when n->infinite) (3) 4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4) 5. It is the key. So write it more detail in series...

Prove:0.9…\=1 1. 1^n=1^10^n=1(even n->infinite) (1) 2. lim(1+1/n)^n=e(when n->infinite) (2) 3. lim(1-1/n)^n=1/e(when n->infinite) (3) 4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4) 5. It is the key. So write it more detail in series...

Prove:0.9…\=1 1. 1^n=1^10^n=1(even n->infinite) (1) 2. lim(1+1/n)^n=e(when n->infinite) (2) 3. lim(1-1/n)^n=1/e(when n->infinite) (3) 4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4) 5. It is the key. So write it more detail in series...