Math Is Fun Forum

Ancient Greeks did not have 0.111... nor 0.999... . What they had are rational numbers. 3/7 is a rational member. 0.1 can be translated into a rational number 1/10 from our decimal system.

but 0.999... is actually defined by 0.9+0.09+0.999+...=translated to=9/10+9/100+9/1000+...
and is it a rational number? At least NO by its BASIC FORM.

in the rational number world, 0.999... is not a number, and by only finite translation, it's only the sum of a INFINITE series.

0.999...?=1 is a question that whether the infinitesmall 1/10^n is equal to zero. And it's a question that infinitesmall?=0

And it had been a tough question for Newton and his following calculus mathematists. It was famous in history as Berkeley's Attack.

Cantor is the Neo who "dealed" this problem. (my point of view) He defined a real number to a infinite set including all the rational numbers ahead of it. And by his definition REAL number 1 is 0.999... to some extent, since 0.999... can represent the latest rational number before 1.

Maybe you've noticed the latest rational number before 1 or any real number cannot be explicitly pointed out. And that's the true problem--he just moved infinity from calculus and infinite series outside number system to inside it! He actually hided infinitesmall into the definition of a real number!

So 0.999... ,can be translated into a sum of rational number series by finite translation, then can be translated into real number 1, by infinite translation.

Real number 1 isn't the integer 1 or the rational number 1. Can a set of rational numbers define a rational number?This is Russel's Attack.

So your problem'sabug of real number system.

Thank you for the link. The link finally reminded me of the proof i once read in Thomas' Calculus book, it uses a neighbor elimation trick:
key thought --  (k+1)³-k³ =3k²+3k+1

     3(1²+2²+3²+...+n²)
     3(1 +2 +3 +... +n)
+)     1 +1 +1 +... +1
___________________

(-1³+2³)+(-2³+3³)+(-3³+4³)+...+(-n³+(n+1)³)
=-1+(n+1)³=n³+3n²+3n

3x + 3n(n+1)/2+n=n³+3n²+3n

3x=n³+3/2 n²+1/2 n= 1/2 n(2n²+3n+1)

x= 1/6 n (2n+1) (n+1)

c:  7 6 5/(3 2 1)=35
d:  right  (diceA diceB diceC) = (1 1 4), (1 4 1), (4 1 1), (1 2 3), (1 3 2), (2 1 3), (2 3 1), (3 1 2), (3 2 1), (2 2 2) ;Prob=(3+6+1)/(6 6 6)

you r welcome

above is the link of an online free math solver

thiat used to be one of my high school excercise.

sin36 = 2 sin18 cos18 = 2 cos72 sin72

2 (sin36)^2=8 (cos72)^2 [1- (cos72)^2]

1 - cos72=8 (cos72)^2  [1- (cos72)^2]

cos72=sin18=x

1-x=8 x^2 (1 - x^2) = 8 x^2 (1-x)(1+x) = (1-x)(8 x^2+8 x^3)

x<1, so 8 x^3+8 x^2-1=0

solve it and get sin18, further get cos18,  sin36 = 2 sin18 cos18

i simplified it and finally made it through!

it really requires trig formula mastering

tan20 tan30 tan40=sin20 sin30 sin40 /( cos20 cos30 cos40) ; tan10=sin10/cos10=cos80/sin80

cos20 cos30 cos40 = (cos20 sin60) cos40 = 1/2 (sin80+sin40) cos40 =1/2 sin80 cos40 + 1/4 sin80

while
sin20 sin30 sin40 = (sin20 sin40) sin30 = 1/2 (cos20-cos60) sin30 = 1/2 cos20 sin30 - 1/2 sin30 sin30 = 1/4 (sin10+sin50) + 1/4 cos60-1/4 = 1/4 sin10 + 1/4cos40-1/8 = 1/4 sin10 + 1/4 (cos40-cos60) = 1/4 sin10 + 1/2 sin10 sin50=1/4 sin10 +1/2 sin10 cos40

so the left side=sin10/sin80=tan10=the right side