math?


but anyway, you don't just use simple algebra on numbers with parts of them heading towards infinity. why? because it won't be accurate, much like this nonsensical argument. want to work with infinity? brush up on calculus.

 

he started with .9999999 (7)
multiplied by 10 it should equal 9.999999 (6) but he has 7

 

It goes to infinity!!!

 

I'm still trying to wrap my head around this one, but I would just say that at a cursory glance, this is a problem of infinite limits, and algebra does not typically deal well with infinite limits.

The correct way to prove this one way or the other is to create an expression for 0.9999..... with respect to a variable and take the limit using calculus of that variable as it approaches infinity.

 

infinity in math/calc is a finite number n

.999 repeating has n number of 9s
multiplied by 10, 9.99 has n-1 number of 9s

 

exactly why you can't just multiply by 10 and expect to say that the decimal point is simply shifted.

the function of multiplying by 10 actually multiplies a number by 10. it does not simply shift the decimal point. that just happens to be the representation in regular numbers.

see my previous post.

 

Somebody should do their doctoral thesis on this

 

This arguement has been around forever, and it's false.

 

So looking at this some more, I think the guy is right (as much of an arrogant prick he is), but his proofs are wrong. Two ways to look at it:
1) Calculus limit
0.99999..... = 1-1/10^x, where x goes to +inf.
lim x->inf (1-1/10^x) = 1

2) Geometric series
0.99999..... = 0.9 + 0.09 + 0.009 + ... + 0.9*1/10^x, where x goes to +inf
sum of a converging geometric series = a/(1-r)
=> 0.9/(1-1/10) = 1

 

i'm convinced h: