Can 9.99 Really Equal 10? An Analysis of the Math"

[_Mayday_]

I was told today that 9.99 = 10? I really don't understand that. Can anyone show me the proof for it.

 

[chroot]

Oh god, not this again.

Since you cannot find a real number between 9.999... (repeating nines) and 10, they are the same number. They're just a different way of representing the same number.

- Warren

 

[HallsofIvy]

Some people do it this way:
If x= 9.9999... then multiplying the equation by 10 gives you 10x= 99.99999... Notice that, since the "9"s in the first number go on indefinitely, the "9"s on the right of the decimal point in both go on indefinitely. Subtracting the first equation from the second, all of the "9"s on the right side canceL: you get 9x= 90 so x= 90/9= 10.

Of course, that assumes that you can multiply infinite numbers of decimal places just like finite ones and subtract the same way. That is true but should be proven. More correct is to argue that an infinite number of decimal place is by definition of "base 10 numeration", an infinite series. In particular, 9.9999... is, by definition, the infinite series
$$9+ 9/10+ 9/100+ 9/1000+ ...= \sum_{n=0}^\infty 9(1/10)^n$$
That is a geometric series and there is an easy formula for the sum of an infinite series:
$$\sum_{n=0}^\infty a r^n= \frac{a}{1- r}$$
as long as -1< r< 1.
For a= 9 and r= 1/10 (and, of course, -1< 1/10< 1)
$$\frac{9}{1- 1/10}= \frac{9}{\frac{9}{10}}= 10$$