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

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The discussion centers on the mathematical equivalence of 9.999... and 10, with participants explaining that they represent the same value. One method involves multiplying 9.999... by 10 and subtracting the original number, leading to the conclusion that x equals 10. Another explanation uses the concept of an infinite geometric series, demonstrating that the sum of the series converges to 10. The debate highlights the understanding of infinite decimal representations in base 10 numeration. Ultimately, 9.999... is mathematically proven to equal 10.
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I was told today that 9.99 = 10? I really don't understand that. Can anyone show me the proof for it.
 
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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
 
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
 
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