Does 10 = 9.99...?

I assume we are working with real numbers. Now assume $10 \neq 9.\overline9$ , then there exists real number $n$ such that $9.\overline9 < n < 10$ . Let $n$ be the mean value of $10$ and $9.\overline9$ , so $n = \dfrac{10 +9.\overline9}2 = \dfrac{19.\overline9}2$ . By performing long division, anyone can see that $\dfrac{19.\overline9}2 = 9.\overline9$ but this contradicts the fact that $9.\overline9 < n$ . Therefore $10 \neq 9.\overline9$ is false which implies $10 = 9.\overline9$ is true.

— n = 9.99...

— 10n = 99.99...

— 10n – n = 99.99... – 9.99...

— 9n = 90

— $\frac{9n}{9}$ = $\frac{90}{9}$

— n = 10

— 10 = 9.99...