Which proof correctly shows that 0.9‾=10.\overline{9} = 10.9=1?
Let x=0.9‾=0.999...x = 0.\overline{9} = 0.999...x=0.9=0.999.... Then 10x=9.999...10x = 9.999...10x=9.999..., so 10x−x=910x - x = 910x−x=9, giving x=1x = 1x=1.
0.9‾=910+9100+91000+...=9/101−1/10=9/109/10=10.\overline{9} = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + ... = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 10.9=109+1009+10009+...=1−1/109/10=9/109/10=1
0.9‾0.\overline{9}0.9 is an infinite series that sums to 1 by the geometric series formula.
All of the above