# Does .999 equal 1?

Using calculus (an infinite series to be more exact), we know that .999 repeating can actually be written as The sum from 0 to infinity of(.9(.1^n)), which is a geometric series of the form a=.9, r=.1. (Meaning that it follows the form a+ar1+ar2+ar3... etc.) Using the geometric series convergence rule, we know that...
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Using calculus (an infinite series to be more exact), we know that .999 repeating can actually be written as The sum from 0 to infinity of(.9(.1^n)), which is a geometric series of the form a=.9, r=.1. (Meaning that it follows the form a+ar1+ar2+ar3... etc.) Using the geometric series convergence rule, we know that any geometric series converges to a/(1-r). So, the series converges to .9/(1-.1), or .9/.9, which obviously equals 1.

This is an infallible proof, using one of the highest levels of math (differential Calculus), that cannot be disproved, and although it conflicts with my sense of logic, .999 repeating equals 1. You can't argue with it in any way, shape, or form. 1 equals .999 repeating and there's not really anything any of us can do about it, unless you want to get mad at Euler and Isaac Newton.

This is an infallible proof, using one of the highest levels of math (differential Calculus), that cannot be disproved, and although it conflicts with my sense of logic, .999 repeating equals 1. You can't argue with it in any way, shape, or form. 1 equals .999 repeating and there's not really anything any of us can do about it, unless you want to get mad at Euler and Isaac Newton.

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