One popular proof for .999… = 1 than I enjoyed back when I was foolish enough to believe it true was through digit manipulation. Observe:
x = .999…
10x = 9.999… (both sides of the equation are multiplied by 10)
9x = 9 (“x” is subtracted from both sides)
x = 1 (both sides of the equation are divided by 9)
Ta da! What could be more foolproof than that? The starting premise is obviously correct, and the same thing is done to both sides of the equation until it spits out a conclusion that even a guy like me would have to admit is true. Right? Right???
Well, obviously not, otherwise I wouldn’t have written this blog. While it appears that the same x is being subtracted from both sides in the third line, it’s actually two different x’s. When you multiply a number by 10, you don’t simply add 9 to it (more on this in part 2)… you get that 9 before the decimal point by moving the decimal place over to the right one space. This is true of any real number… multiplying by 10 moves the decimal place over once. By doing this, there is one less digit to the right of that decimal point. That’s true of every finite number, and should also be considered true of infinite numbers. Therefore, when one subtracts x in the equation above, you’re subtracting an infinite series of 9’s from another infinite series of 9’s that is one digit shorter. If you want to see me do all of the actual math to show you that this checks out, view my YouTube video on the subject.
A final note: While I welcome criticism, please don’t tell me .999… multiplied by 10 still has the same number of 9’s after the decimal point unless you can demonstrate why this is so. I’ve already talked about how mathematicians accept different size infinities, and I explained in my first blog in this series why inductive reasoning suggests that if any real number multiplied by 10 has one less digit after the decimal point then it should still be true when dealing with infinite digits. If you’re going to argue with me, please present a point that I haven’t already refuted.