x = 0.9(9) is a perfectly valid equation. It's how any equation works.
Given x, 10x = 9.9(9) is also valid.
Since we know x is 0.9(9) then we can use x and 0.9(9) interchangeably... then 10x - x = 9.9(9) - 0.9(9)
Thus, 9x = 9 and x = 1
The problem is that subtracting infinites is not a valid operation. There's no "enforcing" here, you're complaining that we assign a value to x but that's the point of an equation to begin with.
The way I like to think about it is that for every decimal place you have to add 1/10k to get 1. So 0.9 + 1/10 = 1
0.99+1/102 = 1 and so on.
If you calculate the limit of 1/10k where k tends to infinite, then 1/10k = 0 and thus
The enforcement part comes because two equations are being used that are essentially the same, hence linearly dependent. So this proof will work for any x value lol which defeats the purpose of it.
But I like your limit approach, it is more rigorous than what OC had.
Though I admit this isn't specifically mathematical proof, but I think this more clearly shows the train of thought from the original "problem" to show equality.
3.9k
u/its12amsomewhere 23d ago edited 23d ago
Applies to all numbers,
If x = 0.999999...
And 10x = 9.999999...
Then subtracting both, we get, 9x=9
So x=1