| Entroper said: The .3333... argument is the easiest to understand. The real "discrete math" proof does say that between any two real numbers are an uncountabily infinite number of real numbers. So you would have to find a number that is between .9999... and 1. jlauro's argument (.9999... + 1)/2 doesn't work, because it assumes that .9999... and 1 are not the same number. You can't assume the thing you are trying to prove! If they are equal, then the average of the two numbers is also equal, and therefore not between them. Happy, .9999... is a rational number. It can be expressed as the ratio between 1 and 1. See the 1/3 argument above. |
Actually the (a + b)/ 2 does work. If it didn't, it wouldn't be math.
The problem is, by itself it doesn't help prove if .999... and 1 are equal or not. It would just be one step. So you could then have to show that result is equal to at least one of the other numbers.
Things can be proven by assuming what you are trying to prove, but it's generally much harder that way to be done correctly. Generally you dissprove stuff by assuming what you want to disprove and then finding the inconsistancies.
So if you want to prove that 1 is equal to .999... then you want to disprove that 1 not equal to .9999.... so the correct way to start the proof is to assume that 1 is not equal to .9999 and so (a+b)/2 is clearly what you want to assume if you were formally proving it from scratch.
