Why? Isn’t .999… an integer? However, the proof was assuming that, given two different numbers, you can find a different number between them. That isn’t true of all number systems, so you need more to the proof than that.
It works for the number system in question. That’s all that is required for proof. You might as well say that 10+11 does not equal 21 because there is a different answer in binary from decimal. This is a question about notation, and the nature of the notation would have to be agreed upon, as it would in the proof of anything. Walla Walla is the capital of Iceland in a system where Walla Walla is the way to spell Reykjavík in the commonly used system, but you will look foolish claiming that the capital of Iceland is Walla Walla without specifying a different terminology than the one almost always used.
OK, so the proof requires .999… to be a real number. How do you know it’s a real number, without some sort of proof involving limits?
And now for something completely different …
I have two U. S. coins in my pocket. The sum of both coins is 55 cents. One of the coins is not a half-dollar. What denomination of coins do I have in my pocket?
A nickel and a half-dollar. The one that is not a half-dollar is the nickel.
The question either assumes, or explicitly states that it is a real number. No proof is necessary.
[(Joke)]
No. The chance is (1+q)/2 where q is the likelihood the coin is double-headed. If you’re still guessing q = 0 after 20 heads, do yourself a favor and stay away from gambling dens. 
How many kids today have even seen a half-dollar coin?
It’d be worse with a problem based on $2 bills. Even a lot of adults are unaware of their existence.
But since the original question doesn’t specify which child was born first, doesn’t that make “boy-girl” and “girl-boy” the same? And by eliminating “boy-boy”, wouldn’t that leave only two possibilities?
There’s been debate about the sematics of this question. Some people claim that the question could be interpreted in different ways allowing the 1 in 2 answer. I don’t know if it’s on this board or elsewhere.
Well lookit that. It’s right there in Wikipedia as theBoy or Girl Paradox.
It sounds like you’re saying that, leaving out the two-boy option, there are two possibilities: 1) two girls, and 2) one of each (which could be either boy-girl or girl-boy). This is how many people (myself included) would intuitively look at this question. The problem is that 2) is twice as likely as 1). It’s a fact that, of the two-child families in the world, there are roughly (but only roughly because there are twins and intersex and so on to consider) twice as many families with one of each as there are with either two girls or two boys. Or two put it another way, there are roughly as many families with one of each as there with two of the same. It’s surprising to a lot of people; I would have initially guessed that such families are roughly 1/3 two boys, 1/3 two girls, and 1/3 one of each. But if you consider that any one child has a .5 chance of being a particular sex, then it’s simple arithmetic that the chance of any two children being a particular sex is .5(.5), or 1/4. It might not seem this should be true, but it is.
It’s clearer if you consider my earlier example; you could also say that there are two possibilities for the outcome of the lottery: 1) you have a winning number, and 2) you have a losing number. It’s obvious that “losing numbers” is a larger group than “winning numbers”, partly because there’s just one winning number and more than one losing one. It’s considerably less obvious that “one of each” is a larger group than “two girls”, but it is.
Now, as TriPolar pointed out, there can be disagreement on the semantics of this question, but that stems from whether you read it as saying:
A) “From all families with two children, at least one of whom is a girl, a family is chosen at random. This would yield the answer of 1/3.”
or:
B) “From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.”
The A) interpretation follows my logic above. The B) interpretaton looks like this:
“From all families with two children…”
100 of Family A: Kid 1 = Boy, Kid 2 = Boy
100 of Family B: Kid 1 = Boy, Kid 2 = Girl
100 of Family C: Kid 1 = Girl, Kid 2 = Boy
100 of Family D: Kid 1 = Girl, Kid 2 = Girl
“…one child is selected at random…”
From half of the 400 families, Kid 1 is selected; from the other half, Kid 2 is selected, for a total of 200 boys and 200 girls.
100 of Family A: Kid 1 = Boy, Kid 2 = Boy
Result = 100 boys
100 of Family B: Kid 1 = Boy, Kid 2 = Girl
Result = 50 boys, 50 girls
100 of Family C: Kid 1 = Girl, Kid 2 = Boy
Result = 50 girls, 50 boys
100 of Family D: Kid 1 = Girl, Kid 2 = Girl
Result = 100 girls
“…and the sex of that child is specified.”
We choose “girl”, which means that we’re now looking at 100 girls who have a brother, and 100 who have a sister.
To put it another way, what’s happening in these scenarios is this:
A) starts with *four *equally likely possibilities (families that have: a boy and then a boy, a boy and then a girl, a girl and then a boy, and a girl and then a girl), and eliminates *one *(boy-boy), so that each of the remaining possibilities has a 1/3 chance.
B) starts with *four *slightly different but also equally likely possibilities (boys who have a brother, boys who have a sister, girls who have a brother, and girls who have a sister) and eliminates two (boys who have a brother, and boys who have a sister), so that each of the remaining possibilities has a 1/2 chance.
But neither of these interpretations involves starting with three possibilities (boy-boy, one of each, and girl-girl) and eliminating one (boy-boy), because those three possibilities aren’t equally likely.