It might help maggenpye if every time you saw the word “infinity” you replaced it by what it really means, “without end.”
1 divided by “without end.” That’s obviously meaningless. So is 2 divided by “without end.” Or 2 multiplied by “without end.”
The only way to use the concept of infinity is to devise a special math with special rules. It has to be kept out of everyday arithmetic.
This works when infinity = aleph-null, but goes to pot when infinity = aleph-one.
I would never describe any aleph as infinity, and I don’t think I’ve seen any mathematicians do so. Probably safest to keep those two ideas separate.
ETA: wasn’t helpful
In my mathematical proofs course, the formal definition of an even number was that an integer n is even if n = 2m for some integer m, and n was odd if n = 2m+1 for some integer m.
if n = 2 and m = 1, n = 2m, so 2 is even.
if n = 0 and m = 0, n = 2m, so 0 is even
if n = 1 and m = 0, n = 2m+1, so 1 is odd.
And if nobody’s beaten me to it by the time I post, I’ll point out that the only context in which 0 is not considered even is on the roulette wheel. If you bet on evens and the ball lands on 0 (or 00, on American wheels), you lose.
I have a feeling I am going to feel very stupid asking this, because there must be some math concept I am not getting and nobody else has questioned it yet:
Isn’t 1 divided by 2 “0.5”?
Not if you are only dealing with integers, which we are. In which case, 2 can go into 1 zero times, leaving you a remainder of 1.
Yes, but you can also think of 3 divded by 2 as either 1.5 or 1 with a remainder of one. They are equivalent: just expressed differently to stress different aspects of their relationship.
I remember reading recently someone saying that if 1 is odd, then 1+1 ought to be even odder. However, ask any number theorist (which I am not) and he will tell you that all primes are odd, 2 is the oddest of all. No question that 2 has many properties unique among primes. But of course, it is even, as is 0, and 1,-1,3,-3, etc. are odd. And it would appear that 2 is likely to be the only even number not the sum of 2 positive primes (Goldbach conjecture).
The statement I bolded is correct only if infinity acts like a regular number.
But one can define self-consistent multiplication and division rules for infinity. (I actually wrote this out earlier this year.)
a * 0 = 0
a * inf = inf
0 * inf = undefined
0 * 0 = 0
inf * inf = inf
a / 0 = inf
a / inf = 0
0 / a = 0
inf / a = inf
0 / inf = 0
inf / 0 = inf
0 / 0 = undefined
inf / inf = undefined
where a is a finite number (not 0 or infinity). You’ll see that zero and infinity have very similar behaviors. Using these rules will produce the common-sense results you’d expect.
Correction: the only positive even number not the sum of 2 primes. You put the “positive” in the wrong place.
And it is possible to define prime negative numbers, or even prime complex numbers, but you need a more sophisticated definition of “prime” than the one you learned in elementary school.
Primes have to be non-units. If you restrict yourself to the Gaussian integers (complex numbers where both the real and imaginary parts are integers), then you can define complex primes, but the minute you allow Gaussian rationals, you don’t have any primes because every non-zero element is a unit.
What happens when you multiply a / inf by inf / a?
It becomes 0 * inf and is undefined, according to the table.
IAAM:
How so? 1 ÷ 2 = 0 Remainder 1.
I put it where I meant it. Ok, the same is probably true of -2. But what I meant is that 2 = 7 + (-5) is not a counterexample to my statement. And yes, -5 is a prime. The usual definition of prime is "A number p is prime if all its divisors are units and associates. A unit is a number with a multiplicative inverse. An associate of a number is that number times an associate. so -5, like 5, is divisible only by 1, -1, 5, and -5 and the first two are units and the second two associates. When you are among Gaussian integers, all numbers of the form a + bi, with a and b ordinary integers, then many ordinary primes (in fact 2 and all primes that leave a remainder of 1 when divided by 4) factor and the rest don’t. So 2 divisible by 1 + i and its associates. The units are 1, -1, i, and -i. 3 does not factor but 5 = (2+i)(2-i) and so on.
So, have we firmly established that 1 divided by 2 equals 0 with a remainder of 1? 
I think we need at least two or three more people jumping in to nail it firmly to the floor.
Sure, but I think you could’ve rested content with the fact that the usual definition of the primes is restricted to the positive integers, absent a specific change of context, such as to discussing prime elements of rings more generally.
That is, in most contexts, to say “Negative five” in response to “Pick a prime number” would elicit the response “No, that’s not a prime number”.