John Gabriel
2015-06-22 17:39:33 UTC
The following link explains:
http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-772.html#post40616
The objects I use, that is, lim (n->oo) [L_n(m), U_n(m)] are *exactly* D. Cuts. The poor idiot Dedekind could not describe them as precisely as I did.
Moron academics hate this proof because it exposes their stupidity.
What's truly hilarious is how academic morons (dullrich especially) are trying to discredit my proof because I claim that for all intents and purposes, the cuts
(-oo,pi) U [pi, oo) and [3,pi) U [pi, 4]
are the same. What's important is that if a number is less than pi, then it will either be in the lower set [3,pi) or it will be in (-oo,3] since [3,pi) has 3 as the greatest lower bound. Similarly, if a number is greater than pi, then it will either be in the upper set [pi, 4] or it will be in [4, oo).
So the definition I use, IS exactly the same as that of Dedekind's original definition.
Definition:
"A cut is a partition of rational numbers into two non-empty sets A and B, such that all elements of A are less than all elements of B, and A contains no greatest element."
"A cut [3,pi) U [pi,4] is a partition of rational numbers into two non-empty sets A = [3, pi) and B = [pi,4], such that all elements of A are less than all elements of B, and A contains no greatest element."
Since [3,4] is a subset of the rationals, whatever applies to it, will also apply to (-oo,pi) U [pi,oo).
The cut (-oo, pi) U [pi, oo) can be written as:
(-oo, 3) U [3, pi) U [pi, 4] U (4, oo)
Now [3, pi) U [pi, 4] is a subset of Q as stated in the previous line. It's easy to see that -1 and -10000000 is in the cut. This is one of the objections by dullrich and EmperorZelos.
As you can see, this objection is null and void, because if k<pi, then k is either in (-oo, 3) or [3, pi), and if k>pi, then k is either in [pi, 4] or (4, oo).
But the monkeys dullrich, emperorzelos and other twatheads can't get past that sticking point in their dysfunctional brains. It's too complex for baboons. Chuckle.
So far, no refutation whatsoever.
Bit by bit, I am dismantling all the myths and fallacies from mythmatics. Who knows, it may some day become mathematics again. Getting rid of all the monkey druids is not an easy task. Their belief in Simos (Simian God) is not easy to shake.
http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-772.html#post40616
The objects I use, that is, lim (n->oo) [L_n(m), U_n(m)] are *exactly* D. Cuts. The poor idiot Dedekind could not describe them as precisely as I did.
Moron academics hate this proof because it exposes their stupidity.
What's truly hilarious is how academic morons (dullrich especially) are trying to discredit my proof because I claim that for all intents and purposes, the cuts
(-oo,pi) U [pi, oo) and [3,pi) U [pi, 4]
are the same. What's important is that if a number is less than pi, then it will either be in the lower set [3,pi) or it will be in (-oo,3] since [3,pi) has 3 as the greatest lower bound. Similarly, if a number is greater than pi, then it will either be in the upper set [pi, 4] or it will be in [4, oo).
So the definition I use, IS exactly the same as that of Dedekind's original definition.
Definition:
"A cut is a partition of rational numbers into two non-empty sets A and B, such that all elements of A are less than all elements of B, and A contains no greatest element."
"A cut [3,pi) U [pi,4] is a partition of rational numbers into two non-empty sets A = [3, pi) and B = [pi,4], such that all elements of A are less than all elements of B, and A contains no greatest element."
Since [3,4] is a subset of the rationals, whatever applies to it, will also apply to (-oo,pi) U [pi,oo).
The cut (-oo, pi) U [pi, oo) can be written as:
(-oo, 3) U [3, pi) U [pi, 4] U (4, oo)
Now [3, pi) U [pi, 4] is a subset of Q as stated in the previous line. It's easy to see that -1 and -10000000 is in the cut. This is one of the objections by dullrich and EmperorZelos.
As you can see, this objection is null and void, because if k<pi, then k is either in (-oo, 3) or [3, pi), and if k>pi, then k is either in [pi, 4] or (4, oo).
But the monkeys dullrich, emperorzelos and other twatheads can't get past that sticking point in their dysfunctional brains. It's too complex for baboons. Chuckle.
So far, no refutation whatsoever.
Bit by bit, I am dismantling all the myths and fallacies from mythmatics. Who knows, it may some day become mathematics again. Getting rid of all the monkey druids is not an easy task. Their belief in Simos (Simian God) is not easy to shake.