Let [1, 2, 3, 4…] be an infinite set called A
Subtract three from it.
[ 4, 5, 6, 7…] and say it is called B
Therefore
B-A=3
You can subtract X numbers and keep on going
[X, X+1, X+2, X+3…]
Thanks again Lincoy.
I don't know what he was talking about. As the reasoning from my posts was logical and demonstrable.
But you must have sympathy for me, because you prevented me from wasting my time, Once Again!
There are an indeterminate from and useless to prove.
What are you saying?
There are 7 indeterminate from in calculus which cannot be find value.
0÷0 , 0⁰ , ∞-∞ , ∞÷∞ , ∞⁰ and 1^∞
What about 0 x infinity?
oh yeah it also indeterminate from. example.
0×∞ = lim x→0 [ x(5÷x) ] = lim x→0 [ x÷x ] × lim x→0 [ 5 ] = 5
0×∞ = lim x→0 [ x²(1÷x) ] = lim x→0 [ x ] = 0
0×∞ = lim x→0 [ x²(1÷x⁴) ] = lim x→0 [ 1/x² ] = ∞
What about 0 x infinity?
oh yeah it also indeterminate from. example.
0×∞ = lim x→0 [ x(5÷x) ] = lim x→0 [ x÷x ] × lim x→0 [ 5 ] = 5
0×∞ = lim x→0 [ x²(1÷x) ] = lim x→0 [ x ] = 0
0×∞ = lim x→0 [ x²(1÷x⁴) ] = lim x→0 [ 1/x² ] = ∞
0^0=0/0, infinity^0=1^infinity=infinity/infinity,
Infinity*0=infinity-infinity
Add X to each side
♾️=♾️+X
Subtract the right infinity to the left side
♾️-♾️=X