Theorem: All positive integers are equal.
Proof: Sufficient to show that for any two positive integers, A and B, A = B.
Further, it is sufficient to show that for all N > 0, if A and B (positive integers) satisfy (MAX(A, B) = N) then A = B.
Proceed by induction.
If N = 1, then A and B, being positive integers, must both be 1. So A = B.
Assume that the theorem is true for some value k. Take A and B with MAX(A, B) = k+1. Then MAX((A-1), (B-1)) = k. And hence (A-1) = (B-1). Consequently, A = B.
Originally posted by aamsThis bit here is nonsense.
Further, it is sufficient to show that for all N > 0, if A and B (positive integers) satisfy (MAX(A, B) = N) then A = B.
MAX(X,Y) returns an array the same size as X and Y with the largest elements taken from X or Y. In this case it would return the higher of A or B.
You could say that if MAX(A,B) = MIN(A,B) then A=B but that won't get you anywhere.
Originally posted by XanthosNZIs it me or is the general forum turning into a place where teenagers get off on algebra?
This bit here is nonsense.
MAX(X,Y) returns an array the same size as X and Y with the largest elements taken from X or Y. In this case it would return the higher of A or B.
You could say that if MAX(A,B) = MIN(A,B) then A=B but that won't get you anywhere.
This is far more worrying than the odd rude word popping up!
let a=Pi, b=3, c=0.141...
a=b+c
(a-b)a=(a-b)(b+c)
a^2-ab=ab+a-a+b^2-b
a(a-b-c)=b(a-b-c)
a=b
thus, Pi is exactly 3 !
while we are on the subject of random proofs-this one is not flase, but more random and quite cool...
let a=0.99999recurring
10a=9.999rec.
10a-a=9.999rec.-a
9a=9
=>a=1
therefore, 0.9999recurring is equal to one... 🙂
Originally posted by geniusThe second one you posted is true.
let a=Pi, b=3, c=0.141...
a=b+c
(a-b)a=(a-b)(b+c)
a^2-ab=ab+a-a+b^2-b
a(a-b-c)=b(a-b-c)
a=b
thus, Pi is exactly 3 !
while we are on the subject of random proofs-this one is not flase, but more random and quite cool...
let a=0.99999recurring
10a=9.999rec.
10a-a=9.999rec.-a
9a=9
=>a=1
therefore, 0.9999recurring is equal to one... 🙂
Originally posted by aamsOf course, silly...we all new that.
Theorem: All positive integers are equal.
Proof: Sufficient to show that for any two positive integers, A and B, A = B.
Further, it is sufficient to show that for all N > 0, if A and B (positive integers) satisfy (MAX(A, B) = N) then A = B.
Proceed by induction.
If N = 1, then A and B, being positive integers, must both be 1. So A = B.
Assume tha ...[text shortened]... with MAX(A, B) = k+1. Then MAX((A-1), (B-1)) = k. And hence (A-1) = (B-1). Consequently, A = B.
