all numbers are equal
Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b
So all numbers are the same, and math is pointless.
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b
So all numbers are the same, and math is pointless.
equal positive integers
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.
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.
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