Proof that 0=1

Requires knowledge of calculus

Given:

• i = √-1
• i^-1 = i³ = -i
• i^-2 = i² = -1
• sin²(y) + cos²(y) = 1
• e^ix = cos(x) + isin(x)
• y = e^ax ∴ dy = ae^ax dx
• y = sin(x) ∴ dy = cos(x) dx
• y = cos(x) ∴ dy = -sin(x) dx
• ∫ku = k∫u where k is a constant
• ∫u dv = uv − ∫v du

Proof

• First, we attempt to find ∫(e^iy sin(y)) dy.
    ∫(e^iy sin(y)) dy = -ie^iy sin(y) − ∫(-ie^iy cos(y)) dy
    u = sin(y), dv = e^iy dy
    du = cos(y) dy, v = -ie^iy
    -i∫(e^iy cos(y)) dy = -e^iy cos(y) + -∫(e^iy sin(y)) dy
    u = cos(y), dv = e^iy dy
    du = -sin(y) dy, v = -ie^iy
    ∫(e^iy sin(y)) dy = -ie^iy sin(y) + e^iy cos(y) + ∫(e^iy sin(y)) dy
• Now, we cancel out and factor.
    0 = -ie^iy sin(y) + e^iy cos(y)
    0 = (e^iy)(-isin(y) + cos(y))
    0 = (e^iy)(cos(y) − isin(y))
• Use Euler's Formula (that e^ix = cos(x) + isin(x)).
    0 = (cos(y) + isin(y))(cos(y) − isin(y))
    0 = cos²(y) − isin(y)cos(y) + isin(y)cos(y) − i²sin²(y)
    0 = cos²(y) − -sin²(y)
    0 = cos²(y) + sin²(y)
    0 = 1. Q.E.D.