User:WBOSITG/Rule-breakers

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User talk:WBOSITG/Border

[edit] 2=1

Let a and b be equal non-zero quantities

a = b \,

Multiply through by a

a^2 = ab \,

Subtract b^2 \,

a^2 - b^2 = ab - b^2 \,

Factor both sides

(a - b)(a + b) = b(a - b) \,

Divide out (a - b) \,

a + b = b \,

Observing that a = b \,

b + b = b \,

Combine like terms on the left

2b = b \,

Divide by the non-zero b

2 = 1 \,

Invalid: Line 5 divides by (a - b) \,, which is zero due to line 1, a = b \,.

[edit] Alternate proof

Consider the function f(x)=x. We write:

x = x
\sum_{i=1}^{x}1=x
\frac{d}{dx}\sum_{i=1}^{x}1=\frac{dx}{dx}
\sum_{i=1}^{x}\frac{d1}{dx}=1 by the derivative sum rule.
\sum_{i=1}^{x}0=1 because the derivative of a constant is zero.
0 = 1

Invalid: The function f(x)=1+1+1+....+1 is not continuous at any point, and thus is not differentiable.

[edit] Alternate alternate proof

\int_ {} u\, dv = u \cdot v - \int_ {} v\, du

Let dv = \sin x \cdot dx

So v = cosx

Let u = secx

du = \sec x \cdot \tan x \cdot dx
\int_ {} \tan x\, dx = \int_ {} \tan x\, dx
\int_ {} \tan x\, dx = \int_ {} \sec x \cdot \sin x\, dx
\int_ {} \tan x\, dx = \sec x \cdot ( - \cos x) - \int_ {} - \cos x \cdot \sec x \cdot \tan x \, dx
\int_ {} \tan x\, dx = - 1 - \int_ {} - \tan x \, dx
\int_ {} \tan x\, dx = - 1 + \int_ {} \tan x \, dx
0 = − 1

Invalid: Subtracting the integral in the penultimate step disregards the different constants of integration.