Can you find the illegal step in the following silly proof? Given: a=b

$$ a^2=ab$$

$$a^2-b^2 = ab-b^2$$

$$(a+b)(a-b)=b(a-b)$$

$$a+b = b$$

$$b+b=b$$

$$2b=b$$

This is mathematics, not real world, isn't it? 

The error, if there is one ;-), is the last step where 'b' should be subtracted from both sides, resulting in the true answer 'b=0; => 0=a', !

See also: dy/dx => $inf * 0, doesn't it? 

(Hence, Are there more infinitesimals than infinities?)

Thinking outside the box ;-)

OK OK, the above problem was pretty easy. Can you find the illegal step in the following?

Theorem: 4=5

Proof:

$$16-36=25-45$$

$$4^2–9*4=5^2–9*5$$

$$4^2–9*4+81/4=5^2–9*5+81/4$$

$$ (4-9/2)^2=(5-9/2)^2$$

$$4-9/2=5-9/2$$

$$4=5$$

Thanks for the mental challenge.  It is tricky. The illegal step is taking only one solution from each side when taking the square root.  There are two factors on each side of the equation.

If $x^2 = y^2$, taking square root on both sides gives $x = +/- y$. In this case accepting the solution $x = -y$ would not have caused a contradiction.

Alternatively, 

$x^2 = y^2$, implies $(x+y)(x-y)=0$, so either $x=y$ or $x=-y$ or both. Given the specific values of $x$ and $y$ in this problem, we know that $x=y$ is not true, so we must have $x=-y$.

Going from step 3 to step 4, you divided by (a-b). Division by 0 is not meaningful.

If you let b = 1, then you get the "classic" homework result:

2 = 1

From this follows the other classic result:

1 = 0. 

Neil

As a corollary, 2 = 0.

Like CharlieRader said, division by 0 is the truly "illegal" step, but to my mind, the worst decision is not factoring out common terms until step 3. The error between steps three and four becomes obvious if the first transformation is to reduce the initial equation to a = b, which also makes the multiplication by zero in step three apparent. 

3rd step into 4th is only correct if a!=b, but the condition was a=b,

therefore proof is inconsistent