Something similar to the concept of Divide by Zero...I think

[SpaceMarine93]

Can anyone resolve this?



Or was the solution painfully obvious?

[Kryten]

That is division by zero; a2-b2 and ab-b2 are both zero, and they were divided when factorised.

[D.Turtle]

a=b means that a-b=0

Going from the fourth to the fifth line, he divides by a-b on both sides.

Ergo, division through zero.

Another way of easily seeing the trick is to replace a with b (or b with a) right from the second line on. Then you can easily see what the trick is.

[Surlethe]

More precisely, since a=b, a-b=0, so on the 4th step you divide by zero. Edit: Damn you, D. Turtle!

For a trickier puzzle, what's wrong with this?

1 = \lim_{k\to-\infty} 1
= \lim_{k\to-\infty} exp(2ik\pi)
= \lim_{k\to-\infty} exp(2k\pi)^i
= [\lim_{k\to-\infty} exp(2k\pi)]^i
= 0^i
= 0.

[Terralthra]

At the 3rd line, \lim_{k\to-\infty} exp(2k\pi)^i, if I'm interpreting your syntax correctly, falls prey to Euler's identity = 0, as the limit approaches e (the +1 is on the other side).

[Eulogy]

That is division by zero; a2-b2 and ab-b2 are both zero, and they were divided when factorised.

Or in other words, since a=b, ab=a2=b2.

If you divide by zero, you literally end up with impossibilities. [/obvious]

[Surlethe]

At the 3rd line, \lim_{k\to-\infty} exp(2k\pi)^i, if I'm interpreting your syntax correctly, falls prey to Euler's identity = 0, as the limit approaches e (the +1 is on the other side).

No, I used that on the second line. The pseudo-exercise is to derive from Euler's identity a contradiction.