Let a = b.
Hence, a^2 = ab.
So, 2a^2-2ab = a^2 - ab
2(a^2-ab) = 1(a^2-ab)
Hence, 2=1?!
I showed this to two maths teachers at my school who couldn't figure this... Let's see how you guys fare.
I have seen ti many times, it goes like this
$\displaystyle a =b$
$\displaystyle a^2 = ab$
$\displaystyle a^2-b^2 = ab-b^2$
$\displaystyle (a-b)(a+b) = b(a-b)$
Zero division here
$\displaystyle a+b = b$
$\displaystyle b+b = b$
$\displaystyle 2b = b$
$\displaystyle 2=1$
This one can be found on Wikipedia if my memory is correct. It is indeed a division by zero.
Indeed. I had a high school teacher. I was interested in his knowledge. I asked him something a bit advanced (modular arithmetic). He looked at me, blinked, and told me to finish my worksheet. Lame. I don't recall having any math teacher that actually was interested in maths.high school teachers are mostly idiots who only know what the school expects them to teach.
Aww, teaching is so much more, so much more effective when the teacher enjoys what he's talking about, it really gives a different feel to the lesson (as I experienced it in biology). Sadly this world is not about a better education but about higher profit (high school teachers are cheap to manufacture)