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.
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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.
The reasonis because
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It is not surprising. I have come to learn over the years, high school teachers are mostly idiots who only know what the school expects them to teach. Which is not much (Rofl)
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.Quote:
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)
Ignoring the last equation and simplifying the second to last one, I got 2A^2-2A^2=A^2-A^2, which is of course 0=0.
I know it's late, but I didn't look at all your guys' answers and solved it on my own, and wanted to check if my solution was correct.