Consider again three arbitrary numbers a, b, c, such that
Then:
Reorganize and simplify:
3(a+b-c) = 2(a+b-c)
3 = 2
QED.
Note that you may use the same scheme to prove x+1=x for any x.
Please, avoid posting spoilers in the comments. For other sophisms check out this list.
You mixed up the signs a bit in the end :-)
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True, thanks for noticing!
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The sophism and puzzle and math eludes me, but I love your informational posts!!!
A real shame that your very first post only garnered $0.17 or something!!
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On the other hand, being the first post of a new user it could have been left unnoticed at all, so perhaps $0.17 is still a fair result.
How can one not appreciate the elegance of a fake yet at first sight convincing proof that 3=2?
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Hahaha...
Because it was 16 years ago since I had Cal2 and before that for earlier math, and if it comes up with 3=2, then my dulled brain just says...."so (3a-2a)+(3b-2b) = (3c-2c) can't be" "it is inherently incorrect"
But I'm waiting to hear back from my 'math friend', he will probably do a good job enlightening me! ✌
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Well, things cannot be "inherently incorrect" for no reason - the question is why and where exactly is the proof incorrect.
Finding the mistake is a nice little exercise for un-dulling the brain, which provides a small "aha!" discovery moment. Such small discoveries are in fact among the nicest things in math and "real science" in general. Here you get a chance to experience them first hand, without the need to be a real mathematician or a scientist.
Don't ask your "math friends", you'll ruin the point. Try finding the bug yourself - in this particular case it is not too hard - just substitute some arbitrary suitable numbers for a, b, c, step through the proof checking whether each equation holds with these, and you'll see what is happening there.
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