Freind of mine showed me this 3= 4?

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The discussion revolves around a mathematical manipulation that leads to the erroneous conclusion that 4 equals 3. The key point is that the manipulation involves dividing by zero, specifically when (a+b-c) equals zero, resulting in an indeterminate form. Participants clarify that the assumptions made in the equations are flawed, as they rely on dividing by zero, which is not permissible in mathematics. The conversation emphasizes the importance of recognizing such mathematical pitfalls and understanding that the conclusion is nonsensical. Ultimately, the consensus is that 4 cannot equal 3 due to the invalid operations performed.
phEight
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what am i missing here? this seems to make sense...

a + b = c



4a - 3a + 4b - 3b = 4c - 3c



4a + 4b - 4c = 3a + 3b - 3c



4(a+b-c) = 3(a+b-c)



4 = 3
 
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You divided by zero.

You can do it even more simply:

a=b
4a = 3a+b
4a - 4b = 3a - 3b
4(a-b) = 3(a-b)
4=3

it's equivalent to saying

3 \cdot 0 = 4 \cdot 0 \Longrightarrow 3 = 4

which is silly
 
4(a+b-c) = 3(a+b-c)

This statement can be true only if (a+b-c) = 0

If you try to divide both sides by ( a+b-c) then it becomes 0/0 i.e. indeterminate form of the function. hence 4 is never equal to 3, at least not for a mathematician.
 
It's even worse. We KNOW a+b-c=0. Note his assumption, a+b = c. Just like in my example, where I assume a=b \Longrightarrow a-b = 0.
 
phEight said:
what am i missing here? this seems to make sense...

a + b = c



4a - 3a + 4b - 3b = 4c - 3c



4a + 4b - 4c = 3a + 3b - 3c

4(a+b-c) = 3(a+b-c)
4*0 = 3*0
ERROR -> 4*0/0 = 3*0/0
4 = 3
 
Thanks for giving me a more "in depth" look at what is most probably basic mathematics to most of you. I'm still learning. :)
 
We all start somewhere.
 

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