Freind of mine showed me this 3= 4?

[phEight]

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

 

[Data]

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

 

[tutor69]

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.

 

[Data]

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.

 

[Alkatran]

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

 

[phEight]

Thanks for giving me a more "in depth" look at what is most probably basic mathematics to most of you. I'm still learning. :)

 

[whozum]

We all start somewhere.