# Freind of mine showed me this 3= 4?

1. Mar 27, 2005

### 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

2. Mar 27, 2005

### 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

3. Mar 27, 2005

### 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.

4. Mar 27, 2005

### 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 $a=b \Longrightarrow a-b = 0$.

5. Mar 27, 2005

### Alkatran

4(a+b-c) = 3(a+b-c)
4*0 = 3*0
ERROR -> 4*0/0 = 3*0/0
4 = 3

6. Mar 28, 2005

### 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. :)

7. Mar 28, 2005

### whozum

We all start somewhere.