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Originally posted by lauseyBut that's not the point. For me to go from 2(a-b)=a-b to 2=1 I have to divide by a-b which is 0 as you already said. But we can't divide by 0. On the other hand 2(a-b)=a-b has nothing wrong with that since as you already noticed it is a valid affirmation.
2(a-b)=a-b
a-b=0
therefore 0=0, not 2=1
Well, you already have 2(a-b) which is already 0.
Therefore you are doing a 0/0 which is 0. 🙂
But 0/0 isn't 0. 0/0 is nothing at all unless your calculating some limit like lim_{x->a}f(x)/g(x) and both functions goes to 0 as x goes to a. That's the only meaningfull way to calculate 0/0.
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Originally posted by adam warlockOk, so 2(a-b)=a-b is just as uncalculable as 0/0.
But that's not the point. For me to go from 2(a-b)=a-b to 2=1 I have to divide by a-b which is 0 as you already said. But we can't divide by 0. On the other hand 2(a-b)=a-b has nothing wrong with that since as you already noticed it is a valid affirmation.
But 0/0 isn't 0. 0/0 is nothing at all unless your calculating some limit like lim_{x->a}f(x)/g ...[text shortened]... d both functions goes to 0 as x goes to a. That's the only meaningfull way to calculate 0/0.
or rather dividing both sides by (a-b) is, considering (a-b) is 0.
EDIT: Assuming uncalculable is a word. 😉
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Originally posted by MathurineFlaw in this one though is doing a square root of both sides of this part:
Well, 5 might equal six. After all, 4=5...
Proof:
-20 = -20
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5
Does 4=6, though?
(4 - 9/2)^2 = (5 - 9/2)^2
Considering the square root of anything is +ve and -ve. In this case, the left being -ve and right being +ve.
This is correct:
-4+9/2 = 5-9/2
