What is wrong with the following proof? Clearly we know that 1 does not equal 2. What rule of algebra did we break that makes this conclusion invalid? I've numbered the lines for you so you can refer to the line numbers in your explanation
An attempt to divide by zero gives a contradictory result
A rule of algebra broken is dividing by zero (leading to a contradiction) and stating a finite result
Reason:
The given calculation is presented as follows;
1. a > 0, b > 0 given
2. a = b given
3. a·b = b²
4. a·b - a² = b² - a²
5. a·(b - a) = (b + a)·(b - a)
6. a = b + a
7. 0 = b
8. b = 2·b
9. 1 = 2
From line 4, the result are;
4. a·b - a² = b² - a² = 0
5. a·(b - a) = (b + a)·(b - a) = 0
On line 6, both sides where divided by (b - a) = 0, which should given an infinite result
Therefore, one rule of algebra broken is dividing by zero to get a finite result
In line 7, we have;
7. 0 = b
From 6. a = b + a, and a = b, we have;
8. b = 2·b
Therefore, line 9 should be;
9. 0 = 2·0; 0 = 0, given that we have;
1 × 0 = 0
2 × 0 = 0
∴ 1 × 0 = 2 × 0
However
In line 9., by dividing by b = 0, again, we have;
Therefore, one rule of algebra that is broken is dividing by zero and having a finite result
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