Encircle the error in the following "proof" that the two arbitrary numbers are equal to each other. Let a and b the arbitrary numbers such that a is not equal to b. Then,
(a - b)^2 = a^2 - 2ab + b^2 = b^2 - 2ab + a^2
(a - b)^2 = (b - a)^2
a - b = b - a
2a = 2b
a = b
There is a + and a - root of each squared term.
Just because (-4)^2 = 4^2 does not mean -4 = 4.
Hi Steve! So with your explanation, am I correct in saying that the error in the above given is (a - b)^2 = (b - a)^2?
no, it's the next line. The squares are equal; it's the assumption that the roots are equal which is wrong.
The error in the "proof" is when the statement (a - b)^2 = a^2 - 2ab + b^2 is concluded as (b - a)^2 = a^2 - 2ab + b^2.
To see why this is an error, let's go through the steps:
1. The proof begins by assuming two arbitrary numbers, a and b, such that a is not equal to b.
2. It then squares the expression (a - b), resulting in (a - b)^2 = a^2 - 2ab + b^2. This step is correct.
3. Next, it rewrites the expression (a^2 - 2ab + b^2) as (b^2 - 2ab + a^2). This step is also correct, as addition is commutative.
4. However, the error occurs when the proof concludes that (a - b)^2 = (b - a)^2. This is incorrect because squaring (a - b) and squaring (b - a) are not equivalent. Squaring a number does not depend on the order of its operands.
In other words, the equality (a - b)^2 = (b - a)^2 is not valid in this case.
Therefore, the conclusion that a - b = b - a is incorrect. It does not imply that 2a = 2b or a = b.
Hence, the error in the "proof" is the incorrect conclusion that (a - b)^2 = (b - a)^2.