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>Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution.

...

>The answer, over the reals, is 7^(1/3).

So if the student did not write 2, but wrote 8^(1/3), it is OK?

As for the method being wrong, no - it isn't. The sequence converges. If you want an exact answer, you should specify that you want an exact answer.

>The solution to the equation in question is a number and not a sequence.

Sorry, but every number is a sequence. You can start with rational numbers, and define every real number as a sequence of rationals. Lots of books actually do this to define what a real number is.




Minor correction: you probably meant that books often define real numbers as equivalence classes of sequences of rational numbers. I've never seen a definition of real numbers as a single sequences in a textbook, not because it is impossible, but because defining arithmetic operations would get really messy.


8^(1/3) is 2 (when interpreted over the reals).

Your method is wrong because it does not produce the answer. One does not, in college algebra, say the solution is:

lim a_n as n->infinity

For one thing you did not prove convergence. It is understood that solutions to algebraic equations over the reals are numbers and not approximations. Any student who knows about Dedekind cuts or infinite sequences knows to take the cube root of both sides.

Most mathematicians consider it silly (strange, wrong) to say that 2 is an element of 3 even though it is.

Anyone solving x^3 = 27 in the method specified does not know any of these finer points of mathematics. The method is bad. It's useful for positive integer solutions but not for the general situation.




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