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If you were teaching a more formal method, I can see marking it wrong. But you're mistaken to think the student can't find the answer that way. If I had to solv

If you were teaching a more formal method, I can see marking it wrong. But you're mistaken to think the student can't find the answer that way. If I had to solve x^3 = 7 in my head, I know that 2^3 = 8, and my next guess would be 1.9 1.9^3 = 6.859 (ok, I'd need a pencil for this) Now I know the answer to two digits, "x is just a bit more than 1.9", and one can keep going if more precision is needed.

yequalsx 10y ago
The student demonstrated that they can do such problems with positive integer solutions. Possibly even negative integer solutions. But certainly they did not show any sort of reasoning along the lines you wrote. But let's take your method. Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. But this isn't the solution to the equation. The solution to the equation in question is a number and not a sequence. The answer, over the reals, is 7^(1/3).
- BeetleB 10y ago
  >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.
  - yequalsx 10y ago
    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.
  - omaranto 10y ago
    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.
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- gizmo686 10y ago
  >The answer, over the reals, is 7^(1/3). Math major here. 7^(1/3) is defined as the solution to x^3 = 7. In this case I would accept it as an answer because 7 does not have a rational cube root, and the question is presumably testing if the students knows fractional exponents [0]. However, in the case of x^3 = 8, I would not accept 8^(1/3) because the students has not actually found the answer; they have merely written the question in a different way. I am also curious what method you would propose the student use to compute 8^(1/3), as all the methods I know degrade to guess and check in the single digit case. You could say that simplifying to x^3 = 8 to x=8^(1/3) is the first step to solving it; to which I would reply that simplifying x=8^(1/3) to x^3 = 8 is the first step to solving it. [0] I could also imagine another math class where I would mark 7^(1/3) as wrong because the student has not actually found the answer, merely written the question in a different way. Presumably we are not talking about such a situation.
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js2 10y ago
This thread is turning into Feynman vs the abacus, is it not? http://www.ee.ryerson.ca/~elf/abacus/feynman.html http://www.ee.ryerson.ca/~elf/abacus/feynman.html
- yequalsx 10y ago
  Yes it is. Thanks for the link.
- paulsutter 10y ago
  funny story, quoting: ``` "Tell me," he said, "how were you able to do that cube-root problem so fast?" I started to explain that it was an approximate method, and had to do with the percentage of error. "Suppose you had given me 28. Now the cube root of 27 is 3 ..." He picks up his abacus: zzzzzzzzzzzzzzz— "Oh yes," he says. I realized something: he doesn't know numbers... Furthermore, the whole idea of an approximate method was beyond him, even though a cubic root often cannot be computed exactly by any method. ```

yequalsx 10y ago

The student demonstrated that they can do such problems with positive integer solutions. Possibly even negative integer solutions. But certainly they did not show any sort of reasoning along the lines you wrote. But let's take your method. Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. But this isn't the solution to the equation. The solution to the equation in question is a number and not a sequence. The answer, over the reals, is 7^(1/3).

BeetleB 10y ago
>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.
- yequalsx 10y ago
  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.
- omaranto 10y ago
  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.
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gizmo686 10y ago
>The answer, over the reals, is 7^(1/3). Math major here. 7^(1/3) is defined as the solution to x^3 = 7. In this case I would accept it as an answer because 7 does not have a rational cube root, and the question is presumably testing if the students knows fractional exponents [0]. However, in the case of x^3 = 8, I would not accept 8^(1/3) because the students has not actually found the answer; they have merely written the question in a different way. I am also curious what method you would propose the student use to compute 8^(1/3), as all the methods I know degrade to guess and check in the single digit case. You could say that simplifying to x^3 = 8 to x=8^(1/3) is the first step to solving it; to which I would reply that simplifying x=8^(1/3) to x^3 = 8 is the first step to solving it. [0] I could also imagine another math class where I would mark 7^(1/3) as wrong because the student has not actually found the answer, merely written the question in a different way. Presumably we are not talking about such a situation.

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js2 10y ago

This thread is turning into Feynman vs the abacus, is it not? http://www.ee.ryerson.ca/~elf/abacus/feynman.html http://www.ee.ryerson.ca/~elf/abacus/feynman.html

yequalsx 10y ago
Yes it is. Thanks for the link.
paulsutter 10y ago
funny story, quoting: ``` "Tell me," he said, "how were you able to do that cube-root problem so fast?" I started to explain that it was an approximate method, and had to do with the percentage of error. "Suppose you had given me 28. Now the cube root of 27 is 3 ..." He picks up his abacus: zzzzzzzzzzzzzzz— "Oh yes," he says. I realized something: he doesn't know numbers... Furthermore, the whole idea of an approximate method was beyond him, even though a cubic root often cannot be computed exactly by any method. ```