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Will Mainland HS ever fix its algebra page?

by Stephen Tashi
Tags: algebra, mainland
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Stephen Tashi
#1
Nov20-11, 07:50 PM
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An algebra tutorial on the Mainland High School Algebra Lab stie
(http://www.algebralab.org/lessons/le...eEquations.xml)

Solves (example ii):

[itex] |2x -3] = x - 5 [/itex]

and obtains solutions [itex] x = -2 [/itex] and [itex] x = 8/3 [/itex]

I've emailed them about this twice and they haven't corrected it yet. Are there algebra texts that actually teach the fallacious method of solution used on that page?
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Hurkyl
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Nov20-11, 10:21 PM
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Quote Quote by Stephen Tashi View Post
Are there algebra texts that actually teach the fallacious method of solution used on that page?
The method is fine; it is definitely true that
|2x-3| = x-5
implies
x=-2 or x=8/3
They've just forgotten to finish the problem from there.
D H
#3
Nov20-11, 10:41 PM
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Quote Quote by Hurkyl View Post
They've just forgotten to finish the problem from there.
That finishing off bit is a bit of an important point with this problem, since this problem has no solutions.

rootX
#4
Nov20-11, 11:10 PM
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Will Mainland HS ever fix its algebra page?

I agree with D H.

If you notice, they have provided incomplete procedure for solving these problems:
To solve |expression| = k, you must solve two separate equations:

exp =k and exp = -k
it's not just one problem they need to fix. They need complete procedure and complete all other problems also.
Stephen Tashi
#5
Nov21-11, 12:18 AM
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The appropriate way to solve the problem would be to use the definition of the absolute value function - namely:

By definition [itex] |a| = a [/itex] if [itex] a \ge 0 [/itex] and [itex] |a| = -a [/itex] if [itex] a < 0 [/itex].

So we have two cases.

Case 1) Assume [itex] 2x - 3 \ge 0 [/itex] This implies [itex] x \ge 3/2 [/itex] and
[itex] | 2x - 3 | = 2x -3 [/itex]. So we solve [itex] 2x -3 = x - 5 [/itex] obtaining [itex] x = -2 [/itex] However, we have assumed [itex] x > 3/2 [/itex], so [itex] x = -2 [/itex] is not a solution.

Case 2) Assume [itex] 2x -3 < 0 [/itex]. This implies [itex] x < 3/2 [/itex] and [itex] |2x -3|= -(2x -3) = -2x + 3 [/itex]. So we solve [itex] -2x + 3 = x - 5 [/itex] obtaining [itex] x = 8/3 [/itex]. But we have assumed [itex] x < 3/2 [/itex], so this is not a solution.

This method has the advantage of teaching the formal definition of absolute value. It trains students to realize that "minus a quantity" need not be a negative number and all the conditions that are required of the solution are deduced.

The method on the page uses some mumbo-jumbo about "opposites" and could be kludged up by telling students that they must "always check your solutions". That's not bad advice with any method, but why use a method that avoids proper deductive reasoning. The correct method can be generalized to solve equations like [itex] |2x - 3| = |x - 5| [/itex] by cases. I don't know how you apply the method of "opposites" to such equations.

In view of the thread http://www.physicsforums.com/showthread.php?t=377783, we might get Bruce Tonkin to weigh in on this matter.
Hurkyl
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Nov21-11, 04:21 AM
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Quote Quote by Stephen Tashi View Post
The appropriate way to solve the problem would be to use the definition of the absolute value function - namely:
Not "The". "A".

The method on the page uses some mumbo-jumbo about "opposites" and could be kludged up by telling students that they must "always check your solutions". That's not bad advice with any method, but why use a method that avoids proper deductive reasoning.
Where do you get the idea it's improper? It is not difficult to conclude "x=y or x=-y" using "proper deductive reasoning" from the equation "|x|=y".

The approach is perfectly good, and more efficient to boot.

The correct method can be generalized to solve equations like |2x−3|=|x−5| by cases. I don't know how you apply the method of "opposites" to such equations.
Both methods are correct. And you apply the method of "opposites" in exactly the same way for such a problem.

(in fact, the "method of opposites" is reversible in this case: |x|=|y| is true if and only if x=y or x=-y)

It is possible there are good arguments that teaching to solve by cases is pedagogically better -- but the one you give is simply not one of them.
D H
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Nov21-11, 05:20 AM
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What's important is to teach that (a) sometimes what appears to be a solution is not a solution; one has to check that it is a solution, and (b) students should always check their work.

In this case I would have used the very advice presented in the next problem: "Before jumping into solving this problem, think about it first." (Their problem iii should have been the very first problem.) "Thinking about it first" in the case of problem ii says that to ensure that the left hand side is not negative we must have x≥5. This in turn means that one need not check the alternative -|2x-3|=x-5 because 2x-3≥8 for all x≥5. So there is at most one solution here, found by solving 2x-3=x-5. The solution, x=-2, is not a solution to the original problem. There are no solutions to this problem!

While a bit verbose for this problem, eliminating entire branches from consideration by "thinking about it first" can save a lot of time on more complex problems.
Stephen Tashi
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Nov21-11, 11:06 AM
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Quote Quote by Hurkyl View Post

Where do you get the idea it's improper? It is not difficult to conclude "x=y or x=-y" using "proper deductive reasoning" from the equation "|x|=y".
If it's not difficult to do then why not do it? And why is that any simpler than showing the cases are x = y or -x = y ?


The approach is perfectly good, and more efficient to boot.
It's efficient in the sense that it's incomplete.


Both methods are correct. And you apply the method of "opposites" in exactly the same way for such a problem.

(in fact, the "method of opposites" is reversible in this case: |x|=|y| is true if and only if x=y or x=-y)
The correct method, using standard mathematical definitions gives the same thing. There is no standard definition for "opposite" in algebra.

It is possible there are good arguments that teaching to solve by cases is pedagogically better -- but the one you give is simply not one of them.
It seems to me that when a problem can be solved by a deductive approach using standard definitions, that the burden of proof for doing otherwise is on person who proposes an alternate technique.

The advantages of the alternate technique in this case are that it does not burden students with learning the definition of the absolute value function, it does not require precise deductive thinking, and it reinforces the advice that one should check answers.
D H
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Nov21-11, 11:29 AM
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Stephen: You are arguing with Hurkyl over the correct way to say "tomato." Both his approach and yours will yield the same answers to any and all questions of this form. It's a bit silly to argue over which approach is the "right" approach since the two are ultimately equivalent.

As for why they won't fix it, perhaps you are being a bit too argumentative in your messages or too adamant in telling them that the very approach they are using is wrong. That's just a perhaps; I don't know the history of your communications with them. Do you really care whether they use your approach, or Hurkyl's, or someone else's, so long as the approach they do use does yield the correct answers?
Stephen Tashi
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Nov21-11, 12:36 PM
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Quote Quote by D H View Post
Stephen: You are arguing with Hurkyl over the correct way to say "tomato." Both his approach and yours will yield the same answers to any and all questions of this form. It's a bit silly to argue over which approach is the "right" approach since the two are ultimately equivalent.
If you are grading a student's paper and he gets the correct answer, but doesn't show work that derives it deductively, does he get the same mark as a student who shows a deductive process? I'm not saying the answer to this question is necessarily "no". But it points out how the two approaches are not equivalent.

A similar debate could be had over whether to teach students to understand the distributive law or just teach them "the FOIL method". On the one hand, given that there are many achievement tests in secondary education nowadays and that teaching precise reasoning is a burdensome task for teachers, there are pragmatic reasons for teaching students how to get the right answers quick at the expense of how to get right answers in some deductive manner. The deductive manner isn't likely to be the subject of a multiple choice question.

Suppose a person is writing a proof and must state the consequences of [itex] |3x - \delta| < |4 \epsilon + 2| [/itex]. What will the correct approach be? To use the "method of opposites" and then try to "check your answer"? Granted that not all students of secondary mathematics are going to proceed into such a situation. And perhaps as students proceed in mathematics they must periodically suffer disappointments when the way they learned some things in high school is not recognized as valid.


The "method of opposites" can be fixed so it uses valid deductions. The procedure on the Mainland site could be solved by introducing a theorem: If |a] =b then a = b if and only if a >= 0 and a = -b if and only if a < 0. Then the cases to work the example would be

Case 1: 2x - 3 = x - 5 and 2x - 3 >= 0
Case 2: 2x -3 = -(x-5) and 2x -3 < 0

I don't see that's any simpler than the cases:

Case 1: 2x - 3 >= 0 and 2x-3 = x-5
Case 2: 2x -3 < 0 and -(2x-3) = x -5

I don't see that introducting the theorem and avoiding the use of the definition of the absolute value function has any pedagogical advantage. In subsequent math courses, the definition of the absolute value function will be the more important concept.



As for why they won't fix it, perhaps you are being a bit too argumentative in your messages or too adamant in telling them that the very approach they are using is wrong. That's just a perhaps; I don't know the history of your communications with them. Do you really care whether they use your approach, or Hurkyl's, or someone else's, so long as the approach they do use does yield the correct answers?
I'm not involved in secondary education now, but when I had contact with it, it always amused me to hear mathematics instructors enjoin students to "think logically" and contrast this with the way that the material is presented. It isn't crucial to me whether or how Mainland HS fixes anything. I do find it an amusing task to see if I can provoke them to do something. It's somewhat like buying a cheap product, having it break and then, just for the "fun" of it, trying to jump through all the hoops to get the manufacturer to honor a warranty. I'm a retired guy. I have time for such inane adventures.

It's silly to argue with a closed mind, but reasonable people can argue profitably.
Borek
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Feb22-12, 08:12 AM
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Quote Quote by Stephen Tashi View Post
An algebra tutorial on the Mainland High School Algebra Lab stie
(http://www.algebralab.org/lessons/le...eEquations.xml)
LOL, "lesson not found".
Stephen Tashi
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Feb22-12, 09:32 AM
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Quote Quote by Borek View Post
LOL, "lesson not found".
Several months ago, I posted about the topic on a woodworking site. A member who happened to be a mathematics instructor took up the cause and was able to get an email response from Mainland, who said they would take the page down and revise the lesson.

Now I know which web sites hold the real power in mathematical affairs!
Astronuc
#13
Feb22-12, 09:51 AM
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The link to the page has been changed.

http://www.algebralab.org/lessons/le...equalities.xml
Stephen Tashi
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Feb22-12, 10:03 AM
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Quote Quote by Astronuc View Post
The link to the page has been changed.
Thanks for the link. The page has been revised.


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