Can you find my mistake?

so should i not answer it? i think i know where the mistake is...but i don't want to spoil it for others trying to figure it out!

Ditto,


This problem highlights a problem with the way we teach algebra.

I am actually working with a group developing an algebra curriculum.

I not only know the mistake, but I can tell you how we can teach students to avoid it.

It's easy to prove 1=2 if you do the undefined.

Rap

Think about the division by a-b very carefully.

good job brandon!

Cancellation of terms eliminates answers, rite?

goodun.

but that's not actually the problem is it? the division by (a-b)?

Edit: I think I know what the problem is too, but you're all confusing me by seemingly suggesting that the problem is that division.

Are we all ready to just come out with the spoiler. I would like to talk algebra theory.

I understand where Brandon and Euler are coming from and it seems like mr_me accepted their hints.

Vengo, why don't you explain the problem in your words (there are a couple of different ways of looking at it) and we can start from there.

Division by zero is undefined. If a=b then a-b=0.

Rap

This problem is interesting-- not for why people get it wrong, but for why people even think this crazy set of operations is right.

The problem is, the way we were all taught algebra is wrong. This is just one example.

First, let's talk about math... clearly the equation "a = b" is not eqivalent to the equation "2 = 0". I know this by looking at the solution set of these two equations.

The solution set of "a = b" has an infinite number of solutions because there are an infinite number of values of 'a' and 'b' that will make this equation true. For example [ a = 1; b = 1] or [a = 2.3 ; b = 2.3] and so on. We can (and often do) represent these solutions as points, that is (1, 1) and (2.3, 2.3). If you plotted the points in the solution set for "a = b", you would get a line.

Of course the solution set "2 = 0" is nothing because there are no values of 'a' and 'b' (or any other variable) that would make this equation true.

What can we do validly in algebra to assure that the result will be the equivalent to the original equation (that is the problem)?

The rule is simple. For an operation on an equation to be algebraicly valid it must not change the solution set. This statement is always correct. If an operation changes the solution set, it is wrong. Period.

This is not what algebra students are generally taught.

The incorrect rule we teach in algebra class
Algebra students are taught "whatever you do to one side of an equation, you must do to the other".

This is clearly often wrong. There are many times when doing the same thing to both sides of the equation changes the solution set-- bad idea. There also times when an operation does different things to both sides of the equation without changing the solution set-- of course this is perfectly OK.

Really this "do the same thing to both sides" thing is just a trick that get's students to kind of do what is right most of the time without understanding why,

Math is a language for understanding and solving problems. Students of algebra should gain the skills to think critically, use reason to solve problems and to understand the underlying math.

Instead, all to often we rely on simple mechanical rules. This case is all the more egregious in that the rule we are teaching students to rely on often leads to bad (incorrect) mathematics.

1.a=b
2.since a=b,
3.b^2=ab
4.now, subtract a(squared)
5.b^2-a^2=ab-a^2
6?.(a+b)(a-b)=a(a-b) <-this is the problem step, isn't it?
6 correct version.b^2-a^2=(b-a)(b+a)=a(b-a)=ab-a^2
7.since (b-a)(b+a)=a(b-a) we're at the same problem... maybe you're right.

Edit: realized it's wrong, fixing it now. well, I'll look over it more anyway. But I guess you were right about the division by zero deal.

Re: Can you find my mistake?


ok call me crazy but i just thought the problem here was that in the first line on the right side you have ab-a(squared) which factored would be

a(b-a) not a(a-b) , thus you would not be able to divide by (a-b)

you would have (a+b)(a-b) = a(b-a) which at this point you can't realy do much else...

maybe i am oversimplifying?

ok never mind, i just saw the flaw in that...

ebrown_p you seem to be saying that the reason the problem is wrong is because it is wrong. I see it, in my training, as doing the same thing to each side of the equal sign. The only problem with the actions id that at some place they set 0=0 and since, by the rules of arithmetic a*0=0 for all a.

So even if I knew a<>b, a*0=b*0 would be true.

As for math and algebra in particular, I wish I'd been taught algebra as an abelian group and linear algebra as one where commutation had been thrown away.

Rap

I just realized there's a simpler way to show this problem.
given, A=B
1. A=B Given
2. A-B=0 Arithmetic
3. (1/(A-B))(A-B)=(1/(A-B))0 Multiply both sides by 1/(a-b)
4. 1=0 Simplification of previous step

No. It is the rule that is wrong.

I am glad that someone else wants to talk theory.

What I am saying is that the rule you propose-- "do the same thing to each side of the equal sign" -- is an incorrect one.

The real rule is "You can do any operation as long as you don't change the solution set". This rule is always correct.

The "do the same thing to each side of the equal sign" at times breaks the real rule.

So why do people teach a rule that is at times incorrect? It is a short cut to the real rule, since in most cases doing the same thing to both sides of the equal sign will not change the solution set, but there are exceptions as this example shows.

My point is that we should teach the theory (i.e. you can't change the solution set) and then talk about why a particular operation does or does not meet it. Teaching the shortcut rule that people use without thinking about the theory is a bad thing... especially when the shortcut rule at times produces the wrong result.

In order to understand why algebraic manipulations work and to be able to prove with confidence that your manipulations are correct, you need to understand about solution sets. The shortcut rule gets in the way of this understanding.

I am currently working on a curriculum to teach solving system of linear equations. This problem gets even more interesting in this case.

My teacher has taught us the correct version, but that was in some weird version of pre-pre-calc or something.

ebrown_p

Linear Algebra is not abelian in multiplication. Consequently, you have to pay attention to multiplying on the left and right. That's a difficult concept to get across when everything you've been taught up to that point is abelian. Then there's the concept of an inverse, which is not really multiplying by the reciprocal (which in a way is an inverse under certain conditions).

BTW is this a kinda a return to new math? The concept of set theory, that is. Or are you introducing a modern algebra curriculum at the secondary level?

If so give me a link, I'm in the process of changing my career to education (math, chemistry, physics). So I've got more than a passing interest in your methods.

Rap