Quod Erat Demonstrandum


[初中] Identities

Filed under: Junior Form Mathematics — johnmayhk @ 4:22 下午

The following is a level zero question in F.2 math textbook.

Determine whether 4(3x + 1) = 2(6x + 2) is an identity.

Being a math teacher, I must suggest the following solution.

= 4(3x + 1)
= 12x + 4

= 2(6x + 2)
= 12x + 4

\because L.H.S. = R.H.S.

\therefore 4(3x + 1) = 2(6x + 2) is an identity.

Everything fine.

But we may have alternative ways.

One student, Lam, gave

= 4(3x + 1)
= 2\times 2(3x + 1)
= 2(6x + 2)
= R.H.S.

Quite good.

For discussion purpose, I gave two more ways.

Method 1

4(3x + 1) = 2(6x + 2)
12x + 4 = 12x + 4
0 = 0

Do you accept this solution? Why or why not?

Method 2

When x = 0,
L.H.S. = 4(3(0) + 1) = 4
R.H.S. = 2(6(0) + 2) = 4
L.H.S. = R.H.S.

When x = 1,
L.H.S. = 4(3(1) + 1) = 16
R.H.S. = 2(6(1) + 2) = 16
L.H.S. = R.H.S.

Hence 4(3x + 1) = 2(6x + 2) an identity.

What is your comment on Method 2? For me, it is OK!

3 則迴響 »

  1. About Method 1,
    from 4(3x+1) = 2(6x+2) to 0 = 0, variable x is eliminated.

    So it means for all values of x, L.H.S. = R.H.S.
    As an identity, the meaning(definition?) is for all values of variables, L.H.S. = R.H.S.
    In the other words, it is an identity if and only if L.H.S. = R.H.S. for all values of the variables.
    Therefore, I accept the Method 1, but needed to add an explanation. (Use this to do M.C. is the best ^^)

    About Method 2, I don’t think it is ok.
    For example, x+x = 2x*x

    When x = 0,
    L.H.S. = (0)+(0) = 0
    R.H.S. = 2(0)*(0) = 0
    L.H.S. = R.H.S.

    When x = 1,
    L.H.S. = (1)+(1) = 2
    R.H.S. = 2(1)*(1) = 2
    L.H.S. = R.H.S.

    So by Method 2, it is an identity but actually it is not.

    When x = 2,
    L.H.S. = (2)+(2) = 4
    R.H.S. = 2(2)*(2) = 8
    L.H.S. not equal to R.H.S.
    So, by definition, it is not an identity.

    So testing by substitution is not the best way,
    as there may be a counter-example but we cannot see it,
    unless you know the counter-example(to prove it is not).

    But be careful of divided by 0 problem.
    For example, (2x*x+x)/x = 2x+1
    L.H.S. = x(2x+1)/x = 2x+1
    R.H.S. = 2x+1
    So L.H.S. = R.H.S.

    There is a mistake. The steps in L.H.S. have an assumption. x not equal to 0.
    When x = 0,
    L.H.S. = (2(0)*(0)+(0))/(0) = undefined (divided by 0)
    R.H.S. = 2(0)+1 = 1
    So L.H.S. is not equal to R.H.S.

    Then, how about this?
    2(x+1)/x = (2x+2)/x
    When x = 0, is undefined = undefined?

    迴響 由 C — 2007/12/02 @ 11:27 下午 | 回覆

  2. Method1: obviously we cannot do the first step since it assumes the proposition is correct.

    Method2: is it a “simplified" MI….?

    John Sir, wanna share something with you:
    When i discuss with my relative who teaches maths in university, he said that the maths syllabus in HK is extremely unhealthy. For example, some university students, who got As in maths subjects, can hardly explain why a/(b/c) = (ac)/b. (In fact I am not very sure about the reason, and therefore I nearly dug a hole and hid in it at that moment!) They just follow what the teachers said and practise for the exam. This learning (or teaching) attitude will put a devastating effect on students’ mind and distort the original meaning of maths training, logic deduction for example.

    P.S. he joked a means to solve the problem: oral exam in maths asking students to explain some math concepts.

    迴響 由 SK — 2007/12/03 @ 12:36 上午 | 回覆

  3. Thank you C and SK for your comments. As in the sharing of SK and my previous post about the talk, many students (or even teachers like me) may ignore the importance of fundamentals in learning mathematics. But the examination-oriented attitude of studying is rooted deeply in our city. I agree to introduce the oral examination in mathematics; it is actually not a new thing. SK, you have a relative teaching math in university, good news! You may ask him or her to clarify some mathematics concepts and know more in this field if you are really interested.

    Talking about my post. Two different opinions about method 1 given by C and SK.

    As a lower form mathematics teacher, I do not accept my students give method 1 and I will say it is not a ‘good’ presentation.

    But, if we know some simple logic, like iff, it is actually OK to give method 1. Explicitly, if I present the same thing as follows.

    0 = 0 (trivially true)
    => 0 + 12x \equiv 0 + 12x (x can assume any value, hence I use \equiv)
    => 12x \equiv 12x
    => 12x + 4 \equiv 12x + 4
    => 4(3x + 1) \equiv 2(6x + 2)


    May be it is clearer when I write

    4(3x + 1) \equiv 2(6x + 2)
    iff 12x + 4 \equiv 12x + 4

    SK claimed that the proposition should not appear at the first step. Urm, if we are sure that every step is connected by “iff", it is OK to write the proposition in the first line. However, to be safe during examinations or quizzes, never write it as the first step. And I will clearly tell my F.2 boys, it is ‘wrong’ to give method 1. Yes, for lower form students, good habbit may be more important.

    For method 2, I think it is OK. Students taking pure mathematics may have come across the following.

    Suppose f(x), g(x) are polynomials of degree n (n \geq 1) over \mathbb{R}, and there are more than n distinct roots satisfying

    f(x) = g(x)


    f(x) \equiv g(x)

    Now, since both L.H.S. and R.H.S. are polynomials of degree 1, and there are two roots, namely 0 and 1, satisfying the equation

    4(3x + 1) = 2(6x + 2)


    4(3x + 1) \equiv 2(6x + 2)

    For the example given by C

    \frac{2(x+1)}{x} = \frac{2x+2}{x}

    generally, we will discuss the situation on the domain, hence we will ignore the case of x = 0.

    迴響 由 johnmayhk — 2007/12/03 @ 2:03 下午 | 回覆

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