WitCH 141: Laws Unto Themselves

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July 23, 2024

9:17 pm

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We’ve done part of this before but it came up again recently and it annoyed us all over again. The excerpts below are from Cambridge‘s coverage of the index laws, at Years 8, 9 and 10. It includes everything we could locate that indicates in any sense why the index laws for general indices are what they are.

Cambridge have known for at least a decade what we think about their index crap, and obviously they don’t care. But, as long as they keep reprinting this crap we’ll keep calling it out for the crap that it is.

YEAR 8

YEAR 9

YEAR 10

 

UPDATE: BOX 1 (07/08/24)

OK, we’ll try to get this WitCH back on track, by updating on the first few boxes, box by box. This has mostly been covered by Red Five and other commenters, but there are a few things to spell out.

The fundamental crap, of course, and why this is really a PoSWW, is part b:

ONE CANNOT APPLY AN INDEX “LAW” UNTIL ONE KNOWS THAT THE LAW IS TRUE.

And, as is the point here, one cannot possibly know a law involving \boldsymbol{5^0} is true until one knows what \boldsymbol{5^0} means. Which is stated to be the entire purpose of the damn exercise.

For the life of me, I cannot understand why writers fail to grasp, or at least give a shit about, this fundamental logical point. Whatever, we’ll keep hammering this until our dying breath if necessary, and we’re ready to have it in engraved on our tombstone.

Now back to the beginning, the preamble to the exercise – but why the Hell is this relegated to an exercise? – is kind of sort of OK, but not really. Yes, the basic point is to determine that \boldsymbol{a^0} “should” equal 1. But how is a kid going to read that “should”? How many kids are going to read it as “how \boldsymbol{a^0} needs to be defined for the index law to remain true in this new context”? Of course it ain’t gonna happen.

So, part a is done, and we’ve discussed the fundamental idiocy of part b, and so it’s on to part c, … But wait a minute. … Where is the punchline to parts a and b?

Idiots.

Anyway, c and d are ok, if as interesting as batshit. But why not have a part requiring the argument for general \boldsymbol{a \neq 0}? Then, part e could focus on where the assumption \boldsymbol{a \neq 0} was used in that argument.

Finally, part e is poorly worded, because it reads as if simply the argument for \boldsymbol{a = 0} breaks down, rather than as a clear statement that, Desmos wanking and Woo wanking notwithstanding, \boldsymbol{0^0} cannot reasonably be defined.

“There, the box is done”, muttered a tired Sisyphus.

 

UPDATE: BOX 2 (07/08/24)

At this stage we can largely just point to the sign: read the red message above. But, there is more to say.

Part a is screwed, because the index “laws” don’t tell us anything about what \boldsymbol{9^{0.5}} is: we don’t yet know the “law” is true for fractional indices, in particular because we don’t yet know (i.e. we haven’t yet decided) what \boldsymbol{9^{0.5}} means.

Note, however, the writers didn’t even bother with the “should” framing here: they just went with “is”. And, once again, why is this all relegated to an exercise? And, using 0.5 rather 1/2 is appalling, but it’s probably too much to expect these calculator-happy decicentric clowns to understand why.

Also, who writes “What number is equal to \boldsymbol{9^{0.5}}?” Who comes up with such a construction?

Part b is ok, except that the wording is again extraordinarily clumsy. Try “What is the positive solution to the equation \boldsymbol{x^2 = 9}?”

Part c, and we finally get our “should”, accompanied by a genuflection to the All-Wise Calculator.

Part d, is boring and not sufficiently general, and the choice of “observation” and “prediction” is lazy and inaccurate.

“Good to have that rock back up on top of that hill.”

 

UPDATE: BOX 3 (07/08/24)

OK, we’re on to Year 9, which somehow got worse. Yes, it is in the central text rather than in the exercise suburbs, but it is worse. Because the vague but at least vaguely correct “should” has been replaced by an absolutely incorrect “implies”.

Also, the second the sentence is ungrammatical.

“OK Rock, now stay there.”

 

UPDATE: BOX 4 (11/08/24)

This WitCH seems kind of dead, so we’ll just get on with updating it (as we should with the other hundred incomplete WitCHes).

The basic idea of the table is fine, if pedestrian: consecutive powers of 3 differ by by a multiple of three. But the second question in Box 4 is nuts. What conclusion do we come to regarding \boldsymbol{3^0}? We, we just filled in the damn table, so we “conclude” whatever we had already “concluded” about \boldsymbol{3^0}. The table itself does nothing.

 

UPDATE: BOX 5 (11/08/24)

The preamble in Box 5 is not great, although it is OK. The problem, of course, pointing once again to the red sign, is that one cannot here “use the index law for division” to “therefore” conclude anything. Beyond that: although it is reasonable to not want to crowd the box, the necessary assumption that a ≠ 0 needs to be mentioned somewhere; the layout is poor, with “for division” seeming to follow “by cancelling”; “therefore dots” is never good, and here it is bad, with clear and proper wording essential.

 

UPDATE: BOX 6 (11/08/24)

This is a continuation of Box 5, is awful in the same manner and in principle no worse: just the red sign again. Still, it is worse.

It is confusing, in one long comma-free sentence, to be determining both the value of \boldsymbol{a^{-m}} and to be concluding that \boldsymbol{a^{m}= \frac1{a^{-m}}}. Also, the appropriate conclusion is that the complicated expression \boldsymbol{\frac1{a^{-m}}} can be rewritten as the simple expression \boldsymbol{a^{m}}, not the other way around. These things matter.

 

UPDATE: BOX 7 (11/08/24)

This is similar, and similarly boring, to Box 4, but it is a lot worse.

To begin, the students do not “explore” the “use” of negative indices; no one is “exploring” or “using” anything, and these are just randomly chosen words to mean “do some stuff”. Next, the row to be filled in is not simply with a “whole number or fraction” but also, later, as a fraction with denominator in positive index form. Then, the student is supposed to “notice” something of the “numbers with negative indices” in the top row “in comparison to the fractions” in the second row, but it is not at all clear what one is supposed to “notice”. That the numbers are equal? Well, duh. That’s what the table is for. Or, that the negative indices become positive in reciprocal form? That’s not what you asked, and of course the kid will “notice” it, because how else would they have, or should they have, computed the numbers to fill in the table. Then the students are asked to “describe this connection formally in words”. Why? For what purpose? Finally, the students conclude with answering what “might” be another way of writing \boldsymbol{2^{-7}} “or” \boldsymbol{5^{-4}}. Even the “or” is annoying, but the entire question as a concluding question is absurd.

This is really appalling stuff. It has the gloss of thoughtfulness and discovery and conclusion, but it is all fake: poorly constructed, poorly written, confusing as all hell and utterly pointless.

 

UPDATE: BOX 8 (11/08/24)

It is on to fractional indices and of course we point, once again, to the red sign. The entire discussion is logical garbage. But, beyond being logical garbage it is garbage garbage.

The discussion is framed around \boldsymbol{\sqrt 9}, which is fine: one needn’t begin with the irrational. If however, one begins with an integral example then you also damn well point out that it is an integral example. The entire box includes not one single mention that \boldsymbol{\sqrt 9} or \boldsymbol{9^{\frac12}} simply equals 3. Yes, they eventually mention it, but way way too late.

Beyond that, it is not great that the argument provided also happens to “prove” that \boldsymbol{9^{\frac12}} equals -3. The argument is of the form \boldsymbol{f(9^{\frac12}) = f(\sqrt 9)} implies \boldsymbol{9^{\frac12}=\sqrt 9}, with the particular function \boldsymbol{f(x)=x^2}. The argument has to be of this form, even if properly framed as justifying the definition, but the argument must also be gently but properly clarified, to avoid proving (more) nonsense.

As well, the typesetting sucks, with the fractional powers way too large. And the layout sucks, with the computations just farting along, with zero emphasis in either the words or the placement.

Finally, and we just noticed it, a straight blooper: the text has written \boldsymbol{9^{\frac12}\times 9^{\frac12}=9^9}.

That’s the thing about a crappy layout: it is almost impossible for the reader to focus properly, to read what it is written.

 

UPDATE: BOX 9 (11/08/24)

This is a continuation of the previous box and is similarly corrupted: the red sign applies as always, and the awful typesetting is there again. But, each box seems to have its own unique treasure.

The box focuses on \boldsymbol{\sqrt[3] 5} and after the standard voodoo, the text concludes:

“This shows that numbers with fractional powers can be written using root signs.”

It is confusing at this late stage to (correctly) focus upon representation rather than equality, but it is backwards: the computation, properly couched, shows that rooty things can reasonably written as fractional powers.

Finally, the last line is a treasure of typesetting, seemingly identifying \boldsymbol{9^{\frac12}} with \boldsymbol{9(\sqrt9)}.

 

UPDATE: BOX 10 (11/08/24)

On to Year 10. The first Year 10 box is better and cleaner. It is complete garbage, of course – the red sign – but it is better laid out than the previous boxes. The phrases “rules for negative powers” is weirdly plural, and “rules” is not the best choice.

 

UPDATE: BOX 11 (11/08/24)

This box, although of course utter garbage – the red sign – is also better laid out, even if the Baby Huey fractions are still there as exponents. It seems a bit odd to only consider integer outputs but at least, and unlike box 8, these integers are properly incorporated into the computations. The “therefore” dots suck, as they always do.

 

UPDATE: BOX 12 (11/08/24)

To end, a complete non sequitur. To what “cases” is the “in most cases” supposed to refer? When do the index laws not apply to fractional indices? What possible use can it be to flag that the laws may not apply without indicating when the laws do or do not apply, let alone indicating why?

As near as we can tell, the text never provides a hint of what they are on about, or even provides a proper summary of the index laws. Which means they don’t know how to write, but it doesn’t mean they’re here on about nothing.

Presumably, the text is alluding to issues that arise with \boldsymbol{a^m} arise when the base \boldsymbol{a} is permitted to be negative. For example, the important and seldom justified identity

\boldsymbol{\sqrt{ab}=\sqrt{a} \times \sqrt{b}}

obviously fails if \boldsymbol{a} and \boldsymbol{b} are negative (modulo ComplexLand fixes). Similarly but more subtle, the identity

\boldsymbol{\left(a^m\right)^n=a^{mn}}

can run into weirdness; see this post.

What to do about this? Sure, one does not want to get too far into the weeds of negative bases and the oddities they contain. But a Year 10 text entirely ignoring negative bases until the Back o’ Bourke “Enrichment” exercises is pretty disgraceful.

In any case, it is not hard to issue a general “be careful with negative bases” warning, and obviously one is obligated to provide some clue as to the issue: the “in some cases” is farcical. But of course the writers and publishers don’t really care: as long as the text appears vaguely mathsy, that’ll do.

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60 Replies to “WitCH 141: Laws Unto Themselves”

  1. I haven’t looked at it in detail, but it seems rather all over the place. I was switched from Pearson to Cambridge this year and was disappointed. I had found the Y11 and 12 Cambridge texts generally well-written, but the Year 9 one not so much – flow issues etc. Don’t they have feedback mechanisms? They make a fortune from the texts!

    Reply

    1. Do they make a fortune? If they do, then I’m not sure the people who work for them do. I looked into working at a textbook publisher earlier this year. I was a bit surprised that editors who plan and manage the creation of textbooks (and are responsible for ensuring they are accurate) get paid a lot less than first-year teachers – at least, that was the offer I got.

      Reply

      1. The publishing companies are definitely raking it in, and I think the lead authors also. Those lower down the totem pole I’m sure do not do as well.

        But the main point is not how much these people make but how little they care. How much should people be paid for that tripe above. Which has been there for over a decade, essentially without change.

        Reply

      2. Hi wst “Today, the academic publishing industry reports profit margins of around 40%. A New Scientist leader article argues it is the most profitable business in the world.10 Oct 2023” – found this on the web. Agree that they would cost cut on their staff too. So much for competition!

        Reply

        1. That’s for higher education and research. It’s not clear how much it translates to school publishing.

          Reply

  2. Often teachers use the phrase “the same thing divided by the same thing equals 1”. It would be not unreasonable for students to think, based on how they have been taught about division before, that 0^2 / 0^2 = 1. So I could see that the first activity about 0^0 would confuse students.

    Reply

    1. If you would want to teach this properly, Wikipedia has a good section on exponent identities that is good place to start.

      Reply

  3. I’ve commented previously on this rubbish (in my opinion the whole thing is backwards, leaves out some very important distinctions between “laws” and “conventions” and the books use examples to “demonstrate” a law which is wrong on a few levels).

    The example used to show that \sqrt{9}=9^\frac{1}{2} is typical of the rubbish but I suspect this is how a lot of students are actually taught.

    That very last sentence though… I missed that gem. In MOST cases…? OK, care to give a counter-example Cambridge…?

    Apart from 0^0 that is.

    Reply

  4. I’m not sure if this aligns with your view, but my main gripe with the usual exposition is assuming x^(rational) is defined in accordance with some ‘index law’, rather than showing that the ‘law’ follows as a direct result of how we defined such expressions. “the index laws apply to rational indices” – of course they do!

    As a related aside, do any of the Y12 subjects/textbooks touch on (informally) how we make sense of things like 2^pi ?

    Reply

    1. I don’t think I’ve seen anything on this in the current Victorian text books. Given we can always find a better rational approximation to pi you can give the kids a sense that such a thing is meaningful. It is a nice teaser for limits. Perhaps it should be in the books.

      Reply

    2. It seems everybody is out of practice with WitCHes: you guys are supposed to be doing the work. I haven’t yet had time to read carefully through the replies, but I don’t see that anyone is properly analysing what is there.

      So, let’s begin with the first box. Is it correct? Is it good?

      Reply

      1. It is not correct, as it still states 0^0 is 1.

        It is not good as it uses the index law about division to establish a^0 =1. It is using inductive reasoning to infer a pattern for a rule. You cannot trust a pattern in mathematics, you need to prove it (take Moser’s circles for example).

        It would be better to use the index law about multiplication to deductively reason why a^0 = 1 (i.e. let m=0 in a^m x a^n and go from there).

        Also, in Core Mathematics (1977, Fitzpatrick and Galbraith) the index division law is proven after a^0 = 1 and a^-n = 1/a^n is establish. Cambridge is doing the reverse.

        Reply

        1. Thanks, Potii. How or where does the first box state that \boldsymbol{0^0 =1}?

          I don’t understand your second sentence. What, if anything, is wrong with or unsatisfying about (b) of the first box?

          Your third sentence is unclear? What argument are you suggesting, and what does this argument establish?

          Reply

    3. Javert, this is the point, but for some reason it seems really really hard for people to get it. So, my intention is to get people to work through the thing, box by box, line by line, and not permitting any fuzziness of expression or argument. Currently, we are on Box 1, Line b …

      Regarding \boldsymbol{2^\pi} and its kin, every text I’ve seen pretends there is no issue. So, the graphs of \boldsymbol{2^x} and \boldsymbol{e^x} are blithely drawn, without a moment’s hesitation. I’m not sure the writers have a whole lot of choice, but it’d be nice if they paused for at least a half-second.

      Reply

  5. When the text says “0^0 should equal 1”. I took it to mean that the above method does not work to show 0^0 = 1 but it should. I can understand, reading it again, that is not what they mean.

    The unsatisfying thing is they use a few examples to show a^0 =1 and students are to assume it works for other values. Part e give the exception for a base of 0. A student could think there might be another counter example. That is what I was trying to get at.

    Though on second thought this doesn’t seem like a good argument from me.

    Reply

    1. Thanks, Potii.

      The last line of the first box is woefully written, and can easily be read, as you did, that simply the method fails for \boldsymbol{0^0 =1}, not the fact.

      But I have larger concerns. Just to get specific, does the first box show that \boldsymbol{5^0 =1}?

      Reply

      1. Is it related to how the expression \frac{4\times{}(2-2)}{(2-2)} does not have a defined value but if you cancel the common factors you get 4?

        Reply

        1. I don’t see how. I’ll try to make it a little easier.

          The first box is screwed, in a number of ways, but in one fundamental way. Why? What is the argument they are making in parts a and b, and why is that argument crap?

          Reply

          1. The argument they are making is that \frac{5^2}{5^2} is equal to two different values so by the transitive property of equality we have 5^0 = 1. However, why 5^0 = 1 is so that powers of zero of a non-zero base are consistent with the index multiplication rule. They are working backwards really from this. So it is not explaining anything.

            Hopefully I’m getting closer.

            Reply

            1. Yes, you’re getting closer, but you’re not spelling it out. What, in detail is the argument they are suggesting box 1, line b?

              Don’t describe the argument, provide the argument, in as much detail as you can. Then, step by step, you can discuss whether the argument is valid.

              Reply

  6. \frac{5^2}{5^2}=\frac{25}{25}=1, OK, fine.

    In order to make the index law \frac{x^a}{x^b}=x^{a-b} consistent with this, it makes sense to define 5^{2-2}=5^0=1.

    But this means 5^0 does equal 1, not should equal 1.

    None of this needs to be demonstrated by an exercise. Just define a^0=1 for non-zero a, this index law is then internally consistent with the others and move on.

    Likewise, define \sqrt{a}=a^\frac{1}{2} (or the other way around, I am assuming the equality sign works both ways… and then show that this is internally consistent with the other index laws and you are done.

    There is no need to “prove” anything, just define it properly, demonstrate that these definitions do not contradict one another and all is good in the world again.

    With regard to the negative indices, I think the Year 9 book actually does a (marginally) better job than the year 10 book because the Year 10 book draws on the zero power and I find this clunky, confusing and a bit weird.

    Define a^{-n}=\frac{1}{a^n} if you want, but simply defining it that way works, too.

    Reply

      1. Assuming all that was intended was the student to write \frac{5^2}{5^2}=5^{2-2}=5^0 then it seems OK.

        However…

        The way the index notation is introduced sort of (kind of) suggests that a^n is a notation that works when n is a positive integer, so…

        Unless 5^0 has been defined, it is possibly meaningless at this stage.

        And I’m sure I have missed a lot here and also that “meaningless” may itself be ambiguous. As you say, I am out of practice.

        Reply

        1. Make up your mind. You gave the argument:

          \boldsymbol{\dfrac{5^2}{5^2}=5^{2-2}=5^0}

          You wrote “it seems ok” and then said “however … it is possibly meaningless at this stage”.

          Is it ok, or is it meaningless at this stage?

          Reply

          1. Fair enough. I think it is backwards.

            If we first define 5^0=1 then we can show that this “agrees” with the index law \frac{5^2}{5^2}=5^{2-2}=5^0.

            I agree with the text that \frac{5^2}{5^2}=1. I do not agree that this “shows” that 5^0=1.

            Reply

            1. Correct, and good.

              How did Cambridge trick themselves into believing this non-sensical argument? Why did they *think* there was substance there? What did they think they were doing?

              Reply

              1. I can only guess what Cambridge writers were thinking (*if* might be better than *what*) in this setting.

                I suspect it was something along the lines of:

                5^{2-2}=5^0. Now, what does 5^0 mean?

                We can make 5^{2-2} by writing \frac{5^2}{5^2} and since \frac{25}{25}=1 it must be true that 5^0=1.

                Unfortunately, the way indices are introduced by many books – the index is the number of times the base is multiplied by itself – any index which is not a positive integer makes no sense (and even a^1 makes minimal sense under this quasi-definition).

                Reply

                  1. Because 2-2=0.

                    Generalising: a-a=0. That I have no issue with.

                    Similarly: \frac{25}{25}=1 I have no issue with. Generalising to \frac{a}{a}=1 is not valid, so it is clear that we cannot simply generalise rules of arithmetic, and so should be careful when it comes to applying rules of arithmetic to index laws.

                    Reply

    1. Ah… I think I see where this might be going…

      Until we define what 5^{-2} means, the notion of 5^{2+(-2)} is unclear/undefined, even though it may seem obvious that because addition is the opposite of subtraction etc…

      Am I getting warmer?

      Edit: For a>b I can see logic in \frac{5^a}{5^b}=5^{a-b} and it could be reasoned thus:

      5^a=5^{a-b+b}
      =5^{a-b}\times 5^b
      Then \frac{5^a}{5^b}=\frac{5^{a-b}\times 5^b}{5^b}}
      And if we assume \frac{5^b}{5^b}=1 which I think is fair enough,
      Then \frac{5^{a-b}\times 5^b}{5^b}=5^{a-b}\times 1

      But I do not think the text is suggesting this at all.

      Reply

      1. You’re warm, but you’re spiralling.

        *) It doesn’t really matter hear what \boldsymbol{5^{-2}} means.

        *) Your argument above, proving that \boldsymbol{5^{a-b}=\dfrac{5^a}{5^b}} for positive integers a > b, is correct.

        *) Obviously 2 – 2 = 0.

        Now let me ask it this way?

        a) Does log(2 – 2) = log(0)? Why or why not?

        b) Does \boldsymbol{5^{2-2}=5^0}? Why or why not?

        c) Does \boldsymbol{5^{2-2}=\dfrac{5^2}{5^2}}? Why or why not?

        Reply

        1. (a) and (b), yes.

          (c)… that is harder to answer, but I would say yes as well, but with a justification:

          I can define 5^0=1 and I know that \frac{5^2}{5^2}=1

          And so, because I believe equality to be transitive, 5^0=\frac{5^2}{5^2}

          Reply

            1. \log{(2-2)}=\log{0} in the sense that neither are defined, but I take your point.

              As to (c) it doesn’t help, because the exercise in the first box has the story backwards.

              Reply

  8. Box 2:

    Minor nitpick – why say 0.5 and not \frac{1}{2}?

    Major nitpick – why say “according to the index laws”? If you have defined a^{\frac{1}{2}} as \sqrt{a} and also defined \sqrt{a} as the inverse of a^{2} for a\geq 0 then the exercise is redundant.

    Minor nit-pick – the question should be “what positive number x makes the equation x^2=9 true but even this is clunkier than asking for the positive solution to x^2=9

    Minor nitpick – why do we need to check with a calculator?

    Major nitpick – the whole thing is backwards. Define x^\frac{1}{2} as \sqrt{x} and then there is no need to “predict” or say what something “should” equal, you can quite quickly say what something DOES equal straight from the definition. Exercise over, move on.

    What did I miss?

    Reply

    1. Thanks, RF.

      1) Yes, using 0.5 instead of 1/2 is pretty bogan.

      2) Well, of course they haven’t defined \boldsymbol{a^{1/2}}.

      3) Yes, the wording is excruciatingly bad.

      4) Because the calculator is God.

      5) Yes, of course.

      All good, and you can move on to box 3. (And I still have to update the post re box 1.) Except one point, for clarity. You make the fundamental point that one must define \boldsymbol{a^{1/2}} to be \boldsymbol{\sqrt{a}}. Where, then, do the “index laws” fit in? If at all.

      Reply

      1. I think Box 3 was pretty well covered by comments on Box 1, but I will add that the ignorance (oversight) of the a=0 case is problematic.

        The use of cancelling (without explaining WHY such a process is valid) is also problematic.

        Reply

        1. Thanks, RF. Yes, most of these boxes are crap for the same basic reason, that the “index laws” are taken as an article of faith rather than truths to be established. The point of posting them all was simply to highlight the fact that Cambridge stuffs it up each and every time. Still, each box has its little special treasures, and we may as well do them.

          I’m fine with them cancelling away like that, and I’m almost fine with ignoring a = 0. That is standard with such algebraic statements, and it can be muddying to declare a ≠ 0 or whatever each time. But, in a situation where one is attempting to make sense of a new power then, yes, more care is required.

          Box 4 whenever you wish (and I’ll catch up with the updating at some point).

          Reply

          1. With your permission… I will leave box 4 and concentrate on the two “proof by table” boxes which I think may have a bit more to them (in terms of crap) than I first thought.

            Reply

  9. OK, I’ve updated the post with comments on the first three boxes. Plenty more in the later boxes, if anybody has the energy.

    Reply

  10. OK, I got bored and updated the WitCH with comments on all the boxes. Commenters can see what they missed. (Plenty, including a gorilla on the basketball court.)

    Reply

  11. That last box about how the index laws apply to rational indices really bothers me.

    I just cannot for the moment work out exactly why nor how.

    Reply

    1. Well, as I wrote in the update, I assume the point is that negative bases can screw things up. But what exactly they had in mind, God knows.

      Reply

      1. I am in full agreement with you on the negative bases issue (remember the Methods 3&4 issue with the \frac{1}{3} vs \frac{2}{6}…?)

        But there is something more there, I am feeling. Perhaps it is the way I suspect in Year 11 Methods texts they are going to somehow make the jump to non-rational, real powers in order to produce graphs…

        Maybe not. I will continue to ponder.

        Reply

  12. Thanks for spelling it out in the updates, Marty. It’s very clear what I was missing in my attempted whacks.

    Reply

  13. Interesting discussion. I think the real point about exponents is that when you get beyond positive integers, it is a matter of definition to get a system that respects the laws of arithmetic. (An analogy would be the definition of multiplication beyond positive integers.) It is easy for children to understand that 5^2 x 5^3 = 5^5 holds by rewriting this as (5 x 5) x (5 x 5 x 5) and that the law of addition of indices corresponds to adding the number of factors. You can go from there to noting that 5 x 5^n = 5^(n+1) so that multiplying by the base increases the exponent by 1, and correspondingly, dividing by the base decreases it by 1. Now you want to give 5^0 a meaning, so you continue this rule by fiat. Then you can continue for negative indices. You also preserve the rules 5^a x 5^b = 5^(a + b) and (5^a)^b = 5^(ab) (which are seen for integers from the definition of positive integer powers). Now what should 5^(1/2) be? You want (5^(1/2))^2 = 5^((1/2)x2) = 5 so it should be the square root of 5.

    Now this might seem arbitrary at first, but like a lot of things in school mathematics, the value of this definition become apparent as the student gets more experience in doing and solving mathematical problems. In short, you provide the situation with the structure that allows you to make progress.

    A teacher in a class can tailor this insight to the situation in front of her; it is much harder to do in the solid print of a textbook and many of them fail at it. Often, the author should probably best just state the rules and let their significance sink in with the children through the examples and exercises.

    Reply

    1. Thank you, Ed. It is obviously difficult to get across to students the role of definition in mathematics, which is why I like my playful BISS approach. But is obviously doesn’t help when the best texts for the Australian Curriculum simply do not give a stuff to get it right.

      Reply

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