Another Fraction Question

OK, following up on our previous post, we have another fraction question. This one is not new, has appeared in the comments of a previous post, and many readers will have heard us bang on about it. Nonetheless, given the discussion on the previous question, and given the possibility that some new readers might not have yet read Marty’s Collected Sermons, it seems worthwhile giving the question its own post. Here it is:

    \[\color{Mulberry}\boldsymbol{(-1)^\frac26 \overset{?}{=} (-1)^\frac13}\]

That is, are the two quantities equal or unequal? Loyal readers of the Sermons might hold off commenting for a little while, to allow others time to ponder.

UPDATE (11/02/22)

Thanks to everyone who has commented. The answer is that the two quantities are equal, as best explained by Sir Humphrey:* since 2/6 equals 1/3, whatever quantity we make from 2/6 will equal the corresponding quantity made from 1/3. That’s it. That’s all there is.

The trouble is, plenty of teachers and students treat “index laws” as if they are commandments from God, and trick themselves into all manner of inequality. This question was born, a decade ago, with a textbook that got it wrong.

*) SIr Humphrey’s perfectly blunt response reminded us of the greatest line of dialogue from the great writer Ring Lardner:

Shut up”, he explained.


75 Replies to “Another Fraction Question”

  1. Well, my first response would be that fractional exponents are not defined when the base is negative, so neither one is defined, and two things that are not defined cannot be equal, so not equal.

    But the real question (pardon the pun) is whether we’re considering these numbers in the set of real numbers or in the set of complex numbers. In the set of complex numbers they are equal. Both are 1/2+(sqrt(3)/2)i

    1. a^b where a and b are complex numbers is equal to exp(ln(a)*b). (-1)^(1/3) is not necessarily 1/2+(sqrt(3)/2)i, unless you take the branch of the complex logarithm where ln(-1) = pi*i, or any of the other branches that lead to such a result.

      With regards to this equation over the real numbers, I personally wouldn’t necessarily define a^b where a<0 and b is not an integer, but I suppose that the convention would be that (-1)^(1/3) over the real numbers is -1.

      Regardless, by the axiom of substitution (for equality), a = b implies that F(a) = F(b). 1/3 = 2/6, and thus letting F(x)=(-1)^x, F(1/3)=F(2/6), or alternatively (-1)^(1/3) = (-1)^(2/6). I suppose it would be possible for one side of the equation to be using a different branch of the complex exponential than the other, which would lead to us not being able to use the substitution property as 1/3 = 2/6 wouldn't necessarily imply that F_a(1/3)=F_b(2/6), but I would take it that the branch used for the complex exponential, unless otherwise stated, would be consistent across both sides of the equation.

      If one chooses to not define (-1)^(1/3), then I suppose (-1)^(1/3) may not necessarily be equal to (-1)^(2/6) in the sense that 0/0 is not necessarily equal to 0/0. Though I must admit that I am not knowledgeable in the area of undefined expressions, so it could in fact be that you would take (-1)^1/3 as being equal to (-1)^2/6 even if they are both undefined.

          1. And unfortunately, many teachers believe what the package tells them. No critical evaluation. They don’t understand that there’s ‘code’ behind the ‘curtain’ and that code is written by someone, based on what textbook *they* learnt from.

          2. Jesus. I didn’t know that aspect, although I guess I’m not surprised. Terry, can you indicate the answers given by R (what is that?) and Excel?

              1. Mathematica gives \displaystyle \frac{1}{2} + i \frac{\sqrt{3}}{2} for both (actually it gives the decimal approximation). Mathematica is assuming the principal root: \displaystyle p^q is defined by exp(q log(p)) using the principal value of the logarithm.

                Which is particularly problematic when plotting power functions such as
                \displaystyle f(x) = x^{\frac{1}{3}}.

                Mathematica users must use the command Surd[x,n] if real values are wanted for \displaystyle f(x) = x^{\frac{1}{n}}.

                Surd[-1,3] gives -1 (because Surd[x,n] is defined to give the real valued nth root of x) but (Surd[-1,6])^2 obviously fails.

                So Mathematica can’t give a real value for \displaystyle (-1)^{\frac{2}{6}} unless you simplify the fraction (and so believe – correctly – in the equality of the two fractions and hence the equality of the two expressions).

                1. There is also a command OddRootPlot but you need to first install a package called ClassroomUtilities (do a search for it – can be useful for things like this – also when plotting Riemann Sums…)

                  Another package I have seen is RealOnly which is probably aimed squarely at school uses.

                2. When interpreting x^{p/q}, Chrystal (p. 185) recommends reducing p/q to its lowest terms to avoid a different paradox.

                  Chrystal, G. (1886/1964). Algebra: An elementary text book for the higher classes of secondary schools and for colleges. Part 1. Chelsea.

                    1. In discussing a paradox that leads to 4 = \pm 4, he writes: “Such difficulties are always apt to arise with x^{p/q} where the fraction is not in its lowest terms”.

              2. Excel says -1^{2}=1 regardless of whether you use brackets.

                I asked a programmer about this and he told me that to an Excel programmer the unary minus is higher in the order of operations.

                So… order of operations is not a fixed set of rules…?

                In which case the answer to the original question is… it depends how you define the operations.

                  1. An extension of your point. Possibly (probably) irrelevant – I just found it speaks to the issue of order of operations when evaluating an expression such as this.

          3. I tried it out in Python 3.9. Using the built-in power function “pow” or the operator **, you get 0.5000000000000001+0.8660254037844386j and the equation evaluates as true. On the other hand, the power function in the math library throws an “math domain error”. The one in the numpy library gives you a warning and returns nan for each side, and therefore evaluates the equality as false.

      1. And unfortunately, many teachers learn what they’re meant to teach from the textbook.

        (Just wait and see what happens with Logic this year… Speaking of which, sit back and watch all the snake oil sellers come slithering out from under their rocks)

  2. My initial reaction is that negative base with non-integer exponent is not well-defined if I follow what I have been taught so far. It seems easiest and most convenient to say \frac{2}{6} = \frac{1}{3} therefore (-1)^{\tfrac{2}{6}} = (-1)^{\tfrac{1}{3}}. I hope that I never have to deal with this in VCE this year, but that appears to be optimistic from what I see here.

  3. Hi,

    assuming you can use Euler’s identity −1=e^iπ and working with RHS say

    let z=re^iθ where r>0 and z is complex

    Then w= z^3=r^3e^3iθ=e^iπ where r is the absolute value (> 0) and θ an argument.

    It follows that r=1 and 3 θ=π (mod 2π ) by equating coefficients

    hence z∈{e^iπ/3,e^3iπ/3,e^5iπ/3….}

    i.e. z∈{1/2+i3√2,−1,1/2−i3√2}

    Steve R

  4. OK… I’ve allowed more than 24 hours for others to comment, so I’m going to give my opinion of the heart of the issue: order of operations.

    If you say that (-1)^{\frac{1}{3}} has only one value because you are working with real numbers then you could say it is equal to -1.

    If you say that (-1)^{\frac{2}{6}} has only one solution then you could… COULD say (-1)^{\frac{2}{6}}=((-1)^{2})^{\frac{1}{6}} =1^{\frac{1}{6}}=1 or you could say (-1)^\frac{2}{6}=((-1)^{\frac{1}{6}})^{\frac{2}}=1^{2}=1

    And this would lead to the conclusion they are NOT equal.

    Please note my repeated use of the word COULD.

    (Marty – I’ve stuffed up the LaTeX here somewhere and my exponents after brackets have gone walkabout… I cannot see where I’ve gone wrong though… delete the comment if it is not an easy fix)

  5. Marty, Do you mind if I ask your original question on another blog with which I am involved? (NCTM – USA)

    Also, the solutions of x^3 = -1 and x^{6/2} = -1 are the same. Does this imply that (-1)^{1/3} = (-1)^{2/6}\,?

    1. Hi, Terry. Sure go for it. I would suggest asking the question as I have, entirely without elaboration, but it’s up to you.

      Your suggestion of undoing the roots seems a reasonable way to maybe convince the recalcitrant. But, Sir Humphrey’s answer is much better, and the message of his answer is much more important.

      1. Thanks Marty; I asked only the question, as you suggested, and it generated considerable discussion quickly; without going through the individual responses, I’d say that most of the respondents had not considered this detail before but were on the ball when they thought about it; clearly they enjoyed the question.

        Marty’s question implies that we should not teach students that x^{ab} = (x^a)^b without further qualification.

      2. One interesting reply from NCTM people was “Larson addresses this problem in his Precalculus with Limits (2nd Ed.) where he defines x^{m/n} to equal the n-th root of x^m only if m/n is fully reduced.”

    2. Thanks, Terry. I think this is a great way of convincing most recalcitrants.

      But I think the hard-core recalcitrant is still going to appeal to

      \displaystyle x^\frac{p}{q} = \left( x^p\right)^\frac{1}{q} = \left( x^\frac{1}{q}\right)^p

      and then argue that for \displaystyle x = -1, \displaystyle p = 2 and \displaystyle q = 6 things (including your argument and the simple statement by Sir H) don’t make sense …

      1. Also, the argument does not go both ways:

        If p=q then x^{p}=x^{q}. I’m fine with this whenever the result is defined (or definable).

        It does not follow (but I can imagine some saying it does) that x^{p}=x^{q} implies p=q.

        This was perhaps a special case of the paradox Terry was referring to (I didn’t read the entirety of the attachment).

        1. So getting into the spirit of Logic:

          Let \displaystyle x, p, q \in R. If \displaystyle p=q, then \displaystyle x^{p}=x^{q}.
          That is, \displaystyle p = q \Rightarrow x^{p}=x^{q}.

          The converse is false (proved by a simple counter-example).
          That is, \displaystyle x^{p}=x^{q} \nRightarrow p = q.

          In other words, \displaystyle p=q \nLeftrightarrow x^{p}=x^{q}.

          1. Now, write it as a circuit composed entirely of NAND gates…

            (If VCAA steals this question from me I want NO ACKNOWLEDGEMENT WHATSOEVER)

            1. \displaystyle A \Rightarrow B \Leftrightarrow \neg A \lor B.

              So you need a digital circuit constructed solely from NAND gates that ‘realises’ OUT\displaystyle = \neg A \lor B.

              Which is a pretty dumb thing to ask (so a great VCAA question: “A student wishes to make a digital circuit to represent \displaystyle \neg A \lor B but only has NAND gates ….”) – I hand it over to a more motivated reader (such a reader may wish to peruse

              PS – RF, VCAA have a poor reputation for acknowledging the IP of others, so you’re safe!

              1. You could use NOR gates as well, but it would take longer…

                The Sheffer Stroke is worth looking up if there are any Year 11 Specialist teachers out there wondering about the logic topic.

                1. Indeed. The Daft Stupid Design gives no details as to which logic gates students are expected to know. All it says is

                  “logic gates and circuits”

                  Now, since Logic and Algebra is (meant to be) a MATHEMATICS subject (NOT a digital electronics subject), a reasonable assumption would be that only the gates corresponding to the operations defined in a Boolean algebra (+ (OR), * (AND) and the negation (NOT)) would be required. But who knows …

                  Maybe NAND, NOR and the whole logic gate gang are meant to be taught …?

                  But who knows? I suppose you could argue a mathematical case for NAND (and NOR …) given that it’s functionally complete (I gots to bait the hook here) … But still … It’s clear from the three major textbooks that the writers are guessing too … And this is a topic that most teachers will have no familiarity with (and hence no ‘intuitive feel’ for what should be taught) …

                  The lack of detail in the Stupid Design is totally pathetic. Its writers should be forced to eat every damned page of it, without salt.

                  1. Part of me also wonders (quite seriously) if they will try to link it to set theory and/or probability somehow.

                    From a logic perspective, this IS possible but not at a VCE level, surely…?

                    If we see any reference to the “symmetric difference of two sets” then the trial exam writers are going to have some tough times ahead trying to second guess VCAA (again).

                    Sorry Marty for getting so off-topic, but it is clearly a sore point for a non-negative number of teachers who are trying to reduce the clues given into their simplest form so as to know whether what the study design says and what is actually possible are equitable by any definition… (I tried to link it to fractions somehow)

                    1. Hi RF.

                      I don’t mind linking set theory to Boolean algebra (in fact, I want it!). I think it can be readily done in VCE, but it requires time, care and teachers doing their job properly in pre- and co-requisite classes (there must be greater communication between the SM12, SM34 and Methods teachers because the VCAA masterplan is that everything outside of SM34 is examinable. So it better get taught!):

                      1) The algebra of sets should have been taught in Yr 10 and in MM12, well before Logic is taught in SM2. (The laws are readily shown to hold using Venn diagrams.) The snag is that the appropriate language (commutative, associative, closure, distributive, Identity) WON’T have been taught (and this is a crime).

                      btw There’s no need for the “symmetric difference of two sets” etc. in my view, so concerns along these lines are unnecessary. (Then again, there’s no VCAA clarity in its Daft Stupid Design, so who knows …)

                      2) The definition of the power set of a set \displaystyle S is easy to teach.

                      3) The power set of \displaystyle S forms a Boolean algebra if the following definitions are made for the binary operations + and * and the negation operator:

                      +: union
                      *: intersection
                      negation: complement

                      4) Let \displaystyle S = \{ x, y \} have two elements. Then the power set of \displaystyle S is easy to state and it’s not difficult to use the algebra of sets to prove 3) for this special case. The only (very) small difficulty is identifying the identities under + and *.

                      I think it’s very important – mathematically – to consider some simple examples of a Boolean algebra (like 4) above) so that students do not think that a two-element Boolean algebra (eg. propositional logic and digital circuits) is \displaystyle the Boolean algebra. (I think its also important that they meet an example that’s not a Boolean algebra …) I guarantee that many teachers will have this misconception.

                      This sort of misconception is the same as the misconception that a graph can never cross an asymptote (held by every student entering Yr 12 maths that I’ve ever taught, as well as most teachers I meet). It’s a lack of appropriate examples (usually caused by a lack of knowledge) that allows these misconceptions to form, grow and become entrenched in students.

                      Ideally VCAA will show some intelligence – perhaps from a New year’s resolution. I’ll be doing my best to force this. Because what’s needed is NOT sample exams – where only 15% of the questions will relate to the new content and the rest will be copied from previous exams. Those sample exams are absolute bullshit. Ordered by someone with no brains simply to tick a box. What VCAA must provide – and what teachers must DEMAND – is a large set of sample questions on the NEW content. A first approximation to this would be making available all the VCE questions on graph theory and logic from the old exams when Logic was a (albeit optional) part of the course back in the late 80’s and early 90’s. Or just publish those exam papers at their website. Old Checkpoints books from that era have suddenly become valuable … (But have probably been thrown out long ago by school libraries as no longer relevant).

                      As for getting off-topic, I’m sure that Marty sees the … logic in this discussion. He might even post a brief blog …

                2. Ah yes … the NOR gate.
                  Pierce’s arrow. The \displaystyle dual of the Sheffer stroke.
                  VCAA’s Daft Stupid Design with its lack of clarity certainly opens up a can of worms.

                  1. I’ve also heard it called the Pierce “dagger”…

                    The attached document may save SM1&2 teachers some time. It “fell off the back of a truck” if anyone asks.

                    1. Started writing on the other post – more people seem to be reading this one.

                      Happy for it to be copied there.

                    2. Thanks SM – will fix and email to Marty.

                      The joys of not having anyone at school check my work…

                      Actually surprised there was only one error.

  6. Thanks to everyone who has commented. Basically everyone is on the right page, except for the occasional overthought to look for a non-existent trick. The trouble, and the reason I raised the question, here and elsewhere, is that the correct, one-word Sir Humphrey answer is extremely uncommon. In particular, most teachers and students give the wrong answer, claiming the two quantities are unequal.

  7. Now we can ask another question. How do I explain to a thoughtful high school student, or a fellow teacher, that (-1)^{1/3} = (-1)^{2/6}?

    I want to stay in the context of real numbers.

    If I give a one line answer (1/3=2/6), my discussant can reply with (-1)^{2/6} = ((-1)^2)^{1/6}....

    1. The two great failings of modern discourse are:

      (i) to pretend that simple things are complicated;

      (ii) To pretend that complicated things are simple.

      (i) is, by far, the greater problem.

      The only proper explanation is Sir Humphrey’s. To give more than a one-line answer is giving the wrong answer.

      1. I recall a colleague, who was a lecturer in engineering, telling me that the job of the mathematics department is to make simple things complicated.

            1. 1. I actually like Cramer’s rule. I don’t teach it, but I do like the idea, especially for 3×3 systems of equations.

              2. Kramer. Infinite wisdom, remains ever relevant today.

            1. I agree with John, I believe that (ii) is the bigger problem. Even if something incredibly simple is made unnecessarily complicated, that doesn’t necessitate that falsehoods must be taught. However in the process of simplifying something more complicated, something necessarily must be lost, and falsehoods or misinterpretations will likely be introduced.

              To me teaching set theory and Dedekind cuts when first introducing irrational numbers, something that is needlessly complicated at the stage in which you first learn about these numbers, is much preferable than to attempt to teach general relativity without requiring basic calculus, something that to my knowledge is planned to be introduced in the next VCE physics study design.

              1. Climate change. Torturing people. Proportional representation. Wearing goddam masks and not having thousand-person parties during a pandemic.

                1. X marks the spot!
                  Supply chains, for example, are a lot more complicated than the simple model our fearless unleaders have clearly assumed.

              2. Sometimes in simplifying a concept we can dispense with ideas that are not essential to understanding the concept.

                1. I agree. Which is the point, and which is why I think your question is way off the point.

                  All that is required for the fraction question is to know, as X (kind of) put it initially:

                  If ZIG = ZAG then f(ZIG) = f(ZAG).

                  To say anything else is to only detract and distract from this straight, simple answer.

                2. The trouble is when you have idiots that dispense with ideas that ARE essential to understanding the concept …

              3. I can see the Stupid Design description now:
                “Space-time tells matter how to move; matter tells space-time how to curve”.
                That way, the more more complex and profound truth can be completely hidden. (With apologies to Wheeler).

                Followed by the dot point: Gravitational waves.

                Because VCAA must make sure the latest shiny toys get added at the expense of the basics. (Just look at VCAA’s plans for Chemistry).

                (When you have to count squares under a graph to estimate the area, you’re just asking for trouble introducing General Relativity. And it’s not so long ago VCAA said that g = 10 because 9.8 was too difficult for physics students to use in calculations).

                We have/had Further Maths (the spirit of the subject has long since been subverted), maybe VCAA should have a Further Physics …

                Anyway, I digress to being off-topic (and have hence made a simple blog complicated).

      2. I raised the question about explaining the issue because … a couple of years ago, I was teaching Year 11 students who had to evaluate 8-6+2. Several students (maybe five in a class of about twenty) were adamant that the answer was 0. They appealed to BODMAS in which they had been drilled in earlier years. Add before you subtract. I suspect that some other students were unsure and happy to listen to the debate.

        I tried several ways, over two lessons, to convince them that the answer was 4. Everything I tried failed. When I suggested they they use a calculator, and they got 4 as the answer, they said that this was because calculators don’t follow BODMAS.

        In the end, one student put it this way: “Sir, I think you need a new approach.” Amen to that!

        Another said “Whatever – can we move on?”

        In addition, I was concerned because in a few days they would have a test and I expected that a question or two like this might be on the test. But then the teacher who had set the test put my mind at rest – they could use calculators in the test.

        Finding the right explanation often requires a lot of thinking.

        1. When I learnt BODMAS it was “… Addition and Subtraction as they come”. The last part of the pneumonic is important.

          In this case, the subtraction comes before addition and so is done first.

          My explanations (I’d offer a choice) would be simple:

          Choice 1: Whoever taught you BODMAS taught you wrong. Admittedly a tough explanation to give if you’re the new kid on the block. But then the calculator also backs you up (even though they claim it’s wrong – confirmation bias on their part?). And if you gain trust quickly and are confident, it can be pulled off.

          (I’m not a big believer in supporting the bullshit taught by others and will call it out without the sugar coating – I take the Greg House approach when it comes to things like this).

          Choice 2 (and maybe you tried this): 8 – 6 + 2 = 8 + (-6) + 2. Admittedly, if they don’t accept the addition of negative numbers, or they don’t accept that a negative number has the same status as a positive number, then you have your work cut out.

          Choice 3: The numbers represent credit and debit. You get 8 dollars, you spend 6 dollars, you get 2 dollars. Appeal to their sense of greed (in this case).

          1. ?? Correct in saying 0 …? If so, why?
            Or correct in the statement “In the end, one student put it this way: “Sir, I think you need a new approach.””

            I’d argue (have argued) very strongly against 0 being the correct answer.

            1. If they had been taught addition comes before subtraction, which is a convention that works just as well, then they correctly got 0.

  8. Hi ,

    Have any other correspondents read the Clements article of T Tao?

    The questions used to evaluate T Tao’s talent at age 7 are quite interesting as are his visual responses

    Eg p221 gives an example of how to solve a continue fractions using a quadratic rather than repeatedly bashing the inverse button on the CAS

    Steve R

  9. Re: Bright young things

    I recall reading an article about a bright young mathematician at Oxford. He was told in physics in school that you must show your working. If you don’t show your working you will get at most 30% of the possible marks. He wrote down just the answers to all the exam questions – and got 30%.

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