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I am no professional educator; I am a student myself! But apparently I come up with useful tricks that help my younger brother do better in maths. I just want to hear your feedback, is all.

My younger brother is in Year $7$ (that is, seventh grade or class in Australia) and was learning the following: $$a=\frac bc\implies b=ac\;\mathrm{and}\;c=\frac ba.$$ He wanted me to help him remember, and I came up with a trick (although I told him that he should know why certain rules are as is, and not find some nifty tricks to help him remember, but he doesn't listen). So here was my trick:

Imagine the vinculum (fraction bar) as the horizon, such that $b$ and $c$ are positions of the sun, sunrise and sundown respectively. Then sunrise multiplies and sundown swaps around.

So if $a=b\div c$ then $b$ (sunrise) is equal to $ac$ (multiplies) and $c$ is equal to $b\div a$ (swaps).

What is your feedback? I think the word sunset is more common than sundown so I am just making sure if this "trick" is actually doing good for my brother. Any thoughts for improvements, perhaps?

Edit:

To anybody who wants to $100\%$ mentally grasp the idea of fractions, I absolutely recommend seeing @Number's (Bill Dubuque's) comment below @inéquation's answer! Please give that an upvote!

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    $\begingroup$ Worth mention is the additive form, i.e. the same sorts of symmetry transformations also apply to difference equalities, e.g. $\,a - d = b - c\,\Rightarrow\, c-d = b-a$. Studying both simultaneously may provide better intuition. Such differences are used when constructing the integers from the naturals - in the same way that fractions are used when constructing the rationals from the integers. $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 16:32
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    $\begingroup$ The additive form has the advantage of using "simpler" arithmetic, but the disadvantage that the terms aren't displayed in a "square" form (making it harder to see / describe things like reflections and diagonal swaps). $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 16:44
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    $\begingroup$ @DanielR.Collins Re: you prior two comments: Those more general techniques don't reveal the richer symmetry here (e.g. see inequation's answer). $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 20:11
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    $\begingroup$ A slick application of $\color{#C00}{\rm diagonal}$ swaps on equal fractions $\!\!\rm\displaystyle\ \ \frac{\color{#C00}A}B = \frac{C}{\color{#C00}D}\, \Rightarrow\, \frac{\color{#C00}D}B = \frac{C}{\color{#C00}A}\,$ is this proof of unique fractionization, i.e. the least denominator of a fraction divides every other denominator. This is equivalent to uniqueness of prime factorizations (the nonvtrial direction of the Fundamental Theorem of Arithmetic). There also is John H. Conway's inline application to prove irrationality of square roots. $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 21:53
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    $\begingroup$ @user477343 Note that there is no atsign in my prior comment, so the comment is not targeted specifically at you (or your brother).. Rather it is intended for readers who may be interested in this and related topics. $\endgroup$
    – Bill Dubuque
    Commented Sep 9, 2018 at 0:12

4 Answers 4

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I do not think that such tricks are helpful: in fact I believe they are deeply damaging.

These types of basic relationships should not be memorized: they should be derived on the fly from an understanding of the meaning of the operations and the meaning of equality.

Namely, you should understand that $a = \frac{b}{c}$ is asserting that the number $a$ is identical to the number $\frac{b}{c}$. Since these numbers are equal, the result of multiplying each by $c$ will also be equal. Thus $ac = \frac{b}{c} c$. Now, if you understand that $\frac{b}{c}$ mean "take $b$ and split it into $c$ equal pieces", and you understand that multiplication by $c$ means "take $c$ of those pieces", then you should instantly recognize that $\frac{b}{c} c = b$. Thus $ac=b$. This reasoning might take a considerable amount of thought the first few times, but it eventually becomes automatic.

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    $\begingroup$ Expert algebraists certainly don't rederive these symmetries on the fly - that would be far too slow. They know by heart the simply (square) transformations (reflections, rotations,.diagonal swaps) that preserve fraction equality (i.e. equality of cross-products). $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 15:46
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    $\begingroup$ @Number Expert algebraists might skip steps, but they would know how to fill it in. They certainly wouldn't think about sunrise and sunset each time they do this. That is more mental energy than just re deriving the identities. If someone is confused about these things, then they should be thinking through them, not memorizing rules. The instantaneous application comes with enough practical use, but that should occur naturally down the road. $\endgroup$
    – Steven Gubkin
    Commented Sep 8, 2018 at 15:51
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    $\begingroup$ Of course experts know how to give formal proofs of the correctness of such transformations. Further they know how to motivate them mathematically - which is lacking in this answer.. In any case, I do agree that the suggested real-world motivation is pedagogically highly detrimental. $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 15:57
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    $\begingroup$ take b and split it into c equal pieces what if $c=\pi$? These types of basic relationships should not be memorized: they should be derived on the fly from an understanding of the meaning of the operations and the meaning of equality Not correct at all. One should be able to memorize some formulas. For example a $(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b)=\ldots=a^2+2ab+b^2$ do you expect a student to rederive it every time? Same for $\sin\alpha\sin\beta-\cos\alpha\cos\beta$. Your claim is unrealistic. $\endgroup$
    – user5402
    Commented Sep 9, 2018 at 9:26
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    $\begingroup$ @user477343 But it is unrealistic Anyone proficient at fraction arithmetic has learned and memorized the basic equality symmetries in inequation's answer just as they have other basic algebra (Binomial Thm, Quadratic Formula, Factor/Remainder Thm). It'd be far too inefficient to re-derive them on-the-fly when (frequently) applied. These simple symmetries can be derived and memorized by anyone who understands the basics of fraction arithmetic. Better to teach this proper mathematical perspective than some attempt at a real-world analogy (often so imprecise that it leads students astray). $\endgroup$
    – Bill Dubuque
    Commented Sep 9, 2018 at 14:28
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$$a=\frac bc\implies \;c=\frac ba.$$

What you propose above is one of the if/thens that students should pick up after seeing it in practice a couple times. The 'long' way has them multiplying both sides by c and dividing by a. It only takes seeing this, with variables or integers a few times before they should be able to do that swap (of a and c) intuitively.

I'd note that you mentioned grade 7. This would be a student who is a bit advanced, as in the US this material is 8th or 9th grade depending on the school. I'd save the 'tricks', the mnemonics, etc, for the things that will really help, such as SOHCAHTOA for the trig identities. While the trick may be cute, it takes more time to think about it than to understand why that variable swap is allowed and just practice it, and embrace it.

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  • $\begingroup$ Yes, I understand your point; this was what @inéquation was talking about, although he/she more explicitly stated that in comments as opposed to the answer. My brother and I live in Australia, not the US. However, I have been taught that SOH-CAH-TOA rule, which made me discover that $\sin^2+\cos^2=1$ when sub-ing the values into the Pythagorean/Pythagoras' Theorem, $a^2+b^2=c^2$, which is pretty cool funnily enough; i.e. people, or students at least, only get the best out of maths when they become a little independent and try playing around with it themselves, methinks. What's your thought? :P $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 12:00
  • $\begingroup$ I mentioned US only to be clear that it seemed to me we have an advanced student. My own observation is there are those who want to stick to the longer process, using a lot of simple manipulations, and the advanced student who is happy to explore, and learn the tricks that are categorized as 'mental math' in grade school, and morph into being able to do most HS math in a fraction of the time it takes other students. $\endgroup$
    – JTP - Apologise to Monica
    Commented Sep 9, 2018 at 12:12
  • $\begingroup$ Well, I know that my brother is definitely not an advanced student, thus I thought helping him with a nifty trick (hopefully) would help him. I understand, now, in accordance with @StevenGubkin 's answer, that tricks serve quite the opposite; though in your scenario, tricks like that I suppose are acceptable, simply because there is more than one equation (or function, in particular) that we are considering :) $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 12:16
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    $\begingroup$ Thank you, and I have reached my daily voting limit and have to wait $2$ more hours before I can upvote again (DVL2). I will upvote when I can :P $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 21:54
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    $\begingroup$ Upvoted. $(+1)$ :D $\endgroup$
    – Mr Pie
    Commented Sep 10, 2018 at 2:17
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I teach Year 7 and I’m with Steven on this.

Sort of.

If a “trick” reminds you of a correct understanding that you’ve already established, then it is a procedural memory aid and shortcut. Nothing tricky about that. If a trick is memorised and applied without an understanding of how it was derived and why it works, then it may prevent getting to that essential fuller understanding later.

This sunrise/sunset metaphor you’ve developed with your brother isn’t necessarily bad. But he does need to understand balancing equations first.

$$ \require{cancel} \begin{align} a&=\frac{b}{c}\\ \Rightarrow a\times c&=\frac{b}{c}\times c\\ \Rightarrow a\times c&=\frac{b}{\cancel{c}}\times\cancel{c}\\ ac&=b \end{align} $$

I teach this explicitly, make sure all my students can do it, and make sure they can explain it (not just describe it).

But once this is understood, I start talking informally about moving $c$ to the top on the left. It is on the bottom (denominator) on the right (RHS), so we can move it to the top (numerator) on the left (LHS). You can figure out what I might say about moving $c$ back, or for $+$/$-$ balancing.

This informal talk describes the end result of the algebraic manipulations. I think your sunrise/sunset talk is figuratively equivalent. But maybe more loosely so.

Steven says these kinds of basic results should be derived on the fly. I would modify that slightly to say that you should be able to derive them on the fly, but that applying a correct memorised result is an appropriate shortcut. Mathematicians do this all the time. For example, identities. I could re-derive $\sin^2 x+\cos^2 x=1$ pretty quickly, but that might distract me from thinking about why I’m selecting this identity when I’m trying to do a difficult integration by parts (senior secondary maths).

Identities are a totally different thing from balancing equations, but the truth is equivalent: mathematics is made easier by applying rules for previously established results—but only in the context of appreciating how those results were obtained.

That being said, encourage your brother to take the time to understand balancing equations, and then to think of a better metaphor/trick which more closely links to the ways terms are allowed to move around in a balanced equation.

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  • $\begingroup$ Well, if I wanted to be extremely technical for "understanding" purposes, $$a\times c = \frac bc\times c = \bigg(b\times \frac 1c\bigg) \times c = b\times \bigg(\frac 1c \times c\bigg) = b\times \bigg(\frac1{\require{cancel}{\cancel c}}\times \cancel{c}\bigg) = b\times 1 = b,$$ to reveal not only the how, but the why, though apart from that, you have raised quite an interesting point of which I strongly agree with; thus far, this is the answer I am looking for (although @StevenGubkin 's came close) :D $\endgroup$
    – Mr Pie
    Commented Sep 10, 2018 at 12:05
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    $\begingroup$ @user477343 Glad to help. My teacherly advice is always to consider your student’s needs. If your brother also still needs to understand canceling in algebraic fractions, then run through isolated examples of that with him first. Once he’s got that, shortcut it with my cancelling notation/working when you’re working on his understanding of balancing equations with algebraic fractions. Focus on one main thing at a time, and avoid belabouring already established aspects when delving into new ones. $\endgroup$
    – lukejanicke
    Commented Sep 10, 2018 at 12:33
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    $\begingroup$ @user477343 Yes, instead of a single rotation / reflection transformation on equal fractions, one can break it down into multiple more primitive transformations which move a factor to the other side of an equation by inverting it (or swapping it from numerator to denominator or vice versa within fraction factors). In the OP this requires twice the effort: move $c$ to the left then $a$ to the right, vs. swap $a$ and $c$ by a diagonal reflection. To be maximally proficient at algebra one should master both methods. $\endgroup$
    – Bill Dubuque
    Commented Sep 10, 2018 at 13:33
  • $\begingroup$ @lukejanicke thank you for your advice :) $\endgroup$
    – Mr Pie
    Commented Sep 10, 2018 at 13:35
  • $\begingroup$ @Number thank you especially for your advice; you have been of more help than some others, so far, and I really appreciate that. (I wish my brother could, too) :P $\endgroup$
    – Mr Pie
    Commented Sep 10, 2018 at 13:35
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The trick is more difficult than the proposition. You can look at the fraction $\dfrac{ a}{b}=\dfrac{c}{d}$ from the right, from the left or from the bottom. For example looking from the right gives $\begin{array}{r} \rightarrow\\\rightarrow\end{array} \dfrac{ a}{b}=\dfrac{c}{d} \Longleftrightarrow \dfrac{a}{c}=\dfrac{b}{d}$. Looking at $a=\dfrac{b}{c}$ from the left gives $c=\dfrac{b}{a}$.

Another way is permuting the diagonals. For example permuting $a$ and $d$ in $\dfrac{ a}{b}=\dfrac{c}{d}$ gives $\dfrac{d}{b}=\dfrac{c}{a}$.

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    $\begingroup$ I downvoted because I believe that students should be thinking through reasons why things are true until they reach this "mechanical reasoning" through enough practice. Training them to perform mechanically first, without the thought, is damaging. $\endgroup$
    – Steven Gubkin
    Commented Sep 8, 2018 at 15:52
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    $\begingroup$ @StevenGubkin One needs both, i.e. the combination of your answer and this answer, along with some motivation (e.g. by symmetry). Your (unexplained) downvote could have left the OP with the wrong impression, e.g. wrongly inferring that the above algebra is incorrect, or it is not how experts perform such transformations, etc. It is usually better to have constructive dialog before downvoting something which is correct but needs elaboration. $\endgroup$
    – Bill Dubuque
    Commented Sep 8, 2018 at 16:04
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    $\begingroup$ @user477343 I'm sure my trick to read an equation from right or from left is easier than the sundown swaps around and "sunrise" since I, as an teacher since 13 years, had to read it more than once to understand what you're saying. $\endgroup$
    – user5402
    Commented Sep 8, 2018 at 16:55
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    $\begingroup$ @user477343 To teach something well you must know it well. Are you sure that you do? I've taught algebra for almost 4 decades, to students at all levels, from elementary to research level (e.g. assisting Artin, Gosper, Knuth, Wolfram with Macsyma). I know from this extensive experience that anyone proficient at fraction arithmetic knows by heart this basic transformation (not "trick") - just as they know well other basic algebra (e.g. Binomial Theorem, Difference of Squares factorization, quadratic formula). We learn them early then commit them to memory for rapid subconscious application. $\endgroup$
    – Bill Dubuque
    Commented Sep 9, 2018 at 13:55
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    $\begingroup$ @user477343 The basic idea is quite simple. If we view the equal fractions as 4 numbers in a square, then fraction equality is equivalent to equality of the diagonal (cross) products. But the diagonals don't change under symmetries of the square (rotations & reflections - see below). They may swap the order of terms in products, or swap sides on the equality, but this doesn't alter the truth of the equality of diagonal products. $$\style{ display: inline-block; background: url(//i.sstatic.net/uJphi.png?s=515&g=1) no-repeat center;}{\phantom{\Rule{515px}{30px}{327px}}} $$ $\endgroup$
    – Bill Dubuque
    Commented Sep 10, 2018 at 1:56

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