5

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!

Bill Dubuque
  • 1,038
  • 1
  • 8
  • 12
Mr Pie
  • 159
  • 6
  • 1
    @TommiBrander thank you :) – Mr Pie Sep 08 '18 at 15:46
  • 2
    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. – Bill Dubuque Sep 08 '18 at 16:32
  • @Number I like that idea. He has told me he is learning about algebra and how if $2x+1=5$ then he must backtrack and solve for $x$, but it gets confusing when he sees something like $(2x+1)\div 3=5$. He has not got the patience for such equations and just wants to find a short simple trick, even when I have countlessly told him otherwise. Perhaps if he focuses on difference equalities (because addition $+$ and subtraction $-$ is noticeable easier for him than $\times$ and $\div$), he might obtain a better understanding, overall. Thank you for your thought; I definitely appreciate it :) – Mr Pie Sep 08 '18 at 16:39
  • 2
    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). – Bill Dubuque Sep 08 '18 at 16:44
  • @Number well, I was thinking that if we redefine $a-b$ as $a+(-b)$ then we can swap these two variables around after realising that $a+(-b)=(-b)+a=-b+a$. This might prove to be a start, though at least when dealing with a single expression. Perhaps the equation, $0=a-b$, might be better off; and we could start letting the subject be another variable, say, $c$, forming $c=a-b$. – Mr Pie Sep 08 '18 at 16:47
  • Consider: How will this trick help later on with identities such as $\sqrt x = a \implies a^2 = x$ and $\ln x = a \implies e^a = x$? Requires new tricks for each, or is there some "main idea" that covers all? – Daniel R. Collins Sep 08 '18 at 17:25
  • 1
    Related: https://math.stackexchange.com/questions/1505354/can-i-think-of-algebra-like-this/ – Daniel R. Collins Sep 08 '18 at 17:26
  • 2
    @DanielR.Collins Re: you prior two comments: Those more general techniques don't reveal the richer symmetry here (e.g. see inequation's answer). – Bill Dubuque Sep 08 '18 at 20:11
  • 3
    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. – Bill Dubuque Sep 08 '18 at 21:53
  • @DanielR.Collins that is beyond my brother. I don't think he will ever want to learn about that. He will go up to Year 10 maths because it is mandatory, and then he will drop it. By that stage, he will probably get used to the idea of maths a little better than currently. – Mr Pie Sep 08 '18 at 23:47
  • @Number that is definitely important, but if you read my comment above, my brother is not interested. This "trick" I have told him is intended to be temporary to help him now, just so he can understand a little better for the time being, because I have told him to actually learn maths the real way. But as you stated earlier, learning these symmetry transformations in additive forms might help; as they say, slow and steady wins the race. – Mr Pie Sep 08 '18 at 23:51
  • 2
    @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. – Bill Dubuque Sep 09 '18 at 00:12
  • @Number well, then, that is very kind of you :) – Mr Pie Sep 09 '18 at 00:13
  • @user477343: It sounds like we all agree that the suggested trick is a dead-end. You'll probably find that most math educators are quite leery of recommending dead-end techniques, having experienced the really bad situation of damaged and hopeless students in later courses. – Daniel R. Collins Sep 09 '18 at 01:40
  • I would also suggest that people are actually not very likely to get a good at math just because they spend future time "dealing" with hard stuff that they don't understand. The more common result is a painful crash-and-burn in a later class. Actually understanding math is more often like a sudden "light switch" when teacher and student find a way to communicate the big idea successfully. – Daniel R. Collins Sep 09 '18 at 01:43
  • @DanielR.Collins well, my maths teacher just teaches the rules but does not actually say why they are that way; e.g., $a^0=1$ for all $a$. Why is that, sir? That's just the way it is. Later on I tried to figure out why, and I noticed that for all $b$, one has that $a^0=b\div b = b^1\div b^1=b^{1-1}=b^0$ and it became easy from there. My point is, before I posted this, I thought tricks were helpful because they appeared to help students cope with maths for the time being, and once they wrap their head around it, they can learn the whys and hows. Why are students not taught about axioms? – Mr Pie Sep 09 '18 at 01:52
  • (My comment is from a high-school perspective.) – Mr Pie Sep 09 '18 at 01:54
  • @JoelReyesNoche hah, thanks for that. I'm glad you picked that up. This "trick" would have otherwise been more confusing... – Mr Pie Sep 09 '18 at 01:56
  • 2
    To respond to your last question, (high-school) students are not (generally) taught axioms because of pedagogical failure, not because it is impossible. Did you know that for many centuries, the mathematicians studied Euclid's Elements, which was a very rudimentary axiomatic treatment of geometry. Of course, we know today that there are a lot of holes in Euclid's Elements, so we can do better! But for some reason the ones creating the syllabuses do not seem to understand that mathematics is not about tricks or computation but about truths and justification. Sad, really. – user21820 Sep 09 '18 at 03:33
  • @user477343: I'd say that you could be getting better instruction. Some high school teachers will argue that students at that level cannot handle "real" math. Based on your questions here, I'd say that you certainly can. Look to your textbook which should usually have reasonable and widely-accepted explanations for things. If you also don't have a good book, then look to free OpenStax books online. E.g., see here and scroll down to the justification for "Exponent of Zero": https://cnx.org/contents/CImQfPDv@3.11:5K97v8bM@2/Divide-Monomials – Daniel R. Collins Sep 09 '18 at 06:49
  • (Cautionary note that the rule $a^0 = 1$ classifies as a definition [not something established by proof], but we are all owed a reasonable motivation for why that definition is worthwhile/makes sense/is consistent with other properties. Also, most formal math people will say it does not hold for all real numbers, in that when $a = 0$ it's undefined.) – Daniel R. Collins Sep 09 '18 at 06:55
  • This is not a trick, this is how division works. – Rusty Core Sep 10 '18 at 07:31
  • @RustyCore I know, but the "sunrise... sundown..." part is just a way of remembering how it works, and that is the part I call the trick. – Mr Pie Sep 10 '18 at 08:24
  • 1
    I'm finding it difficult to imagine a situation where someone would need to know these various equivalent formulations before learning the basics of beginning algebra, such as discussed in this question that @Daniel R. Collins mentioned elsewhere. (continued) – Dave L Renfro Sep 10 '18 at 16:40
  • 2
    Also, despite the enormous amount of arithmetic and algebraic computations I've done over the years (e.g. calculators were not generally available until after I'd learned most of elementary calculus), I don't know these rules. In fact, I found myself having to mentally manipulate equations to see that they were correct (takes about a second). – Dave L Renfro Sep 10 '18 at 16:42
  • @Dave Surely there were plenty of opportunities where you could have done a diagonal swap of $a,c$. But not knowing it you instead broke it down into two separate steps of moving $a$ and $c$ to the opposite sides. Similarly one unfamiliar with fractions can do some of their transformations less efficiently, e.g. instead of $ac/(bc) = a/b$ they can calculate $(ac)(bc)^{-1} = ab^{-1}$ using field axioms. – Bill Dubuque Sep 10 '18 at 20:36
  • 2
    It would be quite cumbersome doing all fraction arithmetic more primitively like that. One should strive to exploit richer structure when it exists. I don't recall anyone claiming that one "needs to know these ... before learning basic of beginning algebra". It's one of many optimizations that can help one simplify algebra - allowing one to focus better on the essence of the matter. – Bill Dubuque Sep 10 '18 at 21:10
  • @Number I agree now, though I don't think "cumbersome" is the right word, but after you showed me your comments, I completely agree :) – Mr Pie Sep 10 '18 at 21:30

4 Answers4

17

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.

Steven Gubkin
  • 25,127
  • 4
  • 61
  • 110
  • 3
    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). – Bill Dubuque Sep 08 '18 at 15:46
  • 1
    "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." $100%$ agree. That's what I meant when I told him not to use tricks. I think this is an axiom, for which we show that $b\div c$ $= b(1\div c)$ and is known as the Multiplicative Cancellation Law. This and this answer should be taught, and not some lame ol' trick. Thank you for your answer, and as complicated as it may seem to my brother, I hope it at some time increases his understanding of algebra at least :D – Mr Pie Sep 08 '18 at 15:47
  • 7
    @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. – Steven Gubkin Sep 08 '18 at 15:51
  • 3
    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. – Bill Dubuque Sep 08 '18 at 15:57
  • 2
    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. – user5402 Sep 09 '18 at 09:26
  • @inéquation The claim is not too unrealistic. Of course $\pi$ is an irrational number, but you can still multiply both sides by $\pi$ and have $(x\div \pi)\pi = x$, so the cancellation laws still apply to that constant. Also, we are talking about students, particularly those in Year 7. The "average" student, say, would not have formulae memorised just for their sake (like a trick, for example), thus deeming reasonable to believe they should often fully derive them to ensure a maximum understanding. That trig rule you mentioned is beyond Year 7 level (apart from gifted students, but off-topic). – Mr Pie Sep 09 '18 at 12:09
  • 2
    @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). – Bill Dubuque Sep 09 '18 at 14:28
  • @user477343 You missed my point yet again. Of course we can multiply and divide by $\pi$ but how do you explain to grade 7 what does splitting $2$ into $\pi$ pieces ($2/\pi$) even mean? Also do you and StevenGubkin expect students to memorize the pythagorean theorem or rederive it every time. I hope your unrealisticness stops at fractions. It seems neither you nor StevenGubkin ever taught in a middle school, middle school$\neq$high school$\neq$university. – user5402 Sep 09 '18 at 15:12
  • +1 but I'll admit that the "split it into $c$ equal pieces" is my least favorite part. I would prefer to point back to the inverse property of real numbers, where presumably we took this as an axiom. For prior motivation with a natural number example, of course, it's a reasonable thing to say. – Daniel R. Collins Sep 09 '18 at 18:22
  • @inéquation look, I can't fully tell how the students in Year 7,8,9 cope with their algebra. I have never done Year 7 maths whilst being in Year 7, I've always been advanced (and that includes 8 and 9). I have some friends who take 10A maths classes every now and then when they can (i.e., when timetables don't clash), but that's why when I referred to average students, I wrote "average", because I do not quite know what to expect. I am no teacher (which is why I came up with tricks) and I have never been average. In my own experience, I have tried to memorise formulae. I guess you are right :\ – Mr Pie Sep 09 '18 at 22:46
  • @inéquation I am generally unconcerned with complete rigor in my teaching. If a student brought up the example of $\pi$, and wanted to dig deeper, we could certainly go down that rabbit hole. Division by rationals can be understood from both "How many groups" and "How many in each group" perspectives. Multiplication and division by irrationals seems to require one to dive pretty deeply into the actual structure of the reals to be able to fully comprehend. Although I think basic limiting ideas and some hand waving can provide some satisfaction for curious minds. – Steven Gubkin Sep 10 '18 at 23:25
3

$$a=\frac bc\implies \;c=\frac ba.$$

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.

JTP - Apologise to Monica
  • 9,770
  • 43
  • 70
  • 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 – Mr Pie Sep 09 '18 at 12:00
  • 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. – JTP - Apologise to Monica Sep 09 '18 at 12:12
  • 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 :) – Mr Pie Sep 09 '18 at 12:16
  • Keep encouraging your brother to work on his math and continue his progress. I've gotten a downvote with no comment why. One of the joys of life. Be well. – JTP - Apologise to Monica Sep 09 '18 at 12:42
  • 1
    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 – Mr Pie Sep 09 '18 at 21:54
  • 1
    Upvoted. $(+1)$ :D – Mr Pie Sep 10 '18 at 02:17
3

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.

lukejanicke
  • 500
  • 3
  • 10
  • 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 – Mr Pie Sep 10 '18 at 12:05
  • 1
    @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. – lukejanicke Sep 10 '18 at 12:33
  • 2
    @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. – Bill Dubuque Sep 10 '18 at 13:33
  • @lukejanicke thank you for your advice :) – Mr Pie Sep 10 '18 at 13:35
  • @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 – Mr Pie Sep 10 '18 at 13:35
-2

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 $\dfrac{d}{b}=\dfrac{c}{a}$.

user5402
  • 1,528
  • 15
  • 23
  • 1
    I do not understand what "proposition" you are referring to. The only equation I am considering is $a=b\div c$ and not any other equation. I can let $a=x\div y$ and have the equation $x\div y = b\div c$ and now I can literally do the same trick from all angles, for there are now two such "horizons" and another pair of "sun positions". I believe you are saying that there are many ways to manipulate the equation, but as far as I am aware, my brother was only concerned with the equation strictly involving $a$, $b$ and $c$. If I show him your answer with $d$, I am sure it will complicate matters. – Mr Pie Sep 08 '18 at 13:48
  • 5
    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. – Steven Gubkin Sep 08 '18 at 15:52
  • 2
    @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. – Bill Dubuque Sep 08 '18 at 16:04
  • @Number I also have no idea what inequation means by "from the left/right/bottom". This answer really says little more than just "do it the right way". – Steven Gubkin Sep 08 '18 at 16:07
  • 1
    @StevenGubkin Your answer is just as lacking - more seriously so in my opinion since it provides no intuition whatsoever. – Bill Dubuque Sep 08 '18 at 16:08
  • @Number I would love if you would answer this question as well, so I can see what intuition I am lacking. – Steven Gubkin Sep 08 '18 at 16:11
  • 1
    @StevenGubkin The innate symmetry that governs the transformations (which at least is hinted above - but completely obfuscated in your answer) – Bill Dubuque Sep 08 '18 at 16:12
  • 1
    @StevenGubkin You're missing the point. The OP ask about a trick, I answered with another (hopefully simpler) trick that I use. Of course one should teach the fundamentals first but that's not what the question is about nor it is how students usually solve equations. – user5402 Sep 08 '18 at 16:18
  • @inéquation well, to be honest, this answer deemed rather partial. I asked for feedback, and you provided a different equation to consider. I know how it works, it is just that my brother does not and cannot stand maths. He hates it, and in his own words, "it is for nerds like me." All he wants to do is pass with something above a $C$ to impress my mum. This is definitely not the right attitude he should have toward his learning, but I do not want him to give up and fail, either. StevenGubkin's answer might just convince my brother to think otherwise, but I am unsure about yours :\ – Mr Pie Sep 08 '18 at 16:33
  • @Number if you read my comment above, I do not think my brother will notice this symmetry you speak of in inéquation's answer. I have posted this for the sake of my brother, not myself. – Mr Pie Sep 08 '18 at 16:44
  • 3
    @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. – user5402 Sep 08 '18 at 16:55
  • @inéquation I wrote $a=b\div c$ instead of $a=\dfrac bc$ to make things look neater, which might have served confusion, but either way, you must be right then; though if your trick was easier, my brother would never have come up to me and asked for a trick. He is learning about what you wrote and finding it difficult; he might just need more time to comprehend it. – Mr Pie Sep 08 '18 at 23:55
  • 3
    @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. – Bill Dubuque Sep 09 '18 at 13:55
  • @Number Yes, I am sure. I am doing Maths Methods currently in my Year 9 class, so I think my knowledge suffices. And by the way, that is some good experience you got there! Well done :P – Mr Pie Sep 09 '18 at 21:57
  • 2
    @user477343 I asked because you wrote in your first comment above that you didn't understand the proposition briefly sketched in this answer. Is it clearer to you now, including the symmetry-based approach that I alluded too? These are not topics typically taught in grade 9 or earlier (though they certainly are capable of comprehension at that level if presented by a talented teacher). Indeed, they are probably independently discovered much more frequently than they are taught. – Bill Dubuque Sep 09 '18 at 22:52
  • 1
    @StevenGubkin Do you really not understand how the new equality is derived by "looking from the left", or do you mean to imply that it might not be clear to some readers so could use further elaboration? Also, the author never claimed that the proofs of these transformations should be omitted. It is truly a shame that this answer have been so heavily downvoted (now -4+3) when in fact it is the only answer that sheds any light on the essence of the matter. – Bill Dubuque Sep 09 '18 at 22:59
  • @Number yes, it is clearer to me, thanks to your help, especially. The maths that I know thus far has mostly been taught and discovered by myself (e.g. in Grade 6, I taught myself how to find $x$ in "typical" algebraic equations such as $2x+1=5$), but I have had some friends who helped me in my high-school years, so I agree with you on how they might be more discovered than taught. Once again, though, thank you for your help. You can combine your comments in one answer, if you like :D – Mr Pie Sep 09 '18 at 23:22
  • 4
    @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. $$\style{ display: inline-block; background: url(//i.stack.imgur.com/uJphi.png?s=515&g=1) no-repeat center;}{\phantom{\Rule{515px}{30px}{327px}}} $$ – Bill Dubuque Sep 10 '18 at 01:56
  • @Number what the — that is all from $\LaTeX$ ??? Where's the answer?! That is a BEAUTIFUL explanation! $(+++++++,1)$ $\ast$takes screenshot$\ast$ – Mr Pie Sep 10 '18 at 02:08
  • 2
    @user477343 It's just an image. Alas, I don't have enough spare time at the moment to compose an answer that does the matter justice. Hopefully the hints in the comments help lead you along the correct path. Maybe you can employ these and other beautiful simple connections between arithmetic and geometry to motivate your brother, e.g. since you enjoy puzzles you could show him some very simple applications of group theory to puzzles, e.g. the fifteen puzzle or Rubik's cube (google them with "group theory"). – Bill Dubuque Sep 10 '18 at 02:32
  • @Number Well, I know both those puzzles you mentioned as examples. Thank you very much!!!!! You have been of great help, and is very noble of you to not use any of your comments as an official answer. Looking at your profile, how you got your Math.SE account suspended is beyond me. Tell 'em mods to put it back :D – Mr Pie Sep 10 '18 at 02:33
  • 2
    @user477343 A simple real world (manipulative) model is a (Flintstone!) square wheel, with 4 spokes representing the diagonally-related numbers in the cross products). The numbers hang on the rim corners at the ends of the spokes. Placed in front of the wheel is a glass pane which displays the fraction bars (vinculums) and the equal sign (in front of the wheel hub), which tells us how to read the numbers as fractions. – Bill Dubuque Sep 10 '18 at 14:30
  • 2
    It is clear that the rigid rotation/reflection transfomations on the wheel don't alter the diagonal connections (spokes), so they preserve the two cross-products (up to order) so the new fractions are also equal. The method in this answer is equivalent, but it instead fixes the square wheel and rotates/reflects the glass pane (= fraction viewpoint, or frame of reference). They are inverse transformations since their product (composition) is the identity (e.g. fix the pane to the wheel and rotate the entire contraption). – Bill Dubuque Sep 10 '18 at 14:30
  • 2
    There are many (web) expositions on these Symmetries of the Square – Bill Dubuque Sep 10 '18 at 14:30
  • @Number Regarding your first comment, the explanation was unnecessary; I understood what the image described, but that's okay. Regarding your second comment, thank you for telling me of the transformation names (I just called 'em "fractions", but not anymore)! And finally, regarding your third comment, thank you for the link, I will check it out! Edit: Wow! That is very in-depth, particularly this other link! Thank you!! :D – Mr Pie Sep 10 '18 at 21:14
  • @Number To answer your question, I still do not understand what the phrase "looking from the left" is trying to get at. Your observation about the symmetries of the square is a nice one for sure, and could certainly be turned into a memorable and enlightening experience for students at this level. – Steven Gubkin Sep 10 '18 at 23:20
  • For some reason, the image in Bill Dubuque's comment above does not display for me on my mobile device. Just in case there are others with this problem, here is the link to the image on Imgur: https://i.stack.imgur.com/uJphi.png – Brahadeesh Jan 30 '23 at 07:49