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I have in my presence a mathematics teacher, who asserts that

$$ \frac{a}{b} = \frac{c}{d} $$

Implies:

$$ a = c, \space b=d $$

She has been shown in multiple ways why this is not true:

$$ \frac{1}{2} = \frac{4}{8} $$

$$ \frac{0}{5} = \frac{0}{657} $$

For me, these seem like valid (dis)proofs by contradiction, but she isn't satisfied. She wants a 'more mathematical' proof, and I can't think of any.

I'm worried that if she isn't convinced, it may be detrimental to some students. Is there another way to systematically demonstrate the untruth of her conjecture?

EDIT: Since the answer which worked was from a comment, but each answer is also very good, I'm upvoting all of them instead of accepting a specific one. Feel free to close this question for being too open if so a moderator desires.

zyx
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user86484
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    I think your examples contradict the claim correctly. – Mikasa Jul 16 '13 at 09:54
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    For a conjecture of form "for all $X$ something is true", a counterexample is a valid proof that this conjecture is false. She should be ashamed, both as a teacher and as a mathematician. – dtldarek Jul 16 '13 at 09:55
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    Is there any other way of disproving it though? – user86484 Jul 16 '13 at 09:56
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    She is clearly missing the concept of the rationals as an equivalence class. – davin Jul 16 '13 at 09:57
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    I think the best thing is that she study an introduction to a course in logic. –  Jul 16 '13 at 09:58
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    @SamiBenRomdhane: or a course in group theory :) – Mikasa Jul 16 '13 at 09:59
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    Try and explain the concept of a fraction. That $\frac a b$ is equal in value to $\frac {pa}{pb}$ for all non-zero $p$, but clearly you're changing the numerator and denominator values themselves, which undermines her equality. – davin Jul 16 '13 at 10:00
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    Her claim is true on $\mathbb{F}_2$. – user1551 Jul 16 '13 at 10:00
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    $\bigl(\forall b \neq 0\bigr)\bigl(\forall a\bigr)\bigl(\forall k \neq 0\bigr)\left(a\cdot(b\cdot k) = b \cdot(a\cdot k) \Rightarrow\frac{a}{b} = \frac{a\cdot k}{b\cdot k}\right)$. Perhaps that's "mathematical enough" for her? – Daniel Fischer Jul 16 '13 at 10:01
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    @DanielFischer Thanks, that did the trick :) – user86484 Jul 16 '13 at 10:06
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    My heart bleeds for that mathematics teacher's students... – DonAntonio Jul 16 '13 at 10:07
  • @DonAntonio Well, I'm one, and I don't think I'm too bad ;) – user86484 Jul 16 '13 at 10:08
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    In short term, I think the best solution is to tell her to read exactly this thread, and post her comments if she still has any disagreement. – user1551 Jul 16 '13 at 10:09
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    Oh, not that you are bad: that "teacher", if what you tell us about him/her is accurate, should be doing anything else but not "teaching" mathematics... – DonAntonio Jul 16 '13 at 10:09
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    @user1551 She saw it, and is now convinced – user86484 Jul 16 '13 at 10:10
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    Then you should be getting his pay check... – DonAntonio Jul 16 '13 at 10:10
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    @user86484 Minor point: these aren't disproofs by contradiction, they're disproofs by counterexample. – Jack M Jul 16 '13 at 11:08
  • @JackM Thanks, I wasn't sure about that – user86484 Jul 16 '13 at 11:30
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    Astonishing, annoying, and depressing. Maybe the school principal and the head of the math department should read this thread, too. I pray that your teacher doesn't hold either of these positions. – bubba Jul 16 '13 at 14:17
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    dear lord, this math teacher should be fired. – acolyte Jul 16 '13 at 14:35
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    Which is worse, this teacher's failure to understand fractions or her apparent belief that "more mathematical" means adding unnecessary complications? – Andreas Blass Jul 16 '13 at 14:58
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    Are you certain that the teacher's claim was not that $ a=c \Rightarrow b=d $ and was simply communicated poorly? – apsillers Jul 16 '13 at 15:00
  • When you jump to the conclusion by using example or counterexample, you skipped the whole "why" part. The teacher did not say that your proof was invalid, or did she? – tia Jul 16 '13 at 15:30
  • Dear @Asal, you are right : I have deleted my comment. – Georges Elencwajg Jul 16 '13 at 16:29
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    @apsillers: This claim is also wrong ($a=c=0$). – Najib Idrissi Jul 16 '13 at 17:10
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    I would ask your math teacher to prove the assertion. That's how it works. You have already disproved it by providing counterexamples. – Chris Cudmore Jul 16 '13 at 17:22
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    Is there really a teacher, or just a stealth delivery for your question? :) – Mihai Danila Jul 16 '13 at 18:01
  • That teacher should be fired. – Stefan Smith Jul 16 '13 at 18:36
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    I met a number of such teacher and I'm not surprised anymore. One of them even claims that algebra is the language of donkeys. – user5402 Jul 16 '13 at 19:19
  • Any chance you can suggest very politely that she ask a coworker? If the other math teaches agree with her, you have a greater issue on your hands. This is 6th grade math. Amazing. – JTP - Apologise to Monica Jul 16 '13 at 20:33
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    @Stefan et al.: A lot of folks are judging the teacher rather harshly here. What about the possibility that the teacher is plenty aware that the notion is preposturous, but is using this problem as an opportunity to teach students to prove things with more rigor? There are several clever proofs already provided in the answers here; maybe her intent was to inspire her students to come up with similar tacks. "She wants a 'more mathematical' proof, and I can't think of any." Sounds like maybe her students might need a little practice in this realm. – J.R. Jul 16 '13 at 21:23
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    To the O.P., RE: I'm worried that if she isn't convinced, it may be detrimental to some students. I'm convinced she's already convinced. Moreover, I'm worried that, if you take one of these proofs to her, you will have missed the larger point. Try coming up with your own, after examining some of these fine examples. As an educator, I'm pretty sure she didn't want to you copy a proof, she wanted you to think of one. – J.R. Jul 16 '13 at 21:41
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    @J.R. Even if the teacher's intent was to get the students to come up with a proof like one of those in the answers here, it doesn't make sense to do so by rejecting a simpler and perfectly correct proof that the student has already given. – Andreas Blass Jul 16 '13 at 22:39
  • Have your teacher read An Introduction to Proof by Contradiction. – brentonstrine Jul 16 '13 at 23:40
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    @Andreas: The O.P. only says: "...she isn't satisfied. She wants a 'more mathematical' proof." Being "unsatisfied" with the obvious proof isn't necessarily the same as "rejecting" it. Is the end goal here to prove the assertion is false, or to help teach students how to construct more formal proofs? I suspect the latter. "Yes, but prove it another way" is a legitimate way to bolster math skills. I don't see why the teacher would ask for another proof if she believed the assertion, yet dozens here seem to think she's daft. I was just offering an alternate theory. – J.R. Jul 17 '13 at 02:52
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    @AndreasBlass According to the OP, the teacher said she "did not satisfied" and demand more "mathematical" proof. It is only OP's conjecture that she was not convinced that the (dis)proof is valid. I think it is like the teacher wanted him to practice running, but he reached there by a skateboard and when the teacher told him to reach there more "physically", he tried to convince the whole StackExchange forum that she rejected the fact that skateboarding is a valid physical method to reach the destination, which she never did. – tia Jul 17 '13 at 02:53
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    Ah! perhaps tia sees what I see. 70 people upvoted the comment by @DonAntonio – "My heart bleeds for that teacher's students..." If my hunch is right (though I'll readily admit, that's not yet proven), I think this teacher could be good for the next generation of mathematicians. Finally a teacher who doesn't spoonfeed, and we call for her pink slip. – J.R. Jul 17 '13 at 03:01
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    @J.R. I generally approve of the "prove-it" prof but why do you need more once you've demonstrated one obvious exception to some supposed rule? – Erik Reppen Jul 17 '13 at 04:32
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    @J.R. I agree that "Yes, but prove it another way" is a legitimate way to bolster math skills, but there's no indication in the question that this teacher said anything like "Yes". And the time to get students to "construct more formal proofs" is not when the student has just given a perfectly correct proof. Wait until the student says or does something that could be improved by formalization. – Andreas Blass Jul 17 '13 at 05:51
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    @J.R., I think that based on the data produced, that maths "teacher" isn't worth the title. Not because she made, apparently, a huge mistake (hey, who hasn't?), but because upon being presented with a perfectly valid and mathematical proof by contradiction that his/her belief is wrong, (s)he still "isn't satisfied with it" and wants a "more mathematical proof" (can we measure more or less mathematical?). This is imo ridiculous, and to think this is a clever, stealth way of the teacher to get the pupils to "think" is even more preposterous. – DonAntonio Jul 17 '13 at 07:30
  • @Don: The way the O.P. presented this, it quickly became an indictment on the teacher's alleged refusal to accept a valid QED. The counterexample is so obviously valid, though, I began to wonder if something might have been misreported by the O.P. I challenge the O.P. to ask the teacher directly: "You do realize this [counterexample] is a valid proof, right? You were just wanting us to come up with something more algebraic?" and see what she says. Shifting from counterexample after counterexample ad infinitum to a proof more like user02138's would be a step forward for a young student. – J.R. Jul 17 '13 at 09:36
  • @J.R. , I think you take things too far away. I thought, and still think now, that the whole OP's story about that teacher maybe an invention to ask a question, and the use of the teacher figure is only to provocate some reaction or whatever. I honestly can't believe there can be a maths teacher anywhere on Earth and 45 light years around that can be so dense as the OP puts it...but I don't care: I just rely on what the OP wrote and I take it on good faith and will, and then based on this I comment/answer. If the OP lied or exaggerated that's his/her problem, not mine. – DonAntonio Jul 17 '13 at 09:41
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    @AndreasBlass: There's no indication that the teacher said, "No," either. That's merely the way the student seems to be interpreting her remarks. That could be a misinterpretation, or she might simply be roleplaying. The O.P. hasn't even said what level math this is, but, given the use of the word "teacher" (vice "professor"), I'm guessing it's high school. I'd be quite surprised to learn that the O.P.'s fears are founded, and this teacher really isn't convinced by the counterexample. (If that ends up being false, though, I'll gladly delete my comments here and upvote some of yours.) – J.R. Jul 17 '13 at 09:43
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    @DonAntonio: Here's another way to say that: If the O.P. had an English teacher who taught as well as the math teacher, maybe we wouldn't be having this discussion ;^) – J.R. Jul 17 '13 at 09:47
  • Actually, I wonder whether the OP misunderstood the teacher, i.e. the teacher meant the implication to go the other way round. – Frerich Raabe Jul 17 '13 at 09:47
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    Iff there is a teacher, and the student misunderstood her enough to give the account above, she either does not know math or does not know how to communicate. I can't see her deserving the title "math teacher" either way. – Muhammad Alkarouri Jul 17 '13 at 09:50
  • @MuhammadAlkarouri: Ever play the telephone game? – J.R. Jul 17 '13 at 10:31
  • Alternatively, instead of disproving her conjecture, you can ask her to add a few conditions like: $a,b$ are co-prime to one another and the same with $c,d$ and, also $a,b,c,d \neq 0$ – Adienl Jul 17 '13 at 10:42
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    I've also met a teacher who argued this. In his opinion fractions should always be simplified. So for him $\frac{2}{4}$ doesn't exist. If you work with simplified fractions, the assertions is true. As it was a low level class and we were very far from understanding what it means (in term of prime decomposition). I always tought he was wrong to teach us this way. – Lucas Morin Jul 17 '13 at 12:11
  • It's already proven. You cannot prove a claim to be right with examples, but you CAN prove a claim to be wrong with a single example. You already did it. – Jake 'Alquimista' LEE Jul 17 '13 at 12:12
  • WOW!! This blew up a LOT! Thanks to everyone, and for those who are doubting that the teacher exists, I can assure you she does, and that the main reason I resorted to asking the question on here is because she doesn't trust my logic over her own. Apart from this issue, she's been a very good teacher, and now she's in full agreement with me. – user86484 Jul 17 '13 at 12:25
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    @user86484: interesting question and answers! Incidentally, are you certain the teacher wasn't using this as a teaching tool, as in "prove it to me another way"? (Possibly saying "I don't believe you, convince me" as a tongue-in-cheek devil's advocate?") – David Robinson Jul 17 '13 at 13:02
  • @DavidRobinson Let us hope so :] – Thomas Jul 17 '13 at 13:13
  • @J.R.: the telephone game assumes long chains of players. We have here a chain of 1. Besides, the whole point of a teacher is to get the point across. Ergo, ... – Muhammad Alkarouri Jul 17 '13 at 14:46
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    Legend has it that Grothendieck once claimed that $57$ was prime. But he would probably have been convinced by the following contradiction: $57=3\cdot 19$. – Julien Jul 17 '13 at 15:23
  • @Muhammad: Each link in the telephone chain is of size 1. If the telephone game results in the inaccurate transmission of information, that can happen anywhere in the chain. Seems like there ought to be a recursive proof for that, but you can just ask anyone who is married – what one person says and another hears doesn't always result in the intended message being successfully conveyed. Also, the O.P. has already called the teacher "very good," so maybe she's more "deserving" than your comment would indicate. In a room full of 11th graders, someone misinterpreted her – so she's no good? – J.R. Jul 17 '13 at 16:33
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    The real problem is convincing the school to get this teacher out of maths classes – wim Jul 17 '13 at 19:26
  • Can I down-vote your teacher? – dav_i Jul 17 '13 at 21:37
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    Accept a specific answer – that's how the site works. The other answerers will not be offended. – slothbear Jul 18 '13 at 03:38
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    Her claim is only true for fully reduced fractions. – ThePiercingPrince Jul 18 '13 at 12:34
  • This question got a huge volume of page views. Was it referenced on another site? – zyx Jul 23 '13 at 19:02
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    Maybe the teacher said the fractions doesn't common factors. Recheck with the teacher. – Felix Marin Oct 24 '13 at 22:09
  • -1 you didn't specify in your edit, which comment answered it. – barlop Jan 09 '14 at 15:41

12 Answers12

108

You can prove that all the numbers are equal ;-)

Let's assume that for all $a,b,c,d \in \mathbb{R}$, $b \neq 0$, $d \neq 0$ we have

$$ \frac{a}{b} = \frac{c}{d}\quad \text{ implies }\quad a = c\ \text{ and }\ b = d. \tag{$\spadesuit$}$$

Now take any two numbers, say $p$ and $q$, and write

$$\frac{p}{p} = \frac{q}{q}.$$

Using claim $(\spadesuit)$ we have $p = q$. For the special case, where one of them equals zero (e.g. $q$), use $$\frac{2p}{2p} = \frac{p+q}{p+q}.$$

I hope this helps ;-)

dtldarek
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Given $a, c \in \mathbb{Z}$ and $b, d \in \mathbb{N}$, suppose that \begin{align} \frac{a}{b} = \frac{c}{d} \Longrightarrow a = c, \, b = d. \end{align} Thus, \begin{align} \frac{a}{b} = 1 \cdot \frac{a}{b} = \frac{2}{2} \cdot \frac{a}{b} = \frac{2a}{2b} \Longrightarrow b = 2 b, \end{align} and $1 = 2$ (as $b$ is non-zero), which is absurd.

user02138
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Say $$\frac { a }{ b } =\frac { c }{ d } =k,$$ then $$a=bk,\\ c=dk.$$ Sum up $$\left( a+c \right) =\left( b+d \right) k.$$ You find $$\\ \frac { a+c }{ b+d } =k=\frac { a }{ b } =\frac { c }{ d }. $$ Which implies that you can find another number which is equal to $\frac { a }{ b } .$

newzad
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If $\frac{a}{b}$ is an integer $n$, then:

$\frac{a}{b}=\frac{n}{1}$.

In other words, if $\frac{a}{b}$ is an integer, we also know $b=1$ if your teacher were correct. However, $\frac{a}{b}$ is an integer if and only if $b$ divides $a$ and we have fractions such as $\frac{4}{2}=2$ for which the denominator is not equal to $1$.

Amitesh Datta
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    Another very nice and simple-to-understand disproof. :) Thanks – user86484 Jul 16 '13 at 10:16
  • @user86484 You're welcome! If you like the other answers as well, then don't forget to upvote them (by clicking the arrow pointing upward on the left of the answer). – Amitesh Datta Jul 16 '13 at 10:19
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I think the implication is that IF $a=c$ THEN $b=d$ which is the mathematical fact.

Thomas
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Dave
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    Yes, I was just about to suggest this! The teacher might just have bungled the statement of the problem in their head, or the OP may have misunderstood the intended statement in this way. Still, we should expect that such a misstatement would have been sorted out rather quickly with these counterexamples... – rschwieb Jul 16 '13 at 15:07
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    I can't believe this got 8 upvotes, as it it is obviously wrong as demonstrated in the second example in the question. – Kevin Jul 16 '13 at 18:15
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    @Kevin I think it's always valuable to be able to see why someone else holds a particular misconception because that can be the key to unlocking them from the rut they are stuck in. – Ben Jackson Jul 16 '13 at 20:00
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    Wha..? Isn't this just restating the teachers wrong belief? – brentonstrine Jul 16 '13 at 23:36
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    Works if you do it the other way around: IF $b = d$ THEN $a = c$. The only place where the other way fails is with 0 in the numerators, but since you can't have 0 in the denominator, this covers all valid cases. – Darrel Hoffman Jul 17 '13 at 02:34
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    This answer statement is true in "spirit" if not in detail. Obviously if $a=c=0$ we know nothing of $b$ and $d$, but that is a specific case in which the concept of a fraction isn't even useful. This reminds me of when I was first told that the limit of a sequence isn't necessarily a limit point (accumulation point) for the sequence and was provided with the sequence $1,1,1,1,\ldots$ as a counter example. – Spencer Jul 17 '13 at 03:12
19

To answer your question:

For me, these seem like valid (dis)proofs by contradiction, but she isn't satisfied. She wants a 'more mathematical' proof, and I can't think of any.

I'm worried that if she isn't convinced, it may be detrimental to some students. Is there another way to systematically demonstrate the untruth of her conjecture?

Your teacher is obviously wrong. Or: there might have been a simple misunderstanding between the two of you. Anyway, as stated, your argument is perfectly fine. If the claim is that: $P(x)$ is true for all $x$, then you can prove with all mathematical precision that the statement is false by just finding one $x$ for which the statement doesn't hold. You give two examples. but you only need one.

Instead of trying to come up with another argument, you could simply ask you teacher about what makes a valid mathematical argument. Ask the teacher to point out exactly she doesn't believe that proof by contradiction is a valid way to argue. You of course want to maintain a good relationship with your teacher, but you could also try to show her a book on mathematical logic. Find a good book about mathematical proofs.

Community
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Thomas
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    Or, the teacher already knows that the claim isn't true, but wants to see if her students can express it in a more mathematical fashion – as opposed to using a simple counterexample. Not every mathematical fallacy has such a ready supply of counterexamples. If a student can't mathematically disprove something so basic, how will those students handle the really tough proofs? This sounds more like a pedagogy strategy than teacher incompetance, especially given the O.P.'s confession. – J.R. Jul 16 '13 at 21:31
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    @J.R.: Or that. – Thomas Jul 16 '13 at 22:01
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    As a (relative) layman, I've got to ask, why is disproving something by counterexample considered not "mathematical"? – mikołak Jul 17 '13 at 11:54
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    @TheTerribleSwiftTomato Generally, it's not. I think what J.R. means is just putting it in 100% precise lingo. Ie, a concise short paragraph full of, "therefore", "there exists", "we can choose" and "such that", possibly ending with "the contradiction establishes the theorem. ▫" Or, in the case here, writing it in terms of $\forall$s and other symbols. –  Jul 17 '13 at 22:48
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It is good to give someone a way to save face. Try to salvage their statement as something true. So you could say:

We know about lowest terms. Let $a,b,c,d$ be positive integers
LEAD IN STATEMENT

then

  • for fractions in lowest terms, IF $\frac{a}{b}=\frac{c}{d}$ THEN $a=c$ and $b=d.$

For the LEAD IN STATEMENT I would use one or both of

Of course IF $a=c$ and $b=d$ THEN $\frac{a}{b}=\frac{c}{d}.$

The correct converse is:....

OR

$\frac{3}{6}$ is not in lowest terms because $\frac{3}{6}=\frac{1}{2}$ and $3 \gt 1, 6 \gt 2.$ However...

Fine points:

  • Alternately, it might be effective to only say : "Oh I see, you meant that for fractions in lowest terms ..." and let her reflect silently that we wouldn't have the concept of lowest terms unless simplification is possible.

  • Really $\frac{0}{1}$ is in lowest terms and $\frac{0}{2}$ is not. However by invoking positive integers early on we avoid discussing that and $\frac{2}{3}=\frac{-2}{-3}$ and we avoid breaking up the flow by saying "provided $b\ne 0$ and $d \ne 0$"

  • The fact about fractions in lowest terms is not trivial. A legitimate proof requires having a certain repertoire of facts about relatively prime integers which is more sophisticated than many bright Calculus students have (because they have not been through them before.)

Aaron Meyerowitz
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    Maybe the lowest term statement is what she meant. As I said, a proof of that is a good exercise. – Aaron Meyerowitz Jul 16 '13 at 21:56
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    I think the usual convention meant by "lowest terms" also stipulates the denominator is positive. –  Jul 16 '13 at 22:00
  • That would do it. If you want to generalize to other domains which have a norm then you might say $a/b$ is in lowest terms if there is no solution of $a/b=c/d$ with $|d| \lt |b|$ and $|c| \lt |a|$ – Aaron Meyerowitz Jul 16 '13 at 22:41
  • "Lowest term" doesn't work for fractions of the form $\frac{0}{n}$, unless I misunderstand the meaning of "lowest term" – Sklivvz Jul 17 '13 at 08:51
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    @Sklivvz The fraction $0/1$ is in its lowest terms; the gcd between $0$ and $n$ is $n$. – egreg Jul 17 '13 at 10:52
  • You don't need any high-powered facts about relatively prime numbers to prove that simplified fractions exist. There are a bunch of ways to write $a/b$ as fraction; just pick the one with the smallest positive denominator. Since $a/b=c/d$ and $c=d$ implies $a=b$, these fractions are uniquely determined. If you're feeling particularly proofy, you'll realize that a smallest positive denominator exists because the naturals are well founded, or alternatively because of induction. – Mario Carneiro Jul 17 '13 at 22:07
  • @Mario So you are saying that IF a/b=c/d with a,b relatively prime and c,d relatively prime AND b is the smallest denominator for such a fraction THEN clearly b=d? That is a true deduction but not clear. – Aaron Meyerowitz Jul 18 '13 at 13:53
  • I suppose a fairly direct proof could use a,b relatively prime implies as+bt=1 for some s,t and that implies that if b|ad then b|d. Since we do have bc=ad we have b|d also d|b etc. unique factorization could be recruited too. – Aaron Meyerowitz Jul 18 '13 at 14:05
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    @Aaron My proof never uses the words "relatively prime", which is why it is so easy to prove. If $x=a/b=c/d$ and $b$ is minimal and $d$ is minimal, then by definition $b=d$, whence $a=c$. The minimum here is with respect to the set ${n\in\Bbb N:\exists m\in\Bbb Z\ x=m/n}$. The characterization of minimal elements as relatively prime is the (not very) hard part. – Mario Carneiro Jul 18 '13 at 15:35
  • @Mario I agree. But what your proof does not address is that it could be a/b=c/d with b minimal and c,d relatively prime yet d greater than b. in fact that can't happen but it needs some work. – Aaron Meyerowitz Jul 18 '13 at 19:16
  • @Aaron As I said, I make no reference to relative primality in my theorem or its proof. I claim that there exists a unique "reduced fraction" equivalent to any fraction, given as the minimal element of the set above. Uniqueness is trivial, and existence follows from wellfoundedness of $\Bbb N$. (con't) – Mario Carneiro Jul 18 '13 at 20:07
  • @Aaron To connect this statement to statements about relative primality takes more work, as follows. Claim: the minimal representatives mentioned above are exactly those such that $a,b$ are coprime. Pf: if $(a,b)>1$, then $a/b=(a/(a,b))/(b/(a,b))$ and $b/(a,b)<b$, so $b$ is not minimal. Conversely, since $a/b=c/d$ implies $a/b=(a,c)/(b,d)$, if $(a,b)=1$, and $c,d$ is the minimal rep. (hence $(c,d)=1$ by the first part), then $(b,d)\le d$ implies $(b,d)=d$ (since $d$ is minimal), so $b$ is a multiple of $d$, and thus $a=c(b/d)$ and $b=d(b/d)$, so $b/d$ divides $(a,b)=1$, hence $b/d=1$. – Mario Carneiro Jul 18 '13 at 20:26
  • @Aaron To sum up, I think the proof above answers your question about "it could be a/b=c/d with b minimal and c,d relatively prime yet d greater than b", but it is irrelevant to the simpler statement that there exist such things as reduced fractions and they are uniquely determined. (In this case I take "in lowest terms" to mean "the denominator is as small as possible", not "$a$ and $b$ are coprime". The difference between these two is that the first needs only the first paragraph, while the second needs the first two paragraphs of this response.) – Mario Carneiro Jul 18 '13 at 20:42
6

It seems more likely to me that the misunderstanding here is not in the mathematics, but in the logic of the argument. This is evident because the teacher still refuses to accept the refutation even in the face of evidence.

The confusion, in the teacher's mind, is probably rooted in the fact that given:

$$ a = c, \space b=d $$

then it follows that: $$ \frac{a}{b} = \frac{c}{d} $$

CSJ
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I would point her to this wiki article, or other reference that explains the use of counterexamples.

A single falsity disproves a supposed proof.

Then show that while 1/2 = 2/4, 1<>2 and 2<>4.

Point proven! (or disproved, as the case may be.)

TecBrat
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    I would hold out a little hope for this teacher. Maybe she KNOWS better, but is challenging people to prove her wrong. – TecBrat Jul 16 '13 at 17:09
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If you use examples where the fractions are equal to integers it becomes impossible to deny. The example in the question becomes

$$ \frac{2}{1} = \frac{8}{4} $$

Now draw pictures or count on fingers or otherwise demonstrate the equality in ways that do not rely on any shared understanding of the algebra.

zyx
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counterexample is the best thing ever to disprove something . i dont know whats wrong with modern teachers , and see they want it in different way is annoying me somehow . counterexample is neat way , there is not such a thing more neat to disprove things like this , if you wanna do a formal proof , then u have to deal with infinitely of contradictions that u can avoid with counterexample . one point for you , 0 point to your teacher .

Bswan
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1

Alternatively, $\frac{a}{b}=\frac{c}{d}$ if and only if $ad=cb$ by the definition of fraction equality. So, according to your teacher, $ad=cb\iff a=c\text{ and }d=b$, i.e., there is only one way of writing the integer $ad=cb$ as the product of two integers. However, in general we can write an integer as the product of two integers in many different ways. For example, $24$ can be written as:

$1\times 24$

$2\times 12$

$3\times 8$

$4\times 6$

$6\times 4$

$8\times 3$

$12\times 2$

$24\times 1$

I hope this helps!

Amitesh Datta
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