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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.

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    $\begingroup$ I think your examples contradict the claim correctly. $\endgroup$
    – Mikasa
    Jul 16, 2013 at 9:54
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    $\begingroup$ Her claim is true on $\mathbb{F}_2$. $\endgroup$
    – user1551
    Jul 16, 2013 at 10:00
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    $\begingroup$ My heart bleeds for that mathematics teacher's students... $\endgroup$
    – DonAntonio
    Jul 16, 2013 at 10:07
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Jul 16, 2013 at 10:09
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    $\begingroup$ Which is worse, this teacher's failure to understand fractions or her apparent belief that "more mathematical" means adding unnecessary complications? $\endgroup$
    – Andreas Blass
    Jul 16, 2013 at 14:58

12 Answers 12

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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 ;-)

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    $\begingroup$ What about if both are equal to 0? $\endgroup$
    – jwg
    Jul 16, 2013 at 13:02
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    $\begingroup$ @jwg Then they are already equal ;-) $\endgroup$
    – dtldarek
    Jul 16, 2013 at 13:18
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    $\begingroup$ "Things which equal the same thing also equal one another." – Euclid, c. 300 BC. $\endgroup$
    – Pål GD
    Jul 16, 2013 at 19:16
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    $\begingroup$ Or in modern terminology, equality is transitive. $\endgroup$
    – Dan
    Jul 17, 2013 at 1:57
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    $\begingroup$ @Dan not to be pedantic (actually, just to be pedantic), but this relation is usually called euclidean. It says that if $R(a,b)$ and $R(a,c)$ then $R(b,c)$. It is equivalent to transitivity if you also have symmetry or reflexivity (which we have in equality). $\endgroup$
    – Pål GD
    Jul 17, 2013 at 9:42
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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.

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  • $\begingroup$ it is not absurd, it is a contradiction. Just saying... $\endgroup$
    – Daniel
    Jul 17, 2013 at 12:27
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    $\begingroup$ A proof by contradiction aims to develop an absurdity. $\endgroup$
    – user02138
    Jul 17, 2013 at 12:46
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    $\begingroup$ Absurdity is actually the correct word. Since it was demostrated by "Reductio ad absurdum". Upvote, I learned something new :) $\endgroup$
    – Daniel
    Jul 17, 2013 at 12:53
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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 $\frac { a }{ b } .$

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  • $\begingroup$ That's a nice way of showing it. Thanks $\endgroup$
    – user86484
    Jul 16, 2013 at 10:06
  • $\begingroup$ Where is contradiction here? 1/2 = 2/4 => 1/2=2/4=3/6, so? $\endgroup$
    – RiaD
    Jul 17, 2013 at 15:28
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    $\begingroup$ @RiaD what do you mean? I am giving an example to show if $a/b=c/d$, it does not mean always $a=b$ and $c=d$. $\endgroup$
    – newzad
    Jul 17, 2013 at 15:32
  • $\begingroup$ Oh, got it, my bad $\endgroup$
    – RiaD
    Jul 17, 2013 at 17:04
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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 $1$.

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    $\begingroup$ Another very nice and simple-to-understand disproof. :) Thanks $\endgroup$
    – user86484
    Jul 16, 2013 at 10:16
  • $\begingroup$ @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). $\endgroup$
    – Amitesh Datta
    Jul 16, 2013 at 10:19
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I think the implication is that IF $a=c$ THEN $b=d$ which is the mathematical fact.

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    $\begingroup$ 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... $\endgroup$
    – rschwieb
    Jul 16, 2013 at 15:07
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    $\begingroup$ I can't believe this got 8 upvotes, as it it is obviously wrong as demonstrated in the second example in the question. $\endgroup$
    – Kevin
    Jul 16, 2013 at 18:15
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    $\begingroup$ @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. $\endgroup$
    – Ben Jackson
    Jul 16, 2013 at 20:00
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    $\begingroup$ Wha..? Isn't this just restating the teachers wrong belief? $\endgroup$
    – brentonstrine
    Jul 16, 2013 at 23:36
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    $\begingroup$ 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. $\endgroup$
    – Darrel Hoffman
    Jul 17, 2013 at 2:34
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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.

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    $\begingroup$ 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. $\endgroup$
    – J.R.
    Jul 16, 2013 at 21:31
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    $\begingroup$ @J.R.: Or that. $\endgroup$
    – Thomas
    Jul 16, 2013 at 22:01
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    $\begingroup$ As a (relative) layman, I've got to ask, why is disproving something by counterexample considered not "mathematical"? $\endgroup$
    – mikołak
    Jul 17, 2013 at 11:54
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    $\begingroup$ @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. $\endgroup$
    – user18862
    Jul 17, 2013 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.)

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    $\begingroup$ Maybe the lowest term statement is what she meant. As I said, a proof of that is a good exercise. $\endgroup$
    – Aaron Meyerowitz
    Jul 16, 2013 at 21:56
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    $\begingroup$ I think the usual convention meant by "lowest terms" also stipulates the denominator is positive. $\endgroup$
    – user14972
    Jul 16, 2013 at 22:00
  • $\begingroup$ 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 $|c| \lt |a|$ $\endgroup$
    – Aaron Meyerowitz
    Jul 16, 2013 at 22:41
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    $\begingroup$ @Sklivvz The fraction $0/1$ is in its lowest terms; the gcd between $0$ and $n$ is $n$. $\endgroup$
    – egreg
    Jul 17, 2013 at 10:52
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    $\begingroup$ @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. $\endgroup$
    – Mario Carneiro
    Jul 18, 2013 at 15:35
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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: $$ \frac{a}{b} = \frac{c}{d} $$

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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.)

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    $\begingroup$ I would hold out a little hope for this teacher. Maybe she KNOWS better, but is challenging people to prove her wrong. $\endgroup$
    – TecBrat
    Jul 16, 2013 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.

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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 .

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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!

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