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

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


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

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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I think your examples contradict the claim correctly. –  B. S. Jul 16 '13 at 9:54
Her claim is true on $\mathbb{F}_2$. –  user1551 Jul 16 '13 at 10:00
My heart bleeds for that mathematics teacher's students... –  DonAntonio Jul 16 '13 at 10:07
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
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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11 Answers

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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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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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


  • 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:....


$\frac{3}{6}$ 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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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
I think the usual convention meant by "lowest terms" also stipulates the denominator is positive. –  Hurkyl 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
@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
@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
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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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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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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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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
@J.R.: Or that. –  Thomas Jul 16 '13 at 22:01
As a (relative) layman, I've got to ask, why is disproving something by counterexample considered not "mathematical"? –  TheTerribleSwiftTomato Jul 17 '13 at 11:54
@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. –  NeuroFuzzy Jul 17 '13 at 22:48
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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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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
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
Wha..? Isn't this just restating the teachers wrong belief? –  brentonstrine Jul 16 '13 at 23:36
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 2:34
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 3:12
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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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What about if both are equal to 0? –  jwg Jul 16 '13 at 13:02
@jwg Then they are already equal ;-) –  dtldarek Jul 16 '13 at 13:18
"Things which equal the same thing also equal one another." – Euclid, c. 300 BC. –  Pål GD Jul 16 '13 at 19:16
Or in modern terminology, equality is transitive. –  Dan Jul 17 '13 at 1:57
@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). –  Pål GD Jul 17 '13 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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it is not absurd, it is a contradiction. Just saying... –  simpleBob Jul 17 '13 at 12:27
A proof by contradiction aims to develop an absurdity. –  user02138 Jul 17 '13 at 12:46
Absurdity is actually the correct word. Since it was demostrated by "Reductio ad absurdum". Upvote, I learned something new :) –  simpleBob Jul 17 '13 at 12:53
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If $\frac{a}{b}$ is an integer $n$, then:


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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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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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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That's a nice way of showing it. Thanks –  user86484 Jul 16 '13 at 10:06
Where is contradiction here? 1/2 = 2/4 => 1/2=2/4=3/6, so? –  RiaD Jul 17 '13 at 15:28
@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 $c=d$. –  newzad Jul 17 '13 at 15:32
Oh, got it, my bad –  RiaD Jul 17 '13 at 17:04
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protected by Michael Greinecker Jul 16 '13 at 18:35

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