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.
You can prove that all the numbers are equal š
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 š
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.
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 $\frac { a }{ b } .$
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 $1$.
I think the implication is that IF $a=c$ THEN $b=d$ which is the mathematical fact.
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.
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 STATEMENTthen
- 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.)
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:
$$ \frac{a}{b} = \frac{c}{d} $$
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.)
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.
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!
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 .