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All People in Canada are the Same Age (1997)

math.toronto.edu

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Re: All People in Canada are the Same Age (1997)

#41

Post by repsilat »

cf-d-ycom wrote: post #33
repsilat wrote: post #31

Earlier quoted context omitted.

It's simpler than that -- the "proof" of the inductive step is just incorrect. It wouldn't be a theorem in a sound logical system.

Potentially. However, I believe the inductive step is correct. I could be wrong though. ie. If you assume S(2) is true, lets prove S(3), consider a set of 3 people, {a, b, c}. apply S(2) to {a, b} are therefore the same age, apply S(2)_ to {b, c} are therefore the same age, this implies a.age == b.age == c.age, there for S(3) is true. The inductive step is done. Thats what I thought made this a mind bender.

Right, it proves S(2)->S(3), but induction asks you to prove S(n)->S(n+1) for n in general, and not just for some n. The inductive proof doesn't work for S(1)->S(2), so it clearly can't work for the more general n->n+1.

Re: All People in Canada are the Same Age (1997)

#42

Post by zarathustreal »

zepto wrote: post #29
MereInterest wrote: post #28

Earlier quoted context omitted.

One part of mathematics that I rather like is figuring out how to phrase a statement as precisely as possible. (For me, this is also what differentiates good philosophy from bad philosophy.) I've been struggling with the best way to interpret the 2016 election results, especially with the 2020 election coming up. The most precise way to phrase it is as follows: For each person who voted for Trump in 2016, at least on…

The primary fallacy is that racism is a Boolean value of some sort.

Wow I’m actually amazed, you replied the exact words that came to my mind when I read the parent comment above you. Kudos.

Re: All People in Canada are the Same Age (1997)

#43

Post by thedufer »

Chrisoaks wrote: post #36

I also think step 10 is wrong because it assumes that P ≠ Q.

There's an imprecision here, but that doesn't break the proof. Note that from step 6 on, all it needs to show is "if P and Q are any members of G, then they have the same age". If P = Q, this is trivial, so we really only need to consider the P ≠ Q. This probably should have been stated, though.

Re: All People in Canada are the Same Age (1997)

#44

Post by cf-d-ycom »

repsilat wrote: post #41
cf-d-ycom wrote: post #33

Earlier quoted context omitted.

Potentially. However, I believe the inductive step is correct. I could be wrong though. ie. If you assume S(2) is true, lets prove S(3), consider a set of 3 people, {a, b, c}. apply S(2) to {a, b} are therefore the same age, apply S(2)_ to {b, c} are therefore the same age, this implies a.age == b.age == c.age, there for S(3) is true. The inductive step is done. Thats what I thought made this a mind bender.

Right, it proves S(2)->S(3), but induction asks you to prove S(n)->S(n+1) for n in general, and not just for some n. The inductive proof doesn't work for S(1)->S(2), so it clearly can't work for the more general n->n+1.

I believe so yeh.

Re: All People in Canada are the Same Age (1997)

#45

Post by Ericson2314 »

coldtea wrote: post #6
smallnamespace wrote: post #3

Step 9 doesn't work because it introduces a third person, and so fails to demonstrate S(2)

Isn't it Step 4, which uses as a premise what the whole thing is supposed to prove? "in every group of k people, everyone has the same age" You can't use your conclusion in your assumption!

This is easier to understand in constructive type theory. You can't pattern match on `k` to recur, but you can pattern match on `succ(k)` to recur on `k`.

Even when doing classic logic, I start with my constructive intuition and the sprinkle in the continuation-passing-style spooky magic when needed to get back the classical craziness.

Everyone should learn constructive first.

Re: All People in Canada are the Same Age (1997)

#46

Post by pansa2 »

repsilat wrote: post #41
cf-d-ycom wrote: post #33

Earlier quoted context omitted.

Potentially. However, I believe the inductive step is correct. I could be wrong though. ie. If you assume S(2) is true, lets prove S(3), consider a set of 3 people, {a, b, c}. apply S(2) to {a, b} are therefore the same age, apply S(2)_ to {b, c} are therefore the same age, this implies a.age == b.age == c.age, there for S(3) is true. The inductive step is done. Thats what I thought made this a mind bender.

Right, it proves S(2)->S(3), but induction asks you to prove S(n)->S(n+1) for n in general, and not just for some n. The inductive proof doesn't work for S(1)->S(2), so it clearly can't work for the more general n->n+1.

The inductive step is fine, but it only works for n >= 2. The issue is a disconnect with the base case n = 1.

However, if it were possible to prove the case n = 2, we would have a valid inductive proof for n >= 2.

Re: All People in Canada are the Same Age (1997)

#48

Post by kortilla »

MereInterest wrote: post #28
dfee wrote: post #13

In 2003, I remember studying fallacies in English class. I literally had an outbreak of laughter during an exercise where the prompt was: “vote for me or admit you’re racist”. It seemed so ridiculous to teenage me that such a thing could be said. In 2020 it has been said. I’m no longer falling out of my seat laughing.

One part of mathematics that I rather like is figuring out how to phrase a statement as precisely as possible. (For me, this is also what differentiates good philosophy from bad philosophy.) I've been struggling with the best way to interpret the 2016 election results, especially with the 2020 election coming up. The most precise way to phrase it is as follows: For each person who voted for Trump in 2016, at least on…

Ah, absolutes, what a wonderful fantasy world where there are so many candidates to choose from that you’re able to “disqualify” a candidate and still vote for someone who will uphold your ideals in office.

There are people who believe abortion is murder and still vote in pro-choice candidates because the other policy positions are important to uphold.

Re: All People in Canada are the Same Age (1997)

#49

Post by thaumasiotes »

kriskrunch wrote: post #25
dfee wrote: post #13

In 2003, I remember studying fallacies in English class. I literally had an outbreak of laughter during an exercise where the prompt was: “vote for me or admit you’re racist”. It seemed so ridiculous to teenage me that such a thing could be said. In 2020 it has been said. I’m no longer falling out of my seat laughing.

I know this as the Kafka trap fallacy. "A Kafka trap is a fallacy where if someone denies being x it is taken as evidence that the person is x since someone who is x would deny being x. The name is derived from the novel The Trial by the Austrian writer Franz Kafka." Source: https://debate.fandom.com/wiki/Kafka_Trap

> A Kafka trap is a fallacy where if someone denies being x it is taken as evidence that the person is x

I think this concept is better known under the name "witch hunt".

Re: All People in Canada are the Same Age (1997)

#50

Post by lalaland1125 »

dfee wrote: post #13

In 2003, I remember studying fallacies in English class. I literally had an outbreak of laughter during an exercise where the prompt was: “vote for me or admit you’re racist”. It seemed so ridiculous to teenage me that such a thing could be said. In 2020 it has been said. I’m no longer falling out of my seat laughing.

I'm pretty sure that "vote for me or admit you're racist" is not necessarily a fallacy. It seems reasonable to conclude that through rather straightforward logic.

Assumption 1. If you are not a racist then you should not support racist policies.

Assumption 2. The opposition wants to enact racist policies while I do not.

Proposition 3. From 2, if you vote for my opposition then you are supporting racist policies.

Theorem 4. Therefore, from 3 and 1, if you don't vote for me then you are racist because you are supporting racist policies.

Assumptions 1 and 2 could be incorrect in certain circumstances (in particular assumption 2), but the reasoning is correct.

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