Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

All People in Canada are the Same Age (1997)

math.toronto.edu

Post reply on HN
Print view View original

61–70 of 102 posts

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

#61

Post by [deleted] »

repsilat wrote: post #31
cf-d-ycom wrote: post #26

So this article shows that you can infer S(n+1) from S(n) for n > 1, and a base case of S(1) is true. However, you can't infer S(2) is true from assuming S(1) is true in the same way, ie. a group of 2, could be represented as two groups of S(1) and S(1). You can't claim these two S(1) groups share the same age. This means that the base case and the inductive step are not connected, which means the proof is invalid.

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.

[deleted]

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

#62

Post by shawnz »

thaumasiotes wrote: post #20
dlkf wrote: post #10

Earlier quoted context omitted.

I don't know why you're getting downvoted, afaict 9 is the problem.

There is a problem in step 9, but the problem is not related to introducing a third person. The existence of a third person makes step 9 correct . The problem is that we establish the base cases S(0) and S(1), and then we establish the inductive step S(n) -> S(n+1) subject to the restriction that n > 1. But this proof never demonstrates that S(1) implies S(2). The proof is completely correct that S(2) implies S(3), S…

I think they mean, introducing a third person into the problem scope. Not into the set G.

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

#63

Post by kerkeslager »

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!

You're right if we were speaking normal English, but "assuming" doesn't mean the same thing in math jargon. In math, you would say, "Assuming A, then B", to mean, "If A is true, then B is true."

It's way more confusing than simply "If A, then B", and I'm not sure why the wording hasn't fallen out of favor, but that's what it is.

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

#65

Post by Aeolun »

cf-d-ycom wrote: post #26

So this article shows that you can infer S(n+1) from S(n) for n > 1, and a base case of S(1) is true. However, you can't infer S(2) is true from assuming S(1) is true in the same way, ie. a group of 2, could be represented as two groups of S(1) and S(1). You can't claim these two S(1) groups share the same age. This means that the base case and the inductive step are not connected, which means the proof is invalid.

I got really confused when they just seemed to assume that. But that doesn’t appear to be what they consider the fallacious step.

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

#66

Post by md224 »

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.

> Thats what I thought made this a mind bender.

I don't know if I'd call it a mind bender, but it does demonstrate that if one holds that all groupings of two individuals must be uniform with regard to an attribute A, then it follows that any grouping of individuals must be uniform with regard to A.

A simple proof by contradiction:

1) Assume some group of more than 2 individuals is not uniform with regard to A

2) There must be at least one pair of individuals for which attribute A differs

3) Create a subgrouping of one of those pairs

4) Now we have a grouping of 2 individuals which is non-uniform with regard to A

5) But we originally held that all groupings of 2 individuals must be uniform with regard to A

6) Contradiction! Hence a non-uniform grouping with regard to A is impossible, Q.E.D.

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

#67

Post by sdenton4 »

thaumasiotes wrote: post #21
nathanwh wrote: post #14

Earlier quoted context omitted.

TFA will inform you of this though, it's intended to be a game of reasoning. Posting the answer is spoiling the fun for people who read the comments first

> it's intended to be a game of reasoning. Posting the answer is spoiling the fun for people who read the comments This is a very common example in math classes, though in my experience usually presented as a proof that "all horses are the same color".

All cats have nine tails...

No cat has eight tails, and every cat has one more tall than no cat. Ergo...

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

#68

Post by rkagerer »

Reading this "proof" gave me a headache, and at first I thought it's because I'm not familiar enough with formal inductive logic to precisely follow the conversion from statements of reasoning to shorthand notation.

Then I realized the aim is to trick you by playing a bit fast and loose with that convention, and hoping you don't notice. e.g. I stumbled at Step 4, and if you click on the details for it the authour admits k is ill-defined. That comes back to bite you when you hit the fallacious step.

I'm fascinated how some very old works by ancient physicists and mathematicians are written using plain (if verbose) language and diagrams, and you didn't need to learn a bunch of shorthand conventions specific to the field in order to participate. Does anyone know any good books on Quantum Mechanics that don't require you to learn Dirac notation first?

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

#69

Post by pcwelder »

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

Also related to the fallacy of false dichotomy. https://en.wikipedia.org/wiki/False_dilemma
Post reply on HN