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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)

#31

Post by repsilat »

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.

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

#33

Post by cf-d-ycom »

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.

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.

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

#34

Post by cf-d-ycom »

cheez wrote: post #30
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.

Not exactly, it was saying that you assumed S(2) implicitly which is wrong.

Yeh, that makes more sense.

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

#35

Post by solids »

cheez wrote: post #30
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.

Not exactly, it was saying that you assumed S(2) implicitly which is wrong.

Why can’t you asume S(2)? Is it not included in the inductive hypothesis S(k)?

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

#37

Post by microcolonel »

andrepd wrote: post #27
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.

Who has said it?

Hillary Clinton, several ways; though most famously in the Basket of Deplorables monologue.

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

#38

Post by dfee »

andrepd wrote: post #27
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.

Who has said it?

Here’s another variant, "If you have a problem figuring out whether you're for me or Trump, then you ain't black," Joe Biden, May 22, 2020

https://www.npr.org/2020/05/22/861007175/biden-pulls-back-on...

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

#40

Post by cf-d-ycom »

solids wrote: post #35
cheez wrote: post #30

Earlier quoted context omitted.

Not exactly, it was saying that you assumed S(2) implicitly which is wrong.

Why can’t you asume S(2)? Is it not included in the inductive hypothesis S(k)?

So the base case that they prove is S(1). And the inductive step is to show: S(1) ->(implies) S(2) -> S(3) -> S(4) -> ....

Or to write it more succinctly, that S(n) -> S(n + 1) assuming S(n) is true.

In this particular problem, the proof that is provided can be used to show S(2) -> S(3), that S(3) -> S(4), etc... are valid and true.

However, the proof could NOT be applied to show that S(1) -> S(2).

So whilst you CAN assume S(2) is true in a proof, the whole inductive chain needs to be attached to a valid base case.

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