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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)

#71

Post by aidenn0 »

rkagerer wrote: post #68

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 adm…

Many books that just give an overview of QM don't use dirac notation; look for books with "Modern Physics" or "Introduction" in their names. However, Dirac notation is used so ubiquitously in QM that your question is a bit like asking if there are linear algebra books that don't require you to learn matrix notation. Sure you can do linear algebra without matrices, but that would be a bit eccentric today.

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

#72

Post by raverbashing »

I disagree with the take here

The fallacy is assuming that a group with more than 1 person has the same age.

It would be ok to assume this if we were looking for a proof of contradiction, but this statement is never challenged nor contradicted

This sounds very much like those "gotchas" that confuse more than help

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

#73

Post by aidenn0 »

lalaland1125 wrote: post #50
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. Theo…

This argument assumes that supporting racist policies is racist. That may seem tautological, but consider: someone holds a gun to my head and threatens to kill me if I don't vote for the opposition. Am I racist for voting for the opposition?

That may seem extreme, but what if candidate #1 is racist, but candidate #2 has vowed to eliminate funding for medical care I need to live and can't afford on my own? Is voting for #1 racist?

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

#74

Post by OJFord »

kerkeslager wrote: post #63
coldtea wrote: post #6

Earlier quoted context omitted.

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.

In what way is that different from 'normal English'?

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

#75

Post by toxik »

nathanwh wrote: post #5

Fun stuff, there's a few more on this page: https://www.math.toronto.edu/mathnet/falseProofs/fallacies.h... The last one in particular I thought was interesting

Hilariously I pinned down all the other problems in one try, except the everybody is the same age one.

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

#77

Post by tsimionescu »

raverbashing wrote: post #72

I disagree with the take here The fallacy is assuming that a group with more than 1 person has the same age. It would be ok to assume this if we were looking for a proof of contradiction, but this statement is never challenged nor contradicted This sounds very much like those "gotchas" that confuse more than help

That is not the fallacy, that is the wrong conclusion.

The fallacy is in step 9 combined with step 1. Step 9 requires at least 3 people to exist - P, Q, and R. So, step 9 only works for k>=2. So, we have proved S(1),S(k>=2) => S(k+1), but we haven't proved S(2).

Of course, S(2) (in any group of 2 people, both people have the same age) is not true, so the whole conclusion is false.

In inductive proofs you always need to prove some rule that says 'for any k [with some property], assuming case k is true, then case k+1 is true as well', and then you also need to prove that, for some k [with the given property], case k is actually true.

Restating the proof in the article in these terms, step 9 correvtly proves that, for any k [greater than or equal to 2], if S(k) then S(k+1). But there is no proof given that there exists some k>=2 for which the statement actually holds, and in fact it can proved that NO such k exists. So overall the proof doesn't hold.

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

#78

Post by raverbashing »

tsimionescu wrote: post #77
raverbashing wrote: post #72

I disagree with the take here The fallacy is assuming that a group with more than 1 person has the same age. It would be ok to assume this if we were looking for a proof of contradiction, but this statement is never challenged nor contradicted This sounds very much like those "gotchas" that confuse more than help

That is not the fallacy, that is the wrong conclusion. The fallacy is in step 9 combined with step 1. Step 9 requires at least 3 people to exist - P, Q, and R. So, step 9 only works for k>=2. So, we have proved S(1),S(k>=2) => S(k+1), but we haven't proved S(2). Of course, S(2) (in any group of 2 people, both people have the same age) is not true, so the whole conclusion is false. In inductive proofs you always need…

I see your point, but I kinda disagree, and again, that's why this is more confusing than helpful.

You're taking a false premise and running with it, then tripping far ahead and saying that's the fallacy.

> but we haven't proved S(2).

Well, not surprising you haven't proven it, because you're already deep down in the mud on steps 7 and 8

> Step 7: Consider everybody in G except P. These people form a group of k people, so they must all have the same age

> Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

Given that

> Let G be an arbitrary group of k+1 people

This is already false

Saying that

> Let R be someone else in G other than P or Q.

Is something completely natural for a group with k+1 elements (with the exception of k < 3), but the "proof" is so deep down in its absurdity at this point calling this the fallacy is almost a technicality

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

#79

Post by leto_ii »

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.

> 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).

Yup, I think this nails it. The more I think about this problem the more it seems meaningless to me. Adding the whole induction stuff just obfuscates this core problem.

It seems to me that if you can prove S(2) you've immediately proven S(n). You don't need the subsequent induction.

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

#80

Post by brazzy »

repsilat wrote: post #53
pansa2 wrote: post #46

Earlier quoted context omitted.

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.

> it only works for n >= 2 Right, so it doesn't work -- either it's a correct proof of a non-sequitur ("true for n implies true for n+1, provided n meets some criteria"), or an incorrect proof of an inductive step ("true for n implies true for n+1"). Because it's claimed to be proof by induction, it's meant to be the latter -- the person doing the proving claimed to have proven the inductive step, and their proof of…

I think a better characterization of the fallacy is that it is indeed a non-sequitur, but it hides the "provided n meets some criteria" by bamboozling you with "arbitrarily chosen" elements from a set of n+1 so that you might fail to spot where you need 3 elements in a set that potentially has a cardinality of 2.
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