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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)

#91

Post by zwaps »

Another simple way to see this is to reverse the usual order of steps, starting with the induction step.

The author shows that S(n)->S(n+1) and that proof is correct if n is at least larger than 1. The hidden assumption of n>1 is the "gotcha" moment, but that part of the proof is still valid as induction step. Imagine starting with this and amending the instructions with "given that n>1".

However, for the purpose of showing that all Canadians are the same age, we now need to find a base case - the first (and now second) step of induction. And here, we see that while n->n+1 holds (for n>1), there simply isn't any n>1 for which S is true! The case of S(1) is irrelevant, since it's not included in the assumption of the induction step.

If we would have started with the induction step, and concluded that our argument holding for n>1 is good enough, we would have then clearly realized that there is no base case and therefore we can not complete the induction proof.

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

#92

Post by istjohn »

trabant00 wrote: post #86

I don't understand some things regarding the resoning: If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho. If you define "P and Q are any members of G" then "everybody in G except P" can only mean to me an empty group. Also "Let R be someone else in G other than P or Q" can only mean R must be outside the g…

> If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho.

Read the explanation on induction at the bottom if you haven't yet[1].

This is induction. The idea is to prove that for any 'n' for which S(n) is true, S(n+1) is true also. I'm guessing Step 4 is where the wording tripped you up?

> Step 4: We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.

Maybe its meaning is more clear stated this way:

Step 4: We can do this by showing that (1) for any k where it is true that "within every group of k people everyone has the same age," then (2) showing that it necessary follows that "within every group of k+1 people everyone has the same age."

[1] https://www.math.toronto.edu/mathnet/falseProofs/sameAge.htm...

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

#93

Post by rahimiali »

aidenn0 wrote: post #73
lalaland1125 wrote: post #50

Earlier quoted context omitted.

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 f…

In both cases you’re still racist, assuming the power gradient flows from you to the underprivileged group. You are willing to put your self interest ahead of others. This is ok. You can just say “I’m sorry, self interest demands I be racist”. Germans joined the Nazi party so they could be eligible for good jobs. They say things like this about their grand parents to this day.

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

#94

Post by trabant00 »

istjohn wrote: post #92
trabant00 wrote: post #86

I don't understand some things regarding the resoning: If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho. If you define "P and Q are any members of G" then "everybody in G except P" can only mean to me an empty group. Also "Let R be someone else in G other than P or Q" can only mean R must be outside the g…

> If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho. Read the explanation on induction at the bottom if you haven't yet[1]. This is induction. The idea is to prove that for any 'n' for which S(n) is true, S(n+1) is true also. I'm guessing Step 4 is where the wording tripped you up? > Step 4: We can do this…

I understand (at some level) what induction is. But the example given in the explanation of induction in the bottom is fundamentally different in that it does not assume anything. It simply calculates the formula for 1 and n+1 and both fit it.

> Step 4: We can do this by showing that (1) for any k where it is true that "within every group of k people everyone has the same age," then (2) showing that it necessary follows that "within every group of k+1 people everyone has the same age."

Try as I might this only looks like circular reasoning to me.

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

#96

Post by trabant00 »

zwaps wrote: post #87
trabant00 wrote: post #86

I don't understand some things regarding the resoning: If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho. If you define "P and Q are any members of G" then "everybody in G except P" can only mean to me an empty group. Also "Let R be someone else in G other than P or Q" can only mean R must be outside the g…

Edit: Let me try to be as clear as possible. Your confusion comes from the language in the question of "ANY SPECIFIC N" versus "ANY, as in ALL N". Say, n=3. Then, we assume S is true for the value n=3, but we don't yet know if S holds for any other n (1,2,4,100 etc.), which we have not assumed. However, we show that IF S is true for n=3, THEN that alone implies it's also true for 4 (n+1). Of course then, we could pro…

> our assumption was that S was true for one specific n, and not all n at the same time

And here is where my problem with the reasoning is. The example in "A Brief Review of the Principle of Induction" makes no such error. It calculates for 1, and it calculates for n and n+1, in which the n can really by any n as in "all n". Also, no assumption there. So in my understanding the fallacies start at step 3.

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

#97

Post by [deleted] »

exmadscientist wrote: post #60
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

The ladder problem is one that I seem to remember was in our first-year classical mechanics text back when I was in grad school, and caused a lot of debate... it's a good one.

[deleted]

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

#98

Post by tsimionescu »

raverbashing wrote: post #89
barrkel wrote: post #83

Earlier quoted context omitted.

The basic principle of induction is to prove k+1 in terms of k. If we can show it's true for k=1, then it's true for all k>1 as well. But assuming that it's true for k is a basic part of inductive proofs. It's like falling dominoes. You show the first domino falls; you show that, if the previous domino falls, the next domino will also fall; and thus all the dominoes fall. The trick in this proof is that the steps for…

I know how induction works, believe it or not > in this proof is that the steps for proving k+1 in terms of k don't work for all k > 1, Yes and those are steps 7 and 8, not step 9. Because that "proof" is already wrong. It comes from a wrong premise, surely, but that's already wrong at this time This is why I'm calling BS on step 9 being the fallacy there. > Step 5: Let G be an arbitrary group of k+1 people > Step 7:…

Why is step 7 wrong? If all groups of k people had the same age, then everyone in G except P would have had the same age. That is a true proposition ('false => true' is true).

The greater point of the excersise is to show a failure mode of an inductive proof. If the property hadn't been so clearly false, it may have been harder to spot the actual mistake ; the excersise is meant to prepare you for those other cases.

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

#99

Post by raverbashing »

tsimionescu wrote: post #98
raverbashing wrote: post #89

Earlier quoted context omitted.

I know how induction works, believe it or not > in this proof is that the steps for proving k+1 in terms of k don't work for all k > 1, Yes and those are steps 7 and 8, not step 9. Because that "proof" is already wrong. It comes from a wrong premise, surely, but that's already wrong at this time This is why I'm calling BS on step 9 being the fallacy there. > Step 5: Let G be an arbitrary group of k+1 people > Step 7:…

Why is step 7 wrong? If all groups of k people had the same age, then everyone in G except P would have had the same age. That is a true proposition ('false => true' is true). The greater point of the excersise is to show a failure mode of an inductive proof. If the property hadn't been so clearly false, it may have been harder to spot the actual mistake ; the excersise is meant to prepare you for those other cases.

> If all groups of k people had the same age, then everyone in G except P would have had the same age. That is a true proposition ('false => true' is true)

The logical formula is right but the premise is wrong. But 7 is considering both.

It's like saying "Person P had 1Mi dollars and got 10% interest last year then now Person P has 1100k dollars". But Person P didn't. The interest calculation is correct, but the premise is wrong.

> The greater point of the excersise is to show a failure mode of an inductive proof.

I see that, but the fact that the other steps are not contributing to the solution makes it harder to argue that the mistake is there. Because that statement needs a qualifier, but it is "not wrong" per se (it's not even affirming anything, it's just saying "pick the person you haven't picked (from a group that might not have anyone else to pick, fair enough)

But everything derived from a false premise can be false. That's how we get the proofs by contradiction, right? We keep going down the wrong path until it obviously blows up

Or we could just prove this whole problem false with a set of two people in Canada with different ages (counterexample to Step 5). Case closed.

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

#100

Post by aidenn0 »

rahimiali wrote: post #93
aidenn0 wrote: post #73

Earlier quoted context omitted.

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 f…

In both cases you’re still racist, assuming the power gradient flows from you to the underprivileged group. You are willing to put your self interest ahead of others. This is ok. You can just say “I’m sorry, self interest demands I be racist”. Germans joined the Nazi party so they could be eligible for good jobs. They say things like this about their grand parents to this day.

If the desire to save one's own life at the expense of an underprivileged group is racist, I'm going to go out on a limb and say nearly everyone is racist.
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