
All People in Canada are the Same Age (1997) - alphanumeric0
https://www.math.toronto.edu/mathnet/falseProofs/sameAge.html
======
cf-d-ycom
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.

~~~
repsilat
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.

~~~
cf-d-ycom
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.

~~~
repsilat
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.

~~~
pansa2
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.

~~~
repsilat
> 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 it was incorrect.

~~~
pansa2
“True for n implies true for n+1, provided n >= 2” is a perfectly good
inductive step.

However, it must be coupled with a base case >= 2, which isn’t the case here.
The only base case proven is 1.

See also “Induction basis other than 0 or 1” [0].

[0]
[https://en.wikipedia.org/wiki/Mathematical_induction#Inducti...](https://en.wikipedia.org/wiki/Mathematical_induction#Induction_basis_other_than_0_or_1)

~~~
repsilat
Sure, their mistaken proof of the n->n+1 implication could be taken for what
it _does_ prove, and such a theorem could be used in other circumstances to
prove different things, but that doesn't really bear on any of the claims in
the original context.

Nobody was trying or claiming to be doing any other kind of induction. They
said they had the base case and the inductive step, and they were clear about
the base case and the inductive step. Their proof of the base case was
correct, and their proof of the inductive step did not prove what they said it
did. It doesn't matter what it _did_ prove.

------
nathanwh
Fun stuff, there's a few more on this page:

[https://www.math.toronto.edu/mathnet/falseProofs/fallacies.h...](https://www.math.toronto.edu/mathnet/falseProofs/fallacies.html)

The last one in particular I thought was interesting

~~~
exmadscientist
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.

------
gweinberg
If it were true that all groups of two people were the same age, then it would
follow that all groups of any number of people are the same age, since any
individual could be pared with any other individual to form a group of 2. But
of course it is not true.

------
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?

~~~
aidenn0
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.

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

~~~
coldtea
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!

~~~
kerkeslager
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.

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

------
solids
I heard this in the “All horses have the same color” version.

~~~
abhgh
The version I had come across was "all billiard balls have the same color" in
Liu's Discrete Mathematics [1]. It is an exercise problem in one of the
chapters.

[1] [https://www.amazon.com/Elements-Discrete-Mathematics-C-
Liu/d...](https://www.amazon.com/Elements-Discrete-Mathematics-C-
Liu/dp/0071005447)

------
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.

------
sandworm101
A similar conjecture:

[https://en.wikipedia.org/wiki/Doomsday_argument](https://en.wikipedia.org/wiki/Doomsday_argument)

"In other words, we could assume that we could be 95% certain that we would be
within the last 95% of all the humans ever to be born."

------
trabant00
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 group.

Can somebody explain these to me?

~~~
istjohn
> 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...](https://www.math.toronto.edu/mathnet/falseProofs/sameAge.html#1)

~~~
trabant00
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.

~~~
istjohn
Let me try again. Think of Step 4 like this:

\---

STEP 4 REDUX RELOADED

Imagine a powerful argument. Let's call it "The Decider." "The Decider" is a
series of logical steps that allows us to go from A to B, where A is a certain
set of assumptions or facts that we know to be true, and B is a conclusion
that we can demonstrate must be true if A is true. We can use the metaphor of
a machine. “The Decider” is a machine that takes material in on one side, and
spits out a product on the other. The input material is facts or assumptions,
and the output is new facts or assumptions that logically follow from the
input. For now, let's not worry about how "The Decider" works. Let's just
imagine that there is in fact some way to fabricate such a logical apparatus
that works the way we need it to work.

Now, suppose that the kind of fact “The Decider” ingests is a very special
kind of fact. If you tell it “the sky is blue,” the metaphorical gears within
“The Decider” grind to a halt. It doesn’t know what to do with this kind of
fact. Instead, it requires a fact of the following form:

IN: “Within any group of k people, everyone has the same age.”

So for example, you can tell “The Decider”:

IN: “Within any group of 1 person, everyone has the same age.”

And “The Decider” will grind away until it spits out a shiny new assertion
that must be true if your input was true. In fact, let’s suppose that the
shiny new assertion “The Decider” would produce in this case would be:

OUT: “Within any group of 2 people, everyone has the same age.”

You could tell “The Decider,”

IN: “Within any group of 9 people, everyone has the same age.”

and it would work, too.

OUT: “Within any group of 10 people, everyone has the same age.”

In fact, “The Decider” follows a pattern. Whatever value k is in the input,
the number in the output will be k+1. In other words, for any k that is a
positive whole number, “The Decider” can turn “within any group of k people,
everyone has the same age” into “within any group of k+1 people, everyone has
the same age.”

If “The Decider” actually existed, if there was a general pattern of reasoning
that could be used to go from A to B in this way, we would accomplish the task
set out for us in Step 3. Namely, we would prove that, “whenever S(n) is true
for one number (say n=k), it is also true for the next number (that is,
n=k+1).”

END STEP 4 REDUX RELOADED

\---

Steps 5 through 13 then proceed to describe how “The Decider” actually works.
Or at least they try to. That is, they describe a general pattern of reasoning
that purportedly can be used to show that if “within any group of 9 people,
everyone has the same age,” then it must also be true that “within any group
of 10 people, everyone has the same age.” It can also show that if it is true
that “within any group of 2 people, everyone has the same age,” then it must
also be true that “within any group of 3 people, everyone has the same age.”
In fact, Steps 5 through 13 endeavor to create a logical argument so powerful
that it can take any statement of the form “within any group of k people,
everyone has the same age” and demonstrate that if that is true, it’s also
true that “within any group of k+1 people, everyone has the same age.” This is
“The Decider” I describe above.

 __ _Spoiler_ __

The argument ultimately falls down because “The Decider” has a fatal flaw.
When k=2, the logical machinery chokes. It implicitly relies on k being
greater than or equal to 3.

Is that helpful at all?

------
thoughtstheseus
This reminds me of a Buddhist game/tradition(?) of debate where you try to get
the other person to admit to a logical fallacy and sway them to your belief.

------
tmabraham
For a mathematical fallacy, see the appendix of Charles Seife's book "Zero",
proving that Winston Churchill is in fact a carrot!

------
jmchuster
The link to the answer:
[https://www.math.toronto.edu/mathnet/falseProofs/guess27.htm...](https://www.math.toronto.edu/mathnet/falseProofs/guess27.html)

It took me a couple visits to realize that each `Step 1` `Step 2` heading is a
link to an explanation for why that step holds true.

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

~~~
thedufer
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.

------
Dolores12
People is a group of more than one. So S(1) is invalid case.

~~~
brazzy
That's not the problem at all.

------
DarkmSparks
They just forgot to define age as:

"born after 1800"

------
dfee
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.

~~~
MereInterest
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 one of the following
statements is true. (A) The person was uninformed as to Trump's character. (B)
The person was actively being misinformed as to Trump's character, in such a
way that correct information was not believed. (C) The person did not believe
that racism was a disqualifying factor for office.

Whether this statement reduces to the "vote for X or admit to being racist"
depends on several additional statements that I don't think can be entirely
stated. First, it asserts that (A) is false, that the amount of media coverage
in an election is enough that no voters are uninformed. Second, it asserts
that (B) is false, that there was no active misinformation being spread during
the 2016 election. Given what we know now, (B) is most certainly true in some
cases. Third, it asserts that (C) is equivalent to a person being racist,
which is a valid position, but one that is harder to discuss without getting
into the nuances of systemic racism.

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

~~~
MereInterest
I agree that racism is a spectrum, spanning anywhere from microagressions that
contribute to systemic racism, to overt dismissals of human rights. Generally,
the term "racist" is used to mean "exists further than X on the scale of
racism". Whether or not somebody is racist by virtue solely of voting for a
blatantly racist candidate is a matter of discussion for where that line of X
is, and is covered under my comment about option (C).

This also gets into the point where racism will continue to exist so long as
"not racist" is seen as one end of the spectrum. Rather, in order to be
appropriately egalitarian, one must be anti-racist wherever society is racist.

------
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

~~~
tsimionescu
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.

~~~
raverbashing
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

~~~
barrkel
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 proving k+1 in terms of k don't
work for all k > 1, which invalidates the generalization that the induction
depends on. The steps obscure an extra predicate on the generalization, so
that it's not fully general.

If the proof (of some other property) didn't have that hidden predicate, or
something like it, it would be fine. The problem really is that far down.

~~~
raverbashing
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: Consider everybody in G except P. These people form a group of k
> people, so they must all have the same age (Right conclusion from wrong
> premises)

> The trick in this proof is that the steps for proving k+1 in terms of k
> don't work for all k > 1,

7 and 8 doesn't work for any k > 1, it's not "for all" it's "for any". That's
why this is so ridiculous

Step 9 fails for k<=1 and k=2 and that's it.

So yeah the induction doesn't work, but it's not only because of step 9, it's
because of 7/8 (which are just erroneous conclusions of a false premise)

~~~
tsimionescu
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.

~~~
raverbashing
> 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.

~~~
tsimionescu
> 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.

No it is not like that. It is like saying "IF person P had 1Mi dollars and got
10% interest last year THEN now Person P would have 1100k dollars". This
statement is true regardless of how much money person P has today. That is how
steps 5-8 work: they are true _regardless_ of the truth of the antecedent ('in
every group of k people, everyone has the same age'). Finding 2 people in
Canada of the same age would NOT prove step 5 wrong (though it would show it
to be pointless, of course).

Probably a much more interesting problem would have found a less obviously
wrong premise to demonstrate this with. I'd love to find a way to build a
similar argument for Fermat's last theorem or some other non-trivial
observation.

