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# You are inductively Calvin

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Are you convinced that induction has proven your name to be Calvin?

Just how powerful is Induction? I'd use it to show that your name is Calvin.

Claim: In a group of $$n$$ people, everyone has the same name.

Proof: Let's show that condition (1) holds. In a group of $$n=1$$ people, all of them have the same name.

Let's show that condition (2) holds. Since the positive integer $$k$$ is in the set, this means that in a group of any $$k$$ people, all of them have the same name. So let's consider a group of $$k+1$$ people. Since the first $$k$$ people have the same name, and the last $$k$$ people also have the same name, it means that all $$k+1$$ of them must have the same name.

Thus, by Mathematical Induction, the set $$S$$ includes all positive integers, so in a group of $$n$$ people, everyone has the same name.

In particular, since my name is Calvin, and you are in the set of all $$N$$ people on Earth, it follows that your name is also Calvin.

The above is adapted from Polya's proof that "No horse is of a different color".

Note by Peter Taylor
4 years, 8 months ago

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Fails at $$k = 1$$, since the first 1 person does not 'intersect' with the last 1 person if there are 2.

- 4 years, 8 months ago

I don't get it... neither the proof nor the anti-proof.

What does the proof by induction, which deals with integer values, have to do with categorical variables such as a name? One has a magnitude the other one doesn't... They're not in the same domain.

- 2 years, 12 months ago

`Hmm...Calvin is pretty famous :D

- 4 years, 8 months ago

let k=1; Because the first 1 person has the same name, and the last one person has the same name, then both people have the same name? Obvious bogus :P

- 4 years, 7 months ago

Yes there is a gap in $$k=1$$. I think this post is meant to remind other Brilliant users about some tricky spots when you are writing a solution.

- 4 years, 8 months ago

Spoiler: To spoil the fun, here's the reason why its false. (Don't see until you are totally sure you cannot make out the goof-up)

Peace.

Faustus

- 4 years, 8 months ago

The mass argument (the 2nd one) does not make sense to me ; i there a typo in it?

- 3 years, 9 months ago

the proof is wrong at the assumption of p(k) , according to this assumption , in a group of any k people, all of them have the same name. lets have two sets of people having k people each , A= {a1, a2, ..........ak}, B={b1 , b2............bk}, by our assumption of p(k), every one in set A has same name , let it be Calvin, and everyone in set B also has same name which may or may not be calvin , suppose let it be abcd, now take a new set C={a1, a2,......ar, b1, b2, ......bq}, such that r+q=k, so the elements of this set has same name so abcd must be same as Calvin, this happens for every set of k people, so all the people in the universal set must have same name(earth), but then we don 't need p(k+1), to prove the result , so this is not proved inductively. as in induction we must have 1) p(1) is true 2) assume p(k) is true and using this prove p(k+1),

but here our assumption itself solves the proof so it cannot be done inductively

- 4 years, 7 months ago

Say two students are $$A$$ and $$B$$.Then according to above proof,$$A$$'s name is $$A$$ ,and $$B$$ at the same time.The same applies to $$B$$,so we have a contradiction.So,if the proof is true,wouldn't everybody's name be Calvin/Zi Song/Abhishek/......... at the same time? :D

- 4 years, 8 months ago

I think this is what we want to prove.

- 4 years, 7 months ago

I agree.Anyway that's just a joke. :D

- 4 years, 7 months ago

This was discussed in Mathematical Induction II-B.

- 4 years, 8 months ago