Main post link -> http://blog.brilliant.org/2013/02/21/you-are-inductively-calvin/

See the Blog for the full story.

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".

## Comments

Sort by:

TopNewestFails at \(k = 1\), since the first 1 person does not 'intersect' with the last 1 person if there are 2. – Zi Song Yeoh · 4 years, 1 month ago

Log in to reply

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. – John Muradeli · 2 years, 5 months ago

Log in to reply

`Hmm...Calvin is pretty famous :D – Arsh Singh · 4 years, 1 month ago

Log in to reply

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 – Daniel Liu · 4 years, 1 month ago

Log in to reply

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. – Yang Conan Teh · 4 years, 1 month ago

Log in to reply

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 – Abhishek De · 4 years, 1 month ago

Log in to reply

The mass argument (the 2nd one) does not make sense to me ; i there a typo in it? – Yash Talekar · 3 years, 3 months ago

Log in to reply

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 – Nidhin Kurian · 4 years, 1 month ago

Log in to reply

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 – Tan Li Xuan · 4 years, 1 month ago

Log in to reply

– Zi Song Yeoh · 4 years, 1 month ago

I think this is what we want to prove.Log in to reply

– Tan Li Xuan · 4 years, 1 month ago

I agree.Anyway that's just a joke. :DLog in to reply

This was discussed in Mathematical Induction II-B. – Zi Song Yeoh · 4 years, 1 month ago

Log in to reply