I was tired solving maths problems , then I stopped for a minute and I thought of doing something meaningless which includes my friend Trevor A (prototype) and the brilliant avatar Trevor B.So what I am going to prove where \(TREVOR\) is an integer is that :

\[\Large min(TREVOR) = \pm 1\]

**Proof:**

Lets assume that \(TREVOR \ A > TREVOR \ B\).

But \(1<2 \Rightarrow A<B\)

By multiplying both the sides by \(TREVOR\) , the inequality sign flips which tells us that:

\[\Large min(TREVOR) = -1\]

Lets consider another case \(TREVOR \ A < TREVOR \ B \\ A < B\)

By multiplying both the sides by \(TREVOR\) , the inequality sign remains the same which tells us that :

\[\Large min(TREVOR) = 1\]

Let us consider one more case \(TREVOR \ A = TREVOR \ B\)

But \(A<B\) makes it impossible hence , \(TREVOR \ A \neq TREVOR \ B\). Hence this case does not hold true.

At last we conclude that \(min(TREVOR) = \pm 1\)

I hope this made you feel nice in your busy study schedule.Cheers!

## Comments

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TopNewest@Nihar Mahajan, as a continued exercise, prove the following:

– Prasun Biswas · 2 years agoLog in to reply

– Nihar Mahajan · 2 years ago

Lol. But i don't know zeta function yet. :(Log in to reply

@Pi Han Goh, I think you should look into this. We have stumbled upon yet another discovery. Shall we uncover it together? – Prasun Biswas · 2 years ago

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– Nihar Mahajan · 2 years ago

LOL...Omg ... if its established , it would be really amazing :PLog in to reply

This might be a new discovery you've had... As it follows the constraint that \(min(TREVOR)=\pm1\) – Trevor Arashiro · 2 years ago

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I also have a maximum upper bound for the A version but I'm lacking a complete proof since I haven't researched on the B version. – Prasun Biswas · 2 years ago

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Kaboobly Doo! – Agnishom Chattopadhyay · 2 years ago

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– Mehul Arora · 2 years ago

Yeah, Kaboobly Doo Stuff :P XDLog in to reply

:3 this actually made me laugh pretty hard. Haha.

But you forgot to consider the case of \(TREVOR=0\). In which case the world implodes. – Trevor Arashiro · 2 years ago

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LOL , What is this? Are you okay? – Nitesh Chaudhary · 2 years ago

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– Nihar Mahajan · 2 years ago

Yes , I am okay. If you don't understand its ok.Its just a time pass stuff.Log in to reply

Well the proof is not complete, you need to prove that A <B ...... 😛😛 – Harsh Shrivastava · 2 years ago

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– Nihar Mahajan · 2 years ago

I have proved it. A=1 , B=2 so , A < B . :PLog in to reply

– Harsh Shrivastava · 2 years ago

How can u assume that A = 1 & B= 1?? :PLog in to reply

@Harsh Shrivastava , He assumed the values of A and B to Be their "Place" values in the alphabet :P. Also, Nice Proof, @Nihar Mahajan XD – Mehul Arora · 2 years ago

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– Nihar Mahajan · 2 years ago

ikr , XD :PLog in to reply

@Trevor Arashiro @Trevor B. Do read this note. :P – Nihar Mahajan · 2 years ago

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