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Lets do something meaningless... :P

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!

Note by Nihar Mahajan
1 year, 10 months ago

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

\[\zeta(\textrm{TREVOR A})=69\]

Prasun Biswas · 1 year, 10 months ago

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@Prasun Biswas Lol. But i don't know zeta function yet. :( Nihar Mahajan · 1 year, 10 months ago

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@Nihar Mahajan Let me enlighten you. The value \(\zeta(\textrm{TREVOR A})\) is a special value, unlike the rest of the zeta function values. I suspect that it might be the key to proving the Trevormann hypothesis since it has a real part of \(69\) and is still a zero of the Trevormann zeta function.

@Pi Han Goh, I think you should look into this. We have stumbled upon yet another discovery. Shall we uncover it together? Prasun Biswas · 1 year, 10 months ago

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@Prasun Biswas LOL...Omg ... if its established , it would be really amazing :P Nihar Mahajan · 1 year, 10 months ago

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@Prasun Biswas So you're telling me that \(\text{TREVOR A}=1.014615241815\)?

This might be a new discovery you've had... As it follows the constraint that \(min(TREVOR)=\pm1\) Trevor Arashiro · 1 year, 10 months ago

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@Trevor Arashiro Should I publish the proof myself in Arxiv?

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 · 1 year, 10 months ago

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Kaboobly Doo! Agnishom Chattopadhyay · 1 year, 10 months ago

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@Agnishom Chattopadhyay Yeah, Kaboobly Doo Stuff :P XD Mehul Arora · 1 year, 10 months ago

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: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 · 1 year, 10 months ago

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LOL , What is this? Are you okay? Nitesh Chaudhary · 1 year, 10 months ago

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@Nitesh Chaudhary Yes , I am okay. If you don't understand its ok.Its just a time pass stuff. Nihar Mahajan · 1 year, 10 months ago

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Well the proof is not complete, you need to prove that A <B ...... 😛😛 Harsh Shrivastava · 1 year, 10 months ago

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@Harsh Shrivastava I have proved it. A=1 , B=2 so , A < B . :P Nihar Mahajan · 1 year, 10 months ago

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@Nihar Mahajan How can u assume that A = 1 & B= 1?? :P Harsh Shrivastava · 1 year, 10 months ago

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@Harsh Shrivastava @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 · 1 year, 10 months ago

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@Mehul Arora ikr , XD :P Nihar Mahajan · 1 year, 10 months ago

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@Trevor Arashiro @Trevor B. Do read this note. :P Nihar Mahajan · 1 year, 10 months ago

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