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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
2 years, 8 months ago

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

$\zeta(\textrm{TREVOR A})=69$

- 2 years, 8 months ago

Lol. But i don't know zeta function yet. :(

- 2 years, 8 months ago

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?

- 2 years, 8 months ago

LOL...Omg ... if its established , it would be really amazing :P

- 2 years, 8 months ago

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 $$min(TREVOR)=\pm1$$

- 2 years, 8 months ago

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.

- 2 years, 8 months ago

Kaboobly Doo!

- 2 years, 8 months ago

Yeah, Kaboobly Doo Stuff :P XD

- 2 years, 8 months ago

: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.

- 2 years, 8 months ago

LOL , What is this? Are you okay?

- 2 years, 7 months ago

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

- 2 years, 7 months ago

Well the proof is not complete, you need to prove that A <B ...... 😛😛

- 2 years, 8 months ago

I have proved it. A=1 , B=2 so , A < B . :P

- 2 years, 8 months ago

How can u assume that A = 1 & B= 1?? :P

- 2 years, 8 months ago

@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

- 2 years, 8 months ago

ikr , XD :P

- 2 years, 8 months ago

@Trevor Arashiro @Trevor B. Do read this note. :P

- 2 years, 8 months ago