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!

#Algebra #Trevor

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

LOL , What is this? Are you okay?

Nitesh Chaudhary - 4 years, 3 months ago

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

Nihar Mahajan - 4 years, 3 months ago

Kaboobly Doo!

Agnishom Chattopadhyay Staff - 4 years, 4 months ago

Yeah, Kaboobly Doo Stuff :P XD

Mehul Arora - 4 years, 4 months ago

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

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

Prasun Biswas - 4 years, 4 months ago

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

Nihar Mahajan - 4 years, 4 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?

Prasun Biswas - 4 years, 4 months ago

@Prasun Biswas – LOL...Omg ... if its established , it would be really amazing :P

Nihar Mahajan - 4 years, 4 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 $min(TREVOR)=\pm1$

Trevor Arashiro - 4 years, 4 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.

Prasun Biswas - 4 years, 4 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.

Trevor Arashiro - 4 years, 4 months ago

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

Harsh Shrivastava - 4 years, 4 months ago

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

Nihar Mahajan - 4 years, 4 months ago

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

Harsh Shrivastava - 4 years, 4 months ago

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

Nihar Mahajan - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$