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|x|<0. Find x. DNE not accepted.

Have fun!

NOTE:

The problem says "absolute value of x is less than zero". This is NOT a matrix.

EDIT:

To all of you saying "DNE", you are wrong (as the title says). Here are the hints to guide you out: 1. Do not try to solve this problem using traditional mathematics. 2. Think outside of the realm that lies outside the box (i.e outside of outside the box). 3. The solution is yet as close to infinity as it is far away from it. For each of the seven days that there is no right solution, I'll put up an additional hint.

HINTS:

Removed.

Note by John Muradeli
3 years, 11 months ago

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No solution for \(x\).

If you want a answer that isn't that, \[ \begin{align}\ |&x|<0 \\ &\uparrow \\ \text{Right}&\text{here} \end{align}\]

Daniel Chiu - 3 years, 11 months ago

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Haha, I know what book you got that from. I love that. :D

Finn Hulse - 3 years, 5 months ago

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Which book is that?

Sharky Kesa - 3 years, 5 months ago

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@Sharky Kesa I don't remember the title, but it's full of stupid responses to test problems, like

Find \(x\):

And then the person draws an arrow towards \(x\), and other stupid stuff like that. It's hilarious. :D

Finn Hulse - 3 years, 5 months ago

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@Finn Hulse F on exams.

Daniel Chiu - 3 years, 5 months ago

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@Daniel Chiu YES! :D

Finn Hulse - 3 years, 5 months ago

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@Daniel Chiu A on creativity :)

Tan Li Xuan - 3 years, 5 months ago

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If something is less than \(0,\) then it is also less than every number greater than \(0.\) Thus, if \(|x|<0,\) then \(|x|<3.\) This shows that the person loves the absolute value function because <3 is a heart. All we need to do is prove that someone loves the absolute value function. I love the absolute value function, so it is indeed possible. \[\mathbb{Q.E.D.}\]

Trevor B. - 3 years, 5 months ago

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To everyone coming up with complex numbers:

For a complex number of the form \(a+bi\), we have that \(|a+bi|=\sqrt{a^2+b^2}\).

Therefore we have that \(|i|=|0+1*i|=\sqrt{0^2+1^2}=\sqrt(1)=1\)

Also, \(i^2=-1\), so \(|i^2|=|-1|=1\)

Ton De Moree - 3 years, 11 months ago

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Here's a solution set where \(|x|<0\) for any \(x\in\mathbb{O}\) (Octonion):

\[\]

\[\Huge \{\}\]

Micah Wood - 2 years, 8 months ago

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Octonions don't work. Octonions is a "normed algebra", i.e., has non-negative norms. Nice try, though. Check out sedenions, though, which don't have this property.

Michael Mendrin - 2 years, 3 months ago

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Hm that looks quite promising! Though I have no idea what Octonions are, despite having read about them briefly though your link, your little "solution" is almost on-point with mine.

However, I'd like to admit that it is merely a clever hypothesis rather than an accepted theorem or solution, but I like to keep it objective to not scare anyone away. If you could perhaps explain a little more what you mean by \({}\), we could be getting on something greater than even I could envision.

John Muradeli - 2 years, 8 months ago

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for me this is peculiar since the other day i had a thought about a number with a smaller magnitude than 0 and thought i'd stumbled upon some new realm of mathematics lol. this is my best attempt at it, though im pretty sure its fraught with logical errors i thought i'd give it a try xD |x|<0 ---> |x|=-k (k is a positive real number) ---> e^|x|=e^-k ---> e^(|x|+k)=1 ---> |x|+k=(2πn)i ---> |x|=-k+(2πn)i ---> x=inverseabs(-k+(2πn)i) ---> |A+Bi|=-k+(2πn)i =sqrt(A^2+B^2) ---> A^2+B^2=k^2-4πkni-(4(π^2)(n^2)) ---> B=sqrt(k^2-4πkni-4(nπ)^2-A^2) --->

x=A+sqrt(k^2-4πkni-4(nπ)^2-A^2)i

there is no way thats right lol

Jord W - 3 years, 11 months ago

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Here's an overview, you've stumbled upon a problem with identities on complex exponentiation and logarithms.

Guillermo Angeris - 3 years, 11 months ago

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It's good you've been thinking of it. Try this app: "The Room". Believe me, you won't regret it ;)

John Muradeli - 3 years, 11 months ago

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Have you finished THE ROOM2 yet?

Mahdi Al-kawaz - 3 years, 5 months ago

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@Mahdi Al-kawaz The Room 3 will be its ultimate conclusion!

John Muradeli - 3 years ago

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ok i will :) thanks

Jord W - 3 years, 11 months ago

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it's been too long and as nobody has solved this i highly doubt it has an established answer among the maths community, i would like to hear what the solution to this is, would be very educational i presume

Jord W - 2 years, 3 months ago

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O-oh...? Did you not get my super-secret 12-hour response a while ago? I explained the concept and told you I will delete it in twelve hours. Whoops. Buut... now this thread is too popular for me to post it again. So, I'll direct you to the guy who found a different type of answer. Erm... an attempt at an answer, rather.

Behold, Mr. Mathopedia: @Michael Mendrin

John Muradeli - 2 years, 3 months ago

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John, it might be of interest to you that no less than Stephen Hawkins has co-authored a paper, first published in 2002, titled, "Living with Ghosts", where he addresses the problem of "ghost states" in mathematical quantum field theory, "which have been though to be a fatal flaw in any quantum field theory". What is such a "ghost state"? Why, states that "spuriously" arise mathematically that have NEGATIVE NORMS. Just like your mystery expression you've posted as the basis of this note. Such things sprout like weeds, much to the annoyance of such professional theoretical physicists as the illustrious Stephen Hawkings.

This link may or may not work

Living with Ghosts

The link is from Stephen's own website

Stephen Hawking

Michael Mendrin - 2 years, 3 months ago

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@Michael Mendrin Oh boy here we go with chicken [scratch] math. I'll bookmark this for until when I graduate college LOL

But GREAT TO HEAR HAH!! ACTUALLY YOU WOULD'VE GIVEN UP IF I TOLD YOU THAT \(x\) ALSO HAS THE FOLLOWING PROPERTIES:

\[|x|<0\]

\[x>\infty\]

\[x<-\infty\]

\[x+a=x, a\neq 0\]

TELL ME, IS STEPHEN TOUGH ENOUGH FOR THIS?? xD

John Muradeli - 2 years, 3 months ago

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@John Muradeli Oh, NaN, NaN, NaN, NaN! I think you're making up stories now.

NaN

also check out this, see "Not a Number"

IEEE 754-1985

Michael Mendrin - 2 years, 3 months ago

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@Michael Mendrin Don't get me wrong, but the answer is DEFINITELY not a number.

John Muradeli - 2 years, 3 months ago

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@John Muradeli NaN is a special binary string used in computers that signify "end of order", and thus can either be greater than infinity or less than negative infinity. And if anything should be added to it, the result is unchanged. It would have all the properties you've described for \(x\).

Michael Mendrin - 2 years, 3 months ago

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@Michael Mendrin END OF ORDER??!? OOOOMG THAT IS SO CLOSE!!! Like... like... as close as \(-2\) is to \(2\) ... But at the same time very incomplete. Hmm...........

Y'know... I'm supposed to write ten books about myself (yeah you heard that right), and in Book X I reveal the answer. Buuut... you may be teh first to hear it after all... wooo...

John Muradeli - 2 years, 3 months ago

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@John Muradeli I think I'll live long enough to see Book X. Maybe I should stop climbing rock?

Michael Mendrin - 2 years, 3 months ago

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@Michael Mendrin Stop climbing?? "Um yeah that's great I'll live long enough for it. Just let me stop living a healthy lifestyle meanwhile." LOL.

(P.S. > you... may not have to worry about "dying" at all... y'know... 2045... wOooOooO....)

(P.S.\({}^2\) Avatars A, B, and C coming way before that!)

John Muradeli - 2 years, 3 months ago

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After decades of effort, physicists using the Large Hadron Collider at CERN have found the \(x\) in that expression. It is inside between the vertical bars.

Michael Mendrin - 2 years, 3 months ago

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clearly its a ruse and there is no answer lol, nice try tho i guess, you at least stimulated some fun discussion with this thread

Jord W - 2 years, 3 months ago

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@Jord W Yeah you're right there is no "answer" - it's rather a philosophical thing. But... it's not something that doesn't make sense at all either. One day I shall reveal the... thing.

Till then, keep doing the lame-old math ;)

John Muradeli - 2 years, 3 months ago

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its been so long and i don't think anybody is close to answering this, could you just explain what it is?

Jord W - 3 years ago

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Since I've been absent for 2 weeks, you get 2 hints.

John Muradeli - 3 years, 11 months ago

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It's not \(i\) since the norm of it is 1.

What is very much possible is to have \(x\) be a matrix with a negative determinant, thus the condition holds. But apart from that, norms are defined to be nonnegative on inner product spaces, so this only happens on very specific cases.

Guillermo Angeris - 3 years, 11 months ago

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This is NOT a matrix.

John Muradeli - 3 years, 11 months ago

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There is no solution.Or if this is a math joke,\( x \) is between the vertical lines.

Tan Li Xuan - 3 years, 11 months ago

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Only solution might be when x represents a matrix but in that case it would represent a determinant symbol

Avinash Iyer - 3 years, 11 months ago

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DNE

Sk Ashif Akram - 3 years, 11 months ago

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To all of you saying "DNE", you are wrong (as the title says). Here are the hints to guide you out: 1. Do not try to solve this problem using traditional mathematics. 2. Think outside of the realm that lies outside the box (i.e outside of outside the box). 3. The solution is yet as close to infinity as it is far away from it. For each of the seven days that there is no right solution, I'll put up an additional hint.

John Muradeli - 3 years, 11 months ago

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How about my answer?

Daniel Chiu - 3 years, 11 months ago

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Nice try, but no.

John Muradeli - 3 years, 11 months ago

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Do not try to solve this problem using traditional mathematics.

Never mind, then. I thought this was an actual problem to solve.

Guillermo Angeris - 3 years, 11 months ago

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"An actual problem"? Any problem that can be solved is an actual problem. There's no such thing as "Actual" and "Not-Actual"

John Muradeli - 3 years, 11 months ago

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The correct answer is i^2. Since i=square root of -1, then i^2 = -1. => imaginary numbers

Joseph Ludwin Marigmen - 3 years, 11 months ago

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\(x=i^2\) gives \[|x|=|i^2|=|-1|=1\ge 0\]

Daniel Chiu - 3 years, 11 months ago

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Comment deleted May 09, 2014

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Is your dad good at math?

Finn Hulse - 3 years, 5 months ago

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x = i

Yash Talekar - 3 years, 11 months ago

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If \(x=i\) then \(|x|=1>0\) as per the definition of modulus of a complex number.

Abhijeeth Babu - 3 years, 11 months ago

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