Have fun!

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

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.

Removed.

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestNo solution for \(x\).

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

Log in to reply

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

Log in to reply

Which book is that?

Log in to reply

Find \(x\):

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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 \(|i^2|=|-1|=1\)

Log in to reply

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. \[\mathbb{Q.E.D.}\]

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

ok i will :) thanks

Log in to reply

Have you finished

THE ROOM2yet?Log in to reply

Log in to reply

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

\[\]

\[\Huge \{\}\]

Log in to reply

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

hypothesisrather 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.Log in to reply

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.

Log in to reply

DNE

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

This is NOT a matrix.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

NaN

also check out this, see "Not a Number"

IEEE 754-1985

Log in to reply

Log in to reply

Log in to reply

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...

Log in to reply

Log in to reply

(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!)

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

How about my answer?

Log in to reply

Nice try, but no.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

x = i

Log in to reply

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

Log in to reply