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.

## Comments

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TopNewestNo 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, 9 months ago

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– Finn Hulse · 3 years, 3 months ago

Haha, I know what book you got that from. I love that. :DLog in to reply

– Sharky Kesa · 3 years, 3 months ago

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 – Finn Hulse · 3 years, 3 months ago

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– Daniel Chiu · 3 years, 3 months ago

F on exams.Log in to reply

– Finn Hulse · 3 years, 3 months ago

YES! :DLog in to reply

– Tan Li Xuan · 3 years, 3 months ago

A on creativity :)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. \[\mathbb{Q.E.D.}\] – Trevor B. · 3 years, 3 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, 9 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, 6 months ago

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– Michael Mendrin · 2 years, 1 month ago

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

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. – John Muradeli · 2 years, 6 months agoLog 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 – Jord W · 3 years, 9 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, 9 months ago

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– John Muradeli · 3 years, 9 months ago

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

THE ROOM2yet? – Mahdi Al-kawaz · 3 years, 3 months agoLog in to reply

– John Muradeli · 2 years, 10 months ago

The Room 3 will be its ultimate conclusion!Log in to reply

– Jord W · 3 years, 9 months ago

ok i will :) thanksLog 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 – Jord W · 2 years, 1 month ago

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Behold, Mr. Mathopedia: @Michael Mendrin – John Muradeli · 2 years, 1 month ago

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This link may or may not work

Living with Ghosts

The link is from Stephen's own website

Stephen Hawking – Michael Mendrin · 2 years, 1 month ago

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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, 1 month ago

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NaN

also check out this, see "Not a Number"

IEEE 754-1985 – Michael Mendrin · 2 years, 1 month ago

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– John Muradeli · 2 years, 1 month ago

Don't get me wrong, but the answer is DEFINITELY not a number.Log in to reply

– Michael Mendrin · 2 years, 1 month ago

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 \(x\).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... – John Muradeli · 2 years, 1 month ago

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– Michael Mendrin · 2 years, 1 month ago

I think I'll live long enough to see Book X. Maybe I should stop climbing rock?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!) – John Muradeli · 2 years, 1 month ago

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– Michael Mendrin · 2 years, 1 month ago

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

– Jord W · 2 years, 1 month ago

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

Till then, keep doing the lame-old math ;) – John Muradeli · 2 years, 1 month 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 · 2 years, 10 months ago

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Since I've been absent for 2 weeks, you get 2 hints. – John Muradeli · 3 years, 8 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, 9 months ago

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– John Muradeli · 3 years, 9 months ago

This is NOT a matrix.Log in to reply

There is no solution.Or if this is a math joke,\( x \) is between the vertical lines. – Tan Li Xuan · 3 years, 9 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, 9 months ago

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DNE – Sk Ashif Akram · 3 years, 9 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, 9 months ago

Log in to reply

– Daniel Chiu · 3 years, 9 months ago

How about my answer?Log in to reply

– John Muradeli · 3 years, 9 months ago

Nice try, but no.Log in to reply

Never mind, then. I thought this was an actual problem to solve. – Guillermo Angeris · 3 years, 9 months ago

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– John Muradeli · 3 years, 9 months ago

"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 – Joseph Ludwin Marigmen · 3 years, 9 months ago

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– Daniel Chiu · 3 years, 9 months ago

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

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– Finn Hulse · 3 years, 3 months ago

Is your dad good at math?Log in to reply

x = i – Yash Talekar · 3 years, 9 months ago

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– Abhijeeth Babu · 3 years, 9 months ago

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