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
5 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 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"
MathAppears 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

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Top Newest

No solution for $$x$$.

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

- 5 years, 2 months ago

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

- 4 years, 8 months ago

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Which book is that?

- 4 years, 8 months ago

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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

- 4 years, 8 months ago

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F on exams.

- 4 years, 8 months ago

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A on creativity :)

- 4 years, 8 months ago

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YES! :D

- 4 years, 8 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 $$|i^2|=|-1|=1$$

- 5 years, 2 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. $\mathbb{Q.E.D.}$

- 4 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

- 5 years, 2 months ago

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

- 5 years, 2 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 ;)

- 5 years, 2 months ago

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ok i will :) thanks

- 5 years, 2 months ago

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Have you finished THE ROOM2 yet?

- 4 years, 8 months ago

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The Room 3 will be its ultimate conclusion!

- 4 years, 3 months ago

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



$\Huge \{\}$

- 3 years, 11 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.

- 3 years, 11 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.

- 3 years, 6 months ago

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DNE

- 5 years, 2 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

- 5 years, 2 months ago

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

- 5 years, 2 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.

- 5 years, 2 months ago

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This is NOT a matrix.

- 5 years, 2 months ago

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

- 5 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?

- 4 years, 3 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

- 3 years, 6 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

- 3 years, 6 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.

- 3 years, 6 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

- 3 years, 6 months ago

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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

- 3 years, 6 months ago

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

also check out this, see "Not a Number"

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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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 $$x$$.

- 3 years, 6 months ago

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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...

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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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!)

- 3 years, 6 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

- 3 years, 6 months ago

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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 ;)

- 3 years, 6 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.

- 5 years, 2 months ago

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How about my answer?

- 5 years, 2 months ago

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Nice try, but no.

- 5 years, 2 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.

- 5 years, 2 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"

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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$$x=i^2$$ gives $|x|=|i^2|=|-1|=1\ge 0$

- 5 years, 2 months ago

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x = i

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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