+19 I have used this website before but it seemed more convenient before just to post anonymously.
I present myself to you now as Mr. Prime-Fred.
One thing I have noticed and am now obviously pointing out is that the more complex a problem is, the less likely it is to left unnoticed while simple unanswered "calculator" problems seem to build in number. I want to figure which of you really are math wizards and which of you keep to the children of mathematics thus the question about to be laid before you is about to be laid before you...
What is the measure of the diagonal of a tesseract with a side length of 8 units?
Would those of you who take a stab at it be so kind as to show work brief or thorough.
Answer correctly and another question will follow. Good Luck and FYI, I don't repost.
+2489 A tesseract has three diagonals and you didn’t specify which diagonal. This is not a problem. I’m trained to read minds and find missing questions and missing parts of questions asked by the lower level primates, like you, that visit this forum. I can do this even if I’ve had too many banana daiquiris. Here’s a sample of my work.
Now, to your question.
I’ll give all three diagonals (each is a unique line connecting two vertices)
for an 8 unit tesseract.
\(\text {In 2-D “the face” (a square.)}\\ d = \sqrt {a^2 + a^2}\\ d = \sqrt {2a^2}\\ d = \sqrt {2}* a\\ d = \sqrt {2}*8\\ \)
\(\text {In 3-D a line connecting two vertices not on the same face (a cube). }\\ d = \sqrt {a^2 + a^2+a^2}\\ d = \sqrt {3a^2}\\ d = \sqrt {3}* a\\ d = \sqrt {3}* 8\\ \)
\(\text {In 4-D a line passing through the center of the “hyperspace” cube }\\ \text {and connecting two vertices (a quadragonal). }\\ d = \sqrt {a^2 + a^2+a^2+a^2}\\ d = \sqrt {4a^2}\\ d = 2a\\ d = 2*(8)\\ d=16 \\ \)
This is quite simple. Isn’t it? Math wizard skill isn’t needed for this. I know because I solved it and math wizard isn’t on my resume (yet). A genetically enhanced chimp taught me how to do this because I was fascinated by the art of hypercubes, and now I’m also fascinated by the math of art.
IF you want to test for math wizard skills, you will need to bump it up a notch or two.
GA
+118677 Hi Mr Prime-Fred,
You have introduced yourself as a new member of the forum and as such you are very welcome here.
However,
We are here to facilitate learning, it does not appear that you are trying to learn anything.
It appears only that you are being a total smart a**e who is trying to get pre-trained trained monkeys to perform tricks of your choice.
Categorize me as you please. Your opinion is of no consequence to anyone!
+2489 A tesseract has three diagonals and you didn’t specify which diagonal. This is not a problem. I’m trained to read minds and find missing questions and missing parts of questions asked by the lower level primates, like you, that visit this forum. I can do this even if I’ve had too many banana daiquiris. Here’s a sample of my work.
Now, to your question.
I’ll give all three diagonals (each is a unique line connecting two vertices)
for an 8 unit tesseract.
\(\text {In 2-D “the face” (a square.)}\\ d = \sqrt {a^2 + a^2}\\ d = \sqrt {2a^2}\\ d = \sqrt {2}* a\\ d = \sqrt {2}*8\\ \)
\(\text {In 3-D a line connecting two vertices not on the same face (a cube). }\\ d = \sqrt {a^2 + a^2+a^2}\\ d = \sqrt {3a^2}\\ d = \sqrt {3}* a\\ d = \sqrt {3}* 8\\ \)
\(\text {In 4-D a line passing through the center of the “hyperspace” cube }\\ \text {and connecting two vertices (a quadragonal). }\\ d = \sqrt {a^2 + a^2+a^2+a^2}\\ d = \sqrt {4a^2}\\ d = 2a\\ d = 2*(8)\\ d=16 \\ \)
This is quite simple. Isn’t it? Math wizard skill isn’t needed for this. I know because I solved it and math wizard isn’t on my resume (yet). A genetically enhanced chimp taught me how to do this because I was fascinated by the art of hypercubes, and now I’m also fascinated by the math of art.
IF you want to test for math wizard skills, you will need to bump it up a notch or two.
GA
+118677 Arr Ginger, you are indeed a well trained chimp :))
Performing on queue.
Your trainer will be proud of you :)
+118677 LOL I meant 'cue'.
I do not think that there was any queue for this performance :))
+2489 “Trained Monkey” is a trigger phrase for me. When I read or hear it, it causes some of my behavior to revert to a state before my genetic enhancement. It’s very hard to resist.
I see my solution and comments triggered a response from the original poster. It seems to have the consistency of fermented troll bait, with high levels of CDD, Batshit-stupid, and
non-!N derangements. Only Chirurgeon Morgan Tud has the skills to treat this advanced case. He probably would think the case is beyond hope and recommend euthanasia.
+118677 Well it would be excellent if Morgan Tud would appear and recomend a fix. That be the truth.
Euthanasia, yes tis possible, the good Doctor often be one to over prescribe, a good sleeping draft be all that is needed. ://
This realm would hardly be recognisable to Churgeon Tud on this 1st day of January in the Lords year two thousand and seventeen. Tis sad but although one can deny the existance of evolution one cannot stop its advance. 
I think that the bridge for which the troll extracted his toll has long since been swept away by the raging floods. Perhaps the troll was also swept to his death. Tis so sad.
+19
This is for you GingerAle...
First of all, don't say hypercube because a cube is not a tesseract as glorified as you want to make said cubes. Nothing you do to a cube will make it a tesseract so don't go there with your hypercubes. (unless of course, this discussion included a large and inconvenient list of shapes across a wide span of dimensions)
Second, of all, you don't read minds. Stop saying you do. Stop making a fool of yourself. Everyone knows it so please stop embarrassing yourself and just stop. Makes you sound childish af.
Third, it is the art of math NOT math of art... know what you are talking about before speaking of it. Okay?
Fourth, great job really... claps* more claps* sarcastic and slow claps* long silence*
Oh and lastly, any hyper-cuboid as you would probably put it has just one diagonal. If you actually know what a diagonal is then it would be silly to say that any shape of any dimension would have more than one diagonal.
This is for you Melody...
I respect what you said. If Gingerale wants to continue then he/she will respond and I will happily continue with Gingerale in private. (btw I am learning... it just is not math from this experience)
As for anyone else...
post HERE in PUBLIC if you would like to attempt my second problem.
